Download Vector Spaces – Chapter 4 of Lay

Vector Spaces – Chapter 4 of Lay
Dr. Doreen De Leon
Math 152, Fall 2014
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Vector Spaces and Subspaces – Section 4.1 of Lay
Goal: Generalize the work done in Chapters 1 and 2.
Definition. A vector space is a nonempty set V of objects called vectors on which are defined
two operations: addition and multiplication by scalars, that satisfy the below properties for all
u, v, w ∈ V and all scalars c and d.
(1) u + v ∈ V (closure under addition)
(2) u + v = v + u
(3) (u + v) + w = u + (v + w)
(4) There is a zero vector 0 ∈ V such that u + 0 = u.
(5) For each u ∈ V , there is a vector −u ∈ V such that u + (−u) = 0.
(6) cu ∈ V (closure under scalar multiplication)
(7) c(u + v) = cu + cv
(8) (c + d)u = cu + du
(9) c(du) = (cd)u
(10) 1u = u
Recall: We discussed in Chapter 1 that Rn is a vector space.
Other examples:
(1) Pn = the set of all polynomials of degree at most n (i.e., all polynomials of the form
p(t) = a0 + a1 t + a2 t2 + · · · + an tn , where ai ∈ R and t ∈ R) is a vector space.
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• Addition: Let p(t) = a0 + a1 t + a2 t2 + · · · + an tn and q(t) = b0 + b1 t + b2 t2 + · · · + bn tn .
Then,
(p + q)(t) = p(t) + q(t)
= (a0 + b0 ) + (a1 + b1 )t + (a2 + b2 )t2 + · · · (an + bn )tn .
• Scalar multiplication: Let p(t) ∈ Pn and let c be a scalar.
Then,
(cp)(t) = cp(t)
= (ca0 ) + (ca1 )t + (ca2 )t2 + · · · + (can )tn .
• The zero vector is the zero polynomial, 0 = 0 + 0t + 0t2 + · · · + 0tn .
The ten axioms can be “easily” verified.
(2) Let F be the set of all real-valued functions defined on R.
• Addition: Let f ∈ F and g ∈ F . Then,
(f + g)(t) = f (t) + g(t).
• Scalar multiplication: Let f ∈ F and let c be a scalar. Then,
(cf )(t) = cf (t).
• The zero vector for F is the zero function, f (t) = 0 for all t.
Properties (1) - (10) can be verified to show that F is a vector space.
Example: Property (2). Let f, g ∈ F . Then,
(f + g)(t) = f (t) + g(t)
= g(t) + f (t)
= (g + f )(t).
(3) Mm×n = the set of all m × n matrices is a vector space.
• Addition: matrix addition
• Scalar multiplication: matrix scalar multiplication
• The zero vector is the zero matrix.
Example: Let V be the set of positive real numbers with addition defined by x + y = xy and
scalar multiplication defined by cx = xc . Is V a vector space?
Solution: Properties (1) - (3) are straightforward to verify, so we will verify properties (4)-(10).
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(4) The zero vector is 1, since for all x ∈ V , x + 1 = x(1) = x.
(5) The additive inverse is 0, since for all x ∈ V , x + 0 = x(0) = 0.
(6) For all x ∈ V and all scalars c, cx = xc ∈ V since a positive number raised to a power is
still positive.
(7)
c(x + y) = (x + y)c
= (xy)c
= xc y c
= (cx)(cy)
= cx + cy.
(8)
(c + d)x = xc+d
= xc xd
= cx + dx.
(9)
c(dx) = cxd
= (xd )c
= xdc
= xcd
= (cd)x.
(10) 1x = x1 = x.
If V is a vector space, then the following properties may also be proven to hold:
• 0u = 0 for each u ∈ V .
• c0 = 0 for each scalar c.
• −u = (−1)u for each u ∈ V .
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Subspaces
If a vector space H is a subset of vectors from a larger vector space V , then H is a subset of V
(i.e., H ⊆ V ).
Definition. A subset H of a vector space V is a subspace of V if H satisfies the following.
(a) The zero vector of V is in H.
(b) For each u, v ∈ H, u + v ∈ H (closure under addition).
(c) For each u ∈ H and each scalar c, cu ∈ H (closure under scalar multiplication).
NOTES: Assume that property (a) above is satisfied.
* To prove that a subset H of a vector space V is a subspace of V , you need to show that
properties (b) and (c) hold for arbitrary vectors in H.
* If you believe that a subset H of a vector space V is not a subspace of V , it suffices to
find one example showing that either (b) or (c) fails.
Examples:
(1) Let H = {(x, y) : x ≥ 0, y ≥ 0}. Is H a subspace of R2 ? Why or why not?
(a) 0 = (0, 0) ∈ H.
(b) Let u ∈ H, so u = (u1 , u2 ), where u1 ≥ 0 and u2 ≥ 0.
Let v ∈ H, so v = (v1 , v2 ), where v1 ≥ 0 and v2 ≥ 0.
Then, u + v = (u1 + v1 , u2 + v2 ) ∈ H since u1 + v1 ≥ 0 and u2 + v2 ≥ 0.
(c) Let u = (1, 1). Then u ∈ H. Let c = −1. Then, cu = (−1, −1) ∈
/ H, so H is not
closed under scalar multiplication.
Therefore, H is not a subspace of R2 .
(2) Let H be the set of all vectors in R3 of the form (x, y, 0). Is H a subspace of R3 ? Why or
why not?
(a) 0 = (0, 0, 0) ∈ H.
(b) Let u ∈ H, so u = (u1 , u2 , 0).
Let v ∈ H, so v = (v1 , v2 , 0).
Then, u + v = (u1 + v1 , u2 + v2 , 0) ∈ H, so H is closed under addition.
(c) Let u ∈ H and let c be a scalar.
Then cu = (cu1 , cu2 , 0) ∈ H, so H is closed under scalar multiplication.
Therefore, H is a subspace of R3 .
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(3) Let H = {p ∈ P2 : p(t) = at2 }. Is H a subspace of P2 ? Why or why not?
(a) The zero polynomial satisfies 0 = 0t2 , so 0 ∈ H.
(b) Let p ∈ H =⇒ p(t) = at2 and let q ∈ H =⇒ q(t) = bt2 . Then,
(p + q)(t) = p(t) + q(t)
= (a + b)t2 ∈ H,
so H is closed under addition.
(c) Let p ∈ H =⇒ p(t) = at2 and let c be a scalar. Then,
(cp)(t) = c · p(t)
= (ca)t2 ∈ H,
so H is closed under scalar multiplication.
Therefore, H is a subspace of P2 .
(4) Let H = {p ∈ P3 : p(0) = 8}. is H a subspace of P3 ? Why or why not?
(a) The zero polynomial in P3 is 0 = 0t3 + 0t2 + 0t + 0, which is 0 for all t. So, 0 ∈
/ H.
Therefore H is not a subspace of P3 .
(5) Let H be the set of all upper triangular 2 × 2 matrices. Is H a subspace of M2×2 ? Why or
why not?
(a) 0 ∈ H since the zero matrix is upper triangular.
a b
d e
(b) Let A ∈ H =⇒ A =
and let B ∈ H =⇒ B =
. Then
0 c
0 f
a+d b+e
A+B =
∈ H,
0
c+f
so H is closed under addition.
a b
(c) Let A ∈ H =⇒ A =
and let k be a scalar. Then
0 c
ka kb
kA =
∈ H,
0 kc
so H is closed under scalar addition.
Therefore, H is a subspace of M2×2 .
(6) C 1 (R) is the set of all real-valued continuous functions with one continuous derivative
defined on R. Is C 1 (R) a subspace of F ? Why or why not?
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(a) The zero function, f (t) = 0 for all t ∈ R, is in C 1 (R), since f (t) is continuous and
f 0 (t) = 0, which is also continuous.
(b) Let g ∈ C 1 (R) and let h ∈ C 1 (R). Then,
(g + h)(t) = g(t) + h(t) is continous on R, and
(g + h)0 (t) = g 0 (t) + h0 (t) is continous on R,
so g + h ∈ C 1 (R), and C 1 (R) is closed under addition.
(c) Let g ∈ C 1 (R) and let c be a scalar. Then,
(cg)(t) = cg(t) is continous on R, and
(cg)0 (t) = cg 0 (t) is continous on R,
so cg ∈ C 1 (R), and C 1 (R) is closed under scalar mutliplication.
Therefore, C 1 (R) is a subspace of F .
Notes:
• C n (R) is the set of all real-valued continuous functions with n continuous derivatives
defined on R.
• C n [a, b] is the set of all functions that are continuous on [a, b] and have n continuous
derivatives on [a, b].
(7) H is the set of all symmetric 12 × 12 matrices. Is H a subspace of M12×12 ? WHy or why
not?
(a) 0 ∈ H since 0T = 0.
(b) Let A ∈ H =⇒ AT = A and let B ∈ H =⇒ B T = B. Then
(A + B)T = AT + B T = A + B,
so A + B ∈ H and H is closed under addition.
(c) Let A ∈ H =⇒ AT = A and let c be a scalar. Then
(cA)T = cAT = cA,
so cA ∈ H and H is closed under scalar multiplication.
Therefore, H is a subspace of M12×12 .
(8) H is the set of all functions on [0, 1] such that f (0)f (1) ≥ 0. Is H a subspace of F ? Why
or why not?
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(a) The zero function, f (t) = 0 for all t ∈ R, satisfies f (0)f (1) = 0, so the zero function
is in H.
(b) Let g ∈ H =⇒ g(0)g(1) ≥ 0 and let h ∈ H =⇒ h(0)h(1) ≥ 0. We must determine
if g + h ∈ H. This does not seem likely. We need only come up with an example
where g, h ∈ H, but g + h ∈
/ H.
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Let g(t) = 1 − t . Then g(0) = 1 and g(1) = 0, so g(0)g(1) = 0 and g ∈ H.
Let h(t) = t2 − 2t. Then h(0) = 0 and h(1) = −1, so h(0)h(1) = 0 and h ∈ H.
But, (g + h)(t) = g(t) + h(t) = 1 − 2t. Since (g + h)(0) = 1 and (g + h)(1) = −1,
[(g + h)(0)][(g + h)(1)] = −1 < 0 and g + h ∈
/ H. So, H is not closed under addition.
Therefore, H is not a subspace of F .
Theorem 1. If v1 , v2 , . . . , vp are in a vector space V , then Span{v1 , v2 , . . . , vp } is a subspace
of V .
So, Span{v1 , v2 , . . . , vp } is the subspace spanned by v1 , v2 , . . . , vp .
Given any subspace H of V , a spanning set for H is a set of vectors v1 , v2 , . . . , vp in H such
that H = Span{v1 , v2 , . . . , vp }.
Example: Let W be the set of all vectors of the form


2b + 3c
 −5b  ,
2c
where b, c are arbitrary. Show that W is a subspace of R3 .
Solution: Let x ∈ W . Then


 
 
2b + 3c
2
3





−5b
x=
= b −5 + c 0 .
2c
0
2
So, W = Span{(2, −5, 0), (3, 0, 2)} =⇒ W is a subspace of R3 .
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Null Spaces, Column Spaces, and Linear Transformations – Section 4.2 of Lay
Recall:
Definition. The null space of an m × n matrix A, Nul A, is the set of all solutions of the
homogeneous equation Ax = 0; i.e.,
Nul A = {x ∈ Rn : Ax = 0}.
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Theorem 2. The null space of an m × n matrix A is a subspace of Rn ; i.e., the set of all
solutions of a system Ax = 0 is a subspace of Rn .
Proof. (a) 0 ∈ Nul A since A0 = 0.
(b) Let u, v ∈ Nul A. Then Au = 0 and Av = 0. So,
A(u + v) = Au + Av
= 0 + 0 = 0.
Therefore, u + v ∈ Nul A and Nul A is closed under addition.
(c) Let u ∈ A and let c be a scalar. Then
A(cu) = c(Au)
= c0 = 0,
so, cu ∈ Nul A and Nul A is closed under scalar multiplication.
Definition. The column space of an m × n matrix A, Col A, is the set of all linear combinations of the columns of A. If A = a1 a2 · · · an , then Col A = Span{a1 , a2 , . . . , an }.
Theorem 3. The column space of an m × n matrix is a subspace of Rm .
Note: If v ∈ Col A for some m × n matrix A, then v = Ax for some x ∈ Rn (because Ax is a
linear combination of the column vectors of A). So,
Col A = {b ∈ Rm : b = Ax for some x ∈ Rn }.
Alternately, Col A is in the range of the linear transformation x 7→ Ax.
Examples:
1 2 3
(1) Consider the matrix A =
.
4 5 6
(a) If Col A is a subspace of Rk , what is k? Why?
(b) If Nul A is a subspace of Rk , what is k? Why?
Solution:
(a) k = 2 since the columns of A are in R2 .
(b) k = 3, because for Ax to be defined, x must be in R3 .
1 2
−2
(2) Let A =
. Is u =
∈ Col A? In Nul A?
4 8
1
8
To determine if u ∈ Nul A, we must solve Ax = 0.
1 2 | 0 r2 →r2 −4r2 1 2 | 0
−−−−−−→
.
4 8 | 0
0 0 | 0
We see that x2 is free and x1 + 2x2 = 0, so x1 = −2x2 and
−2x2
−2
x=
= x2
.
x2
1
So, u ∈ Nul A.
1
We see that Col A = Span
and so u ∈
/ Col A.
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(3) Find a matrix A so that W = Col A, where



 2a − b



3b
W =
: a, b ∈ R .


a+b
We have that


2a − b
x ∈ W =⇒ x =  3b 
a+b
 
 
2
−1
= a 0 + b  3 
1
1
    
−1 
 2
=⇒ W = Span 0 ,  3  .


1
1


2 −1
Therefore, A = 0 3  is one such matrix.
1 1
Kernel and Range of a Linear Transformation
Definition. A mapping T from a vector space V into a vector space W is a linear transformation if
(i) T (u + v) = T (u) + T (v) for all u, v ∈ V , and
(ii) T (cu) = cT (u) for all u ∈ V and all scalars c.
Examples:
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(1) Is T : R2 → R defined by
T (x) = x1 + x2
a linear transformation?
• Let x, y ∈ R2 . Then x + y = (x1 + x2 , y1 + y2 ). So,
T (x + y) = (x1 + y1 ) + (x2 + y2 )
= (x1 + x2 ) + (y1 + y2 )
= T (x) + T (y).
• Let x ∈ R2 and let c be a scalar. Then cx = (cx1 , cx2 ) and
T (cx) = cx1 + cx2
= c(x1 + x2 )
= cT (x).
Therefore, T is a linear transformation.
(2) Is T : C[a, b] → R defined by
Z
b
f (x) dx
T (f ) =
a
a linear transformation?
• Let f, g ∈ C[a, b]. Then
b
Z
T (f + g) =
(f + g)(x) dx
a
b
Z
(f (x) + g(x)) dx
Z b
Z b
=
f (x) dx +
g(x) dx
=
a
a
a
= T (f ) + T (g).
• Let f ∈ C[a, b] and let c be a scalar. Then
Z b
T (cf ) =
(cf )(x) dx
a
Z b
=
cf (x) dx
a
Z b
=c
f (x) dx
a
= cT (f ).
Therefore, T is a linear transformation.
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(3) Is T : M2×2 → M2×2 defined by
T (A) = A + I
a linear transformation?
1 0
0 1
1 1
No. Why? Let A =
and let B =
. Then A + B =
and
0 1
1 0
1 1
2 1
T (A + B) = (A + B) + I =
,
1 2
But, since
2 0
T (A) =
0 2
1 1
3 1
and T (B) =
, T (A) + T (B) =
6 T (A + B).
=
1 1
1 3
Therefore, T is not a linear transformation.
Definition. Let T : V → W be a linear transformation.
(a) The kernel of T , ker T , is defined by
ker T = {v ∈ V : T (v) = 0},
where 0 is the zero vector in W .
(b) The range of T is defined by
range T = {w ∈ W : w = T (v) for some v ∈ V }.
Example: Let D : P3 → P2 be the differentiation operator defined by
D(p) = p0 .
(a) Verify that D is a linear transformation.
(b) Describe the kernel and range of D.
Solution:
(a) We need to verify that D satisfies the requirements for a linear transformation.
• Let p, q ∈ P3 . Then,
D(p + q) = (p + q)0
= p0 + q 0
= D(p) + D(q).
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• Let p ∈ P3 and let c be a scalar. Then,
D(cp) = (cp)0
= cp0 = cD(p).
Since D satisfies the properties of a linear transformation, D is a linear transformation.
(b) The kernel of D is the set of all polynomials in P3 such that p0 = 0. If p ∈ P3 , then
p(t) = a0 + a1 t + a2 t2 + a3 t3 . So, p0 (t) = a1 + 2a2 t + 3a3 t2 . Then, p0 (t) = 0 for all t only if
a1 = a2 = a3 = 0, or p(t) = a0 , a polynomial of degree zero.
The range of D is a bit harder to determine. We know that q ∈ P2 is in the range of T if
we can find p ∈ P3 such that
p0 = q.
Since p ∈ P3 , we know that D(p) ∈ P2 . Similarly, for any q ∈ P2 , we can find a polynomial
p ∈ P3 so that p0 = q. Therefore, the range of D is P2 .
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Linearly Independent Sets; Basis – Section 4.3 of Lay
Definition. A set of vectors {v1 , v2 , . . . , vp } in a vector space V is linearly independent
(LI) if the vector equation
c1 v1 + c2 v2 + · · · + cp vp = 0
(1)
has only the trivial solution c1 = c2 = · · · = cp = 0. The set {v1 , v2 , . . . , vp } is linearly
dependent if (1) has a nontrivial solution, i.e., if there exist some weights c1 , c2 , . . . , cp , not all
zero such that (1) holds. In such a case, (1) is a linear dependence relation among v1 , v2 , . . . , vp .
Theorem 4. An indexed set {v1 , v2 , . . . , vp } of two or more vectors, with v1 6= 0 is linearly dependent if and only if there exists some j such that vj is a linear combination of v1 , v2 , . . . , vj−1 .
Determining if a Set of Vectors is Linearly Independent
Theorem 5. Let x1 , x2 , . . . , xn ∈ Rn , and let X = x1 x2 · · ·
is linearly dependent if and only if X is singular.
xn . Then, {x1 , x2 , . . . , xp }
Proof. The equation c1 x1 + c2 x2 + · · · + cn vn = 0 can be written as
Xc = 0.
This equation has nontrivial solution if and only if X is singular. Thus, {x1 , x2 , . . . , xp } is
linearly dependent if and only if X is singular.
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Example: Are (4, 2, 3), (2, 3, 1), (2, −5, 3) linearly independent?
Solution: To see if the vectors are linearly independent, we first form the matrix whose columns
are the vectors and then determine if that matrix is singular. So, we need to check if


4 2 2
2 3 −5
3 1 3
is singular. Since
4 2 2 2 3 −5 = 0,
3 1 3 the vectors are not linearly independent.
Example: (Review) Are (1, −1, 2, 3), (−2, 3, 1, −2), (1, 0, 7, 7) linearly independent?
Solution: We need to determine if the vector equation
c1 (1, −1, 2, 3) + c2 (−2, 3, 1, −2) + c3 (1, 0, 7, 7) = 0
has only the trivial solution. We form the augmented matrix and perform Gaussian elimination.






1 −2 1 | 0
1 −2 1 | 0
1 −2 1 | 0
−1 3 0 | 0





 −→ 0 1 1 | 0 −→ 0 1 1 | 0 .
2
0 5 5 | 0
0 0 0 | 0
1 7 | 0
3 −2 7 | 0
0 4 4 | 0
0 0 0 | 0
We see that c3 is a free variable, so there are infinitely many nontrivial solutions. Therefore,
the vectors are linearly dependent.
What about Functions? Matrices?
1 0
3 4
0 0
Example: Are
,
,
linearly independent?
2 5
−1 6
1 1
Solution: We need to determine if the equation
1 0
3 4
0 0
0 0
c1
+ c2
+ c3
=
2 5
−1 6
1 1
0 0
has only the trivial solution. We have
c1 + 3c2
4c2
0 0
=
.
2c1 − c2 + c3 5c1 + 6c2 + c3
0 0
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Applying the property of matrix equality gives the system
c1 + 3c2
4c2
2c1 − c2 + c3
5c1 + 6c2 + c3
=0
= 0 =⇒ c2 = 0 =⇒ c1 = 0
= 0 =⇒ c3 = 0
=0
So, c1 = c2 = c3 = 0 is the only solution. Therefore, the matrices are linearly independent.
Example: Are p1 (t) = t2 − 2t + 3, p2 (t) = 2t2 + t + 8, p3 (t) = t2 + 8t + 7 linearly independent?
Solution: We need to determine if the equation c1 p1 (t) + c2 p2 (t) + c3 p3 (t) = 0 has only the
trivial solution. So, we will solve
c1 (t2 − 2t + 3) + c2 (2t2 + t + 8) + c3 (t2 + 8t + 7) = 0
for c1 , c2 , and c3 . Collect terms to obtain
(c1 + 2c2 + c3 )t2 + (−2c1 + c2 + 8c3 )t + 3c1 + 8c2 + 7c3 = 0.
Equate coefficients of like terms to obtain the system of equations
c1 + 2c2 + c3 = 0
−2c1 + c2 + 8c3 = 0
3c1 + 8c2 + 7c3 = 0.
This is equivalent to the equation


1 2 1
−2 1 8 c = 0,
3 8 7
which has only the trivial solution if the coefficient matrix is nonsingular. Since
1 2 1
−2 1 8 = 0,
3 8 7
p1 , p2 , and p3 are not linearly independent.
Functions in C n−1 [a, b]
Let f1 , f2 , . . . , fn ∈ C n−1 [a, b]. We start by searching for a direct way to determine if the
functions are linearly dependent. If the functions are linearly dependent, then there exist
weights c1 , c2 , . . . , cn , not all zero such that
c1 f1 (t) + c2 f2 (t) + · · · cn fn (t) = 0
14
(2)
for each t ∈ [a, b].
Differentiating both sides of (2) with respect to t gives
c1 f10 (t) + c2 f20 (t) + · · · cn fn0 (t) = 0.
We may continue differentiating to obtain a system of n equations in n unknowns (the ci , i =
1, 2, . . . , n).
c1 f1 (t)
c1 f10 (t)
(n−1)
c1 f 1
+ ···
+ ···
..
.
+ c2 f2 (t)
+ c2 f20 (t)
(n−1)
(t) + c2 f2
(t) + · · ·
+ cn fn (t)
+ cn fn0 (t)
(n−1)
+ cn f n
= 0
= 0
..
.
(3)
(t) = 0.
Writing (3) as a matrix equation which for each t ∈ [a, b] has the same solution set as (3):


f1 (t)
f2 (t)
···
fn (t)
 f10 (t)
f20 (t)
···
fn0 (t) 


(4)
 c = 0.

..
..
..


.
.
.
(n−1)
f1
(n−1)
(t) f2
(t) · · ·
(n−1)
fn
(t)
So, if the functions are linearly dependent in C n−1 [a, b], for each t ∈ [a, b], the coefficient matrix
in (4) is singular. In other words, the determinant of the coefficient matix in (4) is 0 for each
t ∈ [a, b]. This leads to the following:
Definition. Let f1 , f2 , . . . , fn ∈ C n−1 [a, b]. The Wronskian of f1 , f2 , . . . , fn on [a, b], denoted
W [f1 , f2 , . . . , fn ](t), is the determinant of the coefficient matrix in (4).
Theorem 6. Let f1 , f2 , . . . , fn ∈ C n−1 [a, b]. If there exists a point t0 ∈ [a, b] such that
W [f1 , f2 , . . . , fn ](t0 ) 6= 0 then f1 , f2 , . . . , fn are linearly independent.
Examples:
(1) Determine if the functions f (x) = cos(πx) and g(x) = sin(πx) are linearly independent
in C 1 [0, 1].
f (x) g(x) cos(πx)
sin(πx) W [f, g](x) = 0
=
f (x) g 0 (x) −π sin(πx) π cos(πx)
= π cos2 (πx) + π sin2 (πx) = π 6= 0.
Therefore, f and g are linearly independent in C 1 [0, 1].
15
(2) Determine if f (t) = 1, g(t) = et + e−t , h(t) = et − e−t are linearly independent.
f (t) g(t) h(t) 1 et + e−t et − e−t W [f, g, h](t) = f 0 (t) g 0 (t) h0 (t) = 0 et − e−t et + e−t f 00 (t) g 00 (t) h00 (t) 0 et + e−t et − e−t t
−t
et + e−t 1+1 e − e
= 1(−1) t
e + e−t et − e−t = (et − e−t )2 − (et + e−t )2
= e2t − 2 + e−2t − (e2t + 2 + e−2t )
= −4 6= 0.
Therefore, the functions are linearly independent.
Basis
Definition. Let H be a subspace of a vector space V . A set B = {b1 , b2 , . . . , bp } in V is a
basis of H if
(i) B is linearly independent, and
(ii) B spans H (i.e., H = Span{b1 , b2 , . . . , bp }).
1 0
0 1
0 0
0 0
Example: Show that the set B =
,
,
,
is a basis of M2×2 .
0 0
1 0
0 1
0 0
Solution: We need to show that B is linearly independent and spans M2×2 . To verify that B
is linearly independent, we need to show that the vector equation
1 0
0 1
0 0
0 0
0 0
c1
+ c2
+ c3
+ c4
=
0 1
0 0
1 0
0 1
0 0
has only the trivial solution. But this is equivalent to
c1 c2
0 0
=
,
c3 c4
0 0
which has as its unique solution c1 = c2 = c3 = c4 = 0. So, B is linearly independent.
a b
We next show that B spans M2×2 . Let A =
∈ M2×2 . Then,
c d
a b
1 0
0 1
0 0
0 0
=a
+b
+c
+d
.
c d
0 0
0 0
1 0
0 1
=⇒ B spans M2×2 .
Therefore, B is a basis of M2×2 .
16
Theorem 7 (The Spanning Set Theorem). Let S = {v1 , v2 , . . . , vp } in vector space V and let
H = Span{v1 , v2 , . . . , vp }.
(a) If vk is a linear combination of the other vectors in S for some k, then the set formed by
removing vk from S spans H.
(b) If H 6= {0}, some subset of S is a basis of H.
Proof. (a) Without loss of generality, assume that vp is a linear combination of v1 , v2 , . . . , vp−1 .
So,
vp = a1 v1 + a2 v2 + · · · ap−1 vp−1
for some constants a1 , a2 , . . . , an . Let x ∈ H. Then,
x = c1 v1 + c2 v2 + · · · cp−1 vp−1 + cp vp
for some scalars c1 , c2 , . . . , cp−1 , cp . Then,
x = c1 v1 + c2 v2 + · · · cp−1 vp−1 + cp (a1 v1 + a2 v2 + · · · ap−1 vp−1 )
= (c1 + a1 )v1 + (c2 + a2 )v2 + · · · + (cp−1 + ap−1 )vp−1 .
Therefore, {v1 , v2 , . . . , vp−1 } spans H.
(b) If S is linearly independent, we are done. If not, one vector can be written as a linear
combination of the others and can be deleted. Repeat this procedure until the set is
linearly independent. If at least one vector, v 6= 0 remains, then the set is linearly
independent and is, thus, a basis of H.
Recall:
• Null space = set of all solutions of Av = 0.
• Column space = span of the columns of the matrix.
Example: Find a basis for the null space and for the column space of


−2
4 −2 −4
1 .
A =  2 −6 −3
−3
8
2 −3
Solution: First, determine the solutions of Ax = 0.






−2
4 −2 −4 | 0
−2
4 −2 −4 | 0
−2
4 −2 −4 | 0
r →r2 +r1
r3 →r3 +r2
 2 −6 −3
1 | 0 −−2−−−
−3−→  0 −2 −5 −3 | 0 −−
−−−−→  0 −2 −5 −3 | 0
r3→r3 − 2 r1
−3
8
2 −3 | 0
0
2
5
3 | 0
0
0
0
0 | 0
17
Since columns 3 and 4 are not pivot columns, x3 and x4 are free variables. The system we now
have is
−2x1 + 4x2 − 2x3 − 4x4 = 0
−2x2 − 5x3 − 3x4 = 0.
Let x3 = r and x4 = s. Then,
1
5
3
−2x2 − 5x3 − 3x4 = 0 =⇒ x2 = (−5x3 − 3x4 ) = − r − s
2
2
2
1
−2x1 + 4x2 − 2x3 − 4x4 = 0 =⇒ x1 = (4x2 − 2x3 − 4x4 ),
2
or
x1 = 2x2 − x3 + 2x4
5
3
= 2 − r − s − r − 2s
2
2
= −6r − 5s.
All solutions x of the equation Ax = 0 take the form

 

 
−6r − 5s
−5
−6
 − 5 r − 3 s
− 3 
− 5 
2
2  = r  2 + s  2 .
x=

 0 

 1 
r
0
1
s
Therefore, a basis of Nul A is
   
−6
−5 


 5   3 

−
−
 2 ,  2 .
 1   0 





0
1
Next, since


−2 4 −2 −4 | 0
A ∼  0 −2 −5 −3 | 0 ,
0
0
0
0 | 0
the pivot columns are columns 1 and 2. Therefore, a basis of Col A is
   
4 
 −2
 2 , −6 .


−3
8
4
Coordinate Systems – Section 4.4 of Lay
A coordinate system is imposed on a vector space V by specifying a basis B.
18
Theorem 8 (The Unique Representation Theorem). let B = {b1 , b2 , . . . , bn } be a basis of a
vector space V . Then, for each x ∈ V , there exists a unique set of scalars c1 , c2 , . . . , cn such
that
x = c1 b1 + c2 b2 + · · · + cn bn .
(5)
Proof. Since B spans V , there exist scalars c1 , c2 , . . . , cn such that (5) holds. Suppose that there
are also scalars d1 , d2 , . . . , dn such that
x = d1 b1 + d2 b2 + · · · + dn bn .
Then,
0=x−x
= (c1 − d1 )b1 + (c2 − d2 )b2 + · · · + (cn − dn )bn
(6)
Since B is linearly independent, (6) only holds if ci − di = 0, for 1 ≤ i ≤ n, or ci = di for
1 ≤ i ≤ n.
Definition. Suppose B = {b1 , b2 , . . . , bn } is a basis of a vector space V and x ∈ V . The
coordinates of x relative to the basis B, or the B-coordinates of x, are the weights
c1 , c2 , . . . , cn such that x = c1 b1 + c2 b2 + · · · + cn bn .
Definition. If c1 , c2 , . . . , cn are the B-coordinates of x, then the vector in Rn
 
c1
 c2 
 
[x]B =  .. 
.
cn
is the coordinate vector of x (relative to B), or the B-coordinate vector of x. The
mapping x 7→ [x]B is the coordinate mapping (determined by B).
Example:
(1) Consider the vectors v1 = (2, 1) and v2 = (1, 4). v1 and v2 are linearly independent and
span R2 , so B = {v1 , v2 } is a basis of R2 .
(a) Write the B-coordinate vector of x = (7, 3).
Solution: We seek c1 and c2 so that
c 1 v1 + c 2 v2 = x
2 1 c1
7
=
.
1 4 c2
3
Solving:
2 1 | 7 r2 ↔r1 1 4 | 3 r2 →r2 −2r1 1 4 | 3
−−−→
−−−−−−→
.
1 4 | 3
2 1 | 7
0 −7 | 1
19
This gives the system
c1 + 4c2 = 3
−7c2 = 1 =⇒ c2 = −
1
7
1
c1 = 3 − 4c2 = 3 − 4 −
7
=
25
.
7
Therefore,
[x]B =
25
7
− 71
.
(b) If [x]B = (1, 3), find x.
Solution:
x = c1 v1 + c2 v2
= 1(2, 1) + 3(1, 4)
= (5, 13).
(2) Let B = {1, t} be a basis of P1 . (Actually, it is the standard basis.)
(a) What is [p]B for p(t) = 3 − 2t?
Solution:
3 − 2t = c1 (1) + c2 (t)
=⇒ c1 = 3, c2 = −2.
so,
3
[p]B =
.
−2
−1
(b) If [p]B =
, what is p(t)?
3
Solution: p(t) = −1 + 3t.
Geometric Interpretation
Example: Consider an ellipse that has been rotated counterclockwise an angle θ in the plane.
If the coordinate x- and y-axes are rotated counterclockwise by θ, the new axes formed allow
the representation of the ellipse using the standard formula.
20
Coordinates in Rn
In Example (1) after the definition of coordinate vectors, we saw that we can easily find the
B-coordinates of a vector x ∈ R2 in standard coordinates. In general, if x ∈ Rn and B =
{b1 , b2 , . . . , bn } is a basis of Rn , then
 
c1
 c2 
 
[x]B =  ..  ,
.
cn
where c1 b1 + c2 b2 + · · · + cn bn = x, or b1 b2 · · ·
bn c = x.
b
b
·
·
·
b
And, given [x]B = c, we can easily
find
the
standard
coordinate
vector
x
by
x
=
c.
1
2
n
Define PB = b1 b2 · · · bn . Then x = PB [x]B . PB is the change-of-coordinates matrix
from B to the standard basis in Rn .
Since the columns of PB form a basis of Rn , PB is invertible, so
PB−1 x = [x]B .
The correspondence x 7→ [x]B , produced here by PB−1 , is the coordinate mapping. Since PB−1
is invertible, the coordinate mapping is a one-to-one linear transformation from Rn to Rn (see
the Invertible Matrix Theorem).
The Coordinate Mapping
Theorem 9. Let B = {b1 , b2 , . . . , bn } be a basis of a vector space V . Then the coordinate
mapping x 7→ [x]B is a one-to-one linear transformation from V onto Rn .
To prove this theorem, we need to show (1) that the coordinate mapping is a linear transformation, (2) that the coordinate mapping is one-to-one, and (3) that the coordinate mapping is
onto.
Definition. A one-to-one linear transformation from a vector space V onto a vector space W
is called an isomorphism from V onto W . If such a mapping exists, then V is isomorphic
to W .
Example: Show that M2×2 is isomorphic to R4 .
Solution: Let B be the standard basis of M2×2 , i.e.,
1 0
0 1
0 0
0 0
B=
,
,
,
.
0 0
0 0
1 0
0 1
21
Any element A ∈ M2×2 has the form
a b
1 0
0 1
0 0
0 0
A=
=a
+b
+c
+d
.
c d
0 0
0 0
1 0
0 1
So,
 
a
b

[A]B = 
c .
d
Thus, the coordinate mapping A 7→ [A]B is an isomorphism from M2×2 onto R4 . We say that
M2×2 is isomorphic to R4 .
In other words, all vector operations in M2×2 correspond to operations in R4 .
We can use coordinate vectors to show that a set of elements in a vector space V is linearly
independent (or spans V ).
Example:
(1) Show that {2t − 1, 2t + 1} is linearly independent. Use the coordinate mapping to the
standard basis of polynomials in P1 , B = {1, t}.
Solution: We see that
−1
1
[2t − 1]B =
= v1 and [2t + 1]B =
= v2 .
2
2
If v1 and v2 are linearly independent, then 2t − 1 and 2t + 1 are also linearly independent.
We know that two vectors x1 and x2 in R2 are linearly independent if the matrix v1 v2
is nonsingular. Since
−1 1
2 2 = −4 6= 0,
v1 and v2 are linearly independent, and so is {2t − 1, 2t + 1}.
Note: We could also have shown that v1 and v2 are linearly independent directly by
definition.
(2) Show that {2t − 1, 2t + 1} spans P1 . Use the coordinate mapping to the standard basis of
polynomials in P1 , B = {1, t}.
Solution: From Example (1), we saw that
−1
1
[2t − 1]B =
= v1 and [2t + 1]B =
= v2 .
2
2
So, we need only show that {v1 , v2 } spans R2 . To show this, we must show that for each
x ∈ R2 , there exist constants c1 and c2 such that c1 v1 + c2 v2 = x, or
−1
1
x
c1
+ c2
= 1 .
2
2
x2
22
Use Gaussian elimination to solve:
−1 1 | x1 r2 →r2 +2r1 −1 1 |
x1
−−−−−−→
.
2 2 | x2
0 4 | x2 + 2x1
Since the last column is not a pivot column, we know that there is at least one solution.
Therefore, {v1 , v2 } spans R2 =⇒ {2t − 1, 2t + 1} spans P1 .
Notes:
(i) The examples above show that {2t − 1, 2t + 1} is a basis of P1 .
(ii) P1 is isomorphic to R2 .
Example: Use coordinate vectors to show that
1 1
−2 3
0 6
S=
,
,
0 1
4 5
7 −4
is linearly independent.
Solution: It is easiest to use the standard basis. Then
 
 
1
−2




1 1
1
−2 3
 3 ,
=
,
=
4 5 B 4
0 1 B 0
1
5
 
0
6
0 6
= 
.
7 −4 B  7 
−4
We next show that the vectors thus obtained are linearly independent. So, we need to show
that the only solution of
   
 
 
0
0
−2
1
3
 6  0
1
 
   

c1 
0 + c2  4  + c3  7  = 0
−4
0
5
1
23
is the trivial solution, c1 = c2 = c3 = 0.



1 −2 0 | 0
1
1 3
 r2 →r2 −r1 0
6
|
0

 −−−−−−→ 
0 4
7 | 0 r4 →r4 −r1 0
1 5 −4 | 0
0

1

r3 →r3 −4r2 0
−−
−−−−→
r4 →r4 −7r2 0
0

1

r4 →r4 −3r3 0
−−
−−−−→ 
0
0
−2 0
5
6
4
7
7 −4
|
|
|
|
−2 0 |
1 −1 |
0 11 |
0
3 |
−2 0 |
1 −1 |
0
1 |
0
0 |



0
1 −2 0 | 0
0
0 1 −1 | 0
r2 →r2 −r3 
−

−−−−−→ 


0
0 4
7 | 0
0
0 7 −4 | 0



1 −2 0 | 0
0
1
r3 → 11
r3 0
1 −1 | 0
0


−
−
−
−
−
→
0 0
1 | 0
0
0
0 0
3 | 0

0
0

0
0
There are no free variables, so the vector equation has only the trivial solution. Therefore the
coordinate vectors are linearly independent =⇒ S is linearly independent.
NOTE: We had no real savings in work here, since the system of equations obtained is identical
to the one obtained by directly showing that the matrices are linearly independent.
5
Dimension of a Vector Space – Section 4.5 of Lay
Since the coordinate mapping from a basis B = {b1 , b2 , . . . , bn } of a vector space V is one-toone and onto Rn , then V is isomorphic to Rn . This leads to the following theorem.
Theorem 10. If a vector space V has a basis B = {b1 , b2 , . . . , bn }, then any set in V containing
more than n vectors must be linearly dependent.
Proof. Let {u1 , u2 , . . . , up } ∈ V , where p > n. Then {[u1 ]B , [u2 ]B , . . . , [up ]B } is linearly dependent in Rn because p > n. So, there exist c1 , c2 , . . . , cp not all zero such that
c1 [u1 ]B + c2 [u2 ]B + · · · + cp [up ]B = 0n
=⇒ [c1 u1 + c2 u2 + · · · + cp up ]B = 0n
=⇒ c1 u1 + c2 u2 + · · · + cp up = 0V .
Therefore, {u1 , u2 , . . . , up } is linearly dependent.
Theorem 11. If a vector space V has a basis consisting of n vectors, then every basis of V
contains exactly n vectors.
Definition. If V is spanned by a finite set, then V is finite-dimensional and the dimension
of V, dim V , is the number of vectors in a basis of V . The dimension of the zero vector space,
{0} is defined to be zero. If V is not spanned by a finite set, then V is infinite-dimensional.
24
Example:
1. dim R2 = 2, dim R3 = 3, . . . , dim Rn = n
2. dim P1 = 2, dim P2 = 3, . . . , dim Pn = n + 1
3. The set P of all polynomials is infinite-dimensional.
4. The set C n (R) is infinite-dimensional.
Examples:
(1) Find the dimension of the subspace



 a + 2b

H =  b − c  : a, b, c ∈ R .


5c
Solution:


 
 
 
a + 2b
1
2
0







v ∈ H =⇒ v = b − c = a 0 + b 1 + c −1 = av1 + bv2 + cv3 .
5c
0
0
5
Therefore, a basis of H is {v1 , v2 , v3 } =⇒ dim H = 3.
1
−2
−3
(2) Find dim H, where H = Span
,
,
.
−5
10
15
1 −2 −3
Solution: First, we need to find a basis of Col A, where A =
.
−5 10 15
1 −2 −3 r2 →r2 +5r1 1 −2 −3
−−−−−−→
.
−5 10 15
0 0
0
1
The first column is the only pivot column, so
is a basis of Col A, so H =
−5
1
Span
. Therefore, dim H = 1.
−5
(3) Find dim S, where S is the set of 2 × 2 symmetric matrices.
Solution:
a b
1 0
0 1
0 0
A ∈ S =⇒ A =
=a
+b
+c
.
b c
0 0
1 0
0 1
Therefore, a basis of S is
1 0
0 1
0 0
,
,
,
0 0
1 0
0 1
and dim S = 3.
25
(4) Find dim H, where H = {p ∈ P2 : a1 = a2 }.
Solution: p ∈ H =⇒ p(t) = a0 + a1 t + a1 t2 = a0 + a1 (t + t2 ). Therefore, a basis of H is
{1, t + t2 }, and dim H = 2.
Subspaces of Finite-Dimensional Vector Spaces
Theorem 12. Let H be a subspace of a finite-dimensional vector space V . Any linearly
independent set in H can be expanded, if necessary, to be a basis of H. Also, H is finitedimensional and
dim H ≤ dim V.
Why? Idea: If H = {0}, then dim H = 0 ≤ dim V . Otherwise, if S = {u1 , u2 , . . . , up } is a
linearly independent set of vectors in H, then either H = Span S or we can find up+1 ∈
/ Span S
such that {u1 , u2 , . . . , up , up+1 } is linearly independent. Call this new set S, as well. Keep
doing this until H = Span S. The number of vectors in S cannot exceed dim V or the vectors
will be linearly dependent. Therefore, dim H ≤ dim V .
Theorem 13 (The Basis Theorem). Let V be a p-dimensional vector space with p ≥ 1. Any
linearly independent set of p vectors in V is a basis of V . Any set of p vectors in V that spans
V is a basis of V .
The Basis Theorem follows directly by applying Theorem 12 and the Spanning Set Theorem
to V .
Example: Show that {1 + 2t, t2 − t, t − 1} is a basis of P2 .
Solution: Since dim P2 = 3 and the set contains three vectors, it is a basis of P2 if it is linearly
independent.
c1 (1 + 2t) + c2 (t2 − t) + c3 (t − 1) = 0
c2 t2 + (2c1 − c2 + c3 )t + c1 − c3 = 0.
Equate coefficients of like terms.
c2 = 0
2c1 − c2 + c3 = 0
c1 − c3 = 0.
Solve using Gaussian elimination.



0 1
0 | 0
2 −1 1 |
r1 ↔r2
2 −1 1 | 0 −

0 |
−−→ 0 1
1 0 −1 | 0
1 0 −1 |

1 −1 2
r3 →r3 −r1

0
−−−−−−→ 0 1
0 1 −3
26


0
1 −1
r1 →r1 −r3


0 −−−−−−→ 0 1
0
1 0


| 0
1
r3 →r3 −2r2


| 0 −−−−−−→ 0
| 0
0

2 | 0
0 | 0
−1 | 0

−1 2 | 0
1
0 | 0 .
0 −3 | 0
Back substitution gives us
−3c3 = 0 =⇒ c3 = 0
c2 = 0
c2 − c2 + 2c3 = 0 =⇒ c1 = 0.
Since c1 = c2 = c3 = 0 is the only solution, the set is linearly independent. Therefore, it is a
basis of P2 .
Note: From previous work, we see that:
• dim Nul A = number of free variables; and
• dim Col A = number of pivot columns.
6
Rank – Section 4.6 of Lay
The Row Space
Definition. The set of all linear combinations of the row vectors of an m × n matrix A is the
row space of A, Row A.
Theorem 14. If two matrices A and B are row equivalent, then their row spaces are the same.
If B is in echelon form, then the nonzero rows of B form a basis of both Row A and Row B.
Proof. (Skipped in class.)
If B is obtained from A by elementary row operations, any linear combination of the rows of
B is a linear combination of the rows of A. Thus, Row B ⊆ Row A. Since elementary row
operations are reversible, Row A ⊆ Row B. Therefore, Row B = Row A. If B is in echelon
form, its nonzero rows are linearly independent, so the nonzero rows of B form a basis of Row A
and Row B.
Example: Find a basis of the row space, column space, and null space of


1 −4
9 −7
2 −4
1 .
A = −1
5 −6 10
7
Solution:

 

1 −4
9 −7
1
0 −1
5
−1
2 −4
1 ∼ 0 −2
5 −6 .
5 −6 10
7
0
0
0
0
27
Therefore, a basis of Row A is {(1, 0, −1, 5), (0, −2, 5, −6)}. Note that dim Row A = 2.
Also, a basis of Col A is
   
−4 
 1
−1 ,  2 .


5
6
Now, we will find a basis of the nullspace of A. Since A | 0 ∼ B | 0 , we have


1
0 −1
5 | 0
0 −2
5 −6 | 0 .
0
0
0
0 | 0
We see that columns 3 and 4 are not pivot columns, so x3 and x4 are free variables. We thus
have
x1 − x3 + 5x4 = 0 =⇒ x1 = x3 − 5x4
5
−2x2 + 5x3 − 6x4 = 0 =⇒ x2 = x3 − 3x4 .
2
Let x3 = 2r and x4 = s. Then
x2 = 5r − 3s
x1 = 2r − 5s.
So,

 
 
2r − 5s
2
−5
5r − 3s
5
−3

 
 
x=
 2r  = r 2 + s  0  .
s
0
1

Therefore, a basis of Nul A is
   
2
−5 


   

5
  , −3 .
2  0 





0
1
Thus, dim Nul A = 2.
Definition.
1. The rank of A, rank A, is dim Col A.
2. The nullity of A is dim Nul A.
Theorem 15 (Rank-Nullity Theorem). The dimensions of the column space and the row space
of an m × n matrix are equal. This common dimension, rank A, also equals the number of
pivot positions in A and satisfies
rank A + dim Nul A = n.
28
(7)
Proof.
rank A = number of pivot columns in A
= number of pivots in an echelon form B of A.
Since B has a nonzero row for each pivot, and these rows form a basis of Row A, rank A =
dim Row A. Now,
dim Nul A = number of free variables in Ax = 0
= number of columns of A that are not pivot columns.
But, {number of pivot columns} + {number of nonpivot columns} = {number of columns}.
So, (7) is true.
Example: Suppose a 4 × 7 matrix A has rank 2. Find dim Nul A, dim Row A, and rank AT .
Is Col A = R2 ?
Solution:
dim Col A = 2 =⇒ dim Row A = 2.
rank A + dim Nul A = 7 =⇒ dim Nul A = 5.
rank AT = dim Col AT = dim Row A = 2.
Col A 6= R2 because each column in A is in R4 .
Example: Suppose the solutions of a homogeneous system of five equations in six unknowns
are all multiples of one nonzero solution. Will the system necessarily have a solution for every
possible right-hand side? Why or why not?
Solution: Let A be the 5 × 6 coefficient matrix of the system. Then dim Nul A = 1. So,
dim Col A = 6 − 1 = 5. Since R5 is the only subspace of R5 whose dimension is 5, Col A = R5 .
Therefore, every nonhomogeneous system of equations has a solution.
What if A were 6 × 5? Then dim Col A = 5 − 1 = 4, so the columns of A do not span R6 .
7
Change of Basis – Section 4.7 of Lay
In some applications, a problem is initally described using a basis B, but the problem’s solution
is made easier by changing B to a new basis, C. So, each vector is assigned a new C-coordinate
vector.
Goal: Do this systematically.
29
Example: Consider two bases, B = {b1 , b2 } and C = {c1 , c2 } for a vector space V such that
b2 = 3c1 − 2c2 .
b1 = 2c1 + c2 ,
Suppose x = 5b1 + 2b2 . Find [x]C .
Solution:
[x]C = [5b1 + 2b2 ]C
= 5[b1 ]C + 2[b2 ]C
5
= [b1 ]C [b2 ]C
.
2
Since
2
[b1 ]C =
and
1
3
[b2 ]C =
,
−2
2 3
5
16
[x]C =
=
.
1 −2 2
1
This example is generalized in the following theorem.
Theorem 16. Let B = {b1 , b2 , . . . , bn } and C = {c1 , c2 , . . . , cn } be bases of vector space V .
Then there is a unique n × n matrix, P such that
C←B
[x]C = P [x]B .
C←B
The columns of P are the C-coordinate vectors of the vectors in the basis B; i.e.,
C←B
P = [b1 ]C [b2 ]C · · ·
C←B
[bn ]C .
The matrix P is called the change-of-coordinates matrix from B to C. Multiplication
C←B
by P converts B-coordinate vectors into C-coordinate vectors.
C←B
The columns of P are linearly independent because they are coordinate vectors of the lienarly
C←B
independent set B. Since P is square, it is invertible (by the Invertible Matrix Theorem). So,
C←B
[x]C = P [x]B
C←B
−1
=⇒ ( P ) [x]C = [x]B .
C←B
Therefore,
P
C←B
−1
converts C-coordinate vectors into B-coordinate vectors, or
P
−1
C←B
= P .
B←C
30
Change of Basis in Rn
7
−3
1
−2
Example: Let B =
,
and C =
,
.
5
−1
−5
2
(a) Find P .
C←B
(b) Find P .
B←C
2
(c) If [x]B =
, find [x]C .
5
Solution:
x1
y
(a) Let [b1 ]C =
and [b2 ]C = 1 . Then, by definition
x2
y2
x1
y1
c1 c2
= b1 and c1 c2
= b2 .
x2
y2
c
c
x
y
Since
the
coefficient
matrix
for
both
systems
is
the
same,
instead
solve
=
1
2
b1 b2 . So, we form the augmented matrix c1 c2 | b1 b2 and solve using GaussJordan elimination.
1 −2 | 7 −3 r2 →r2 +5r1 1 −2 | 7 −3 r2 →− 18 r2 1 −2 | 7 −3
−−−−−−→
−−−−−→
−5 2 | 5 −1
0 −8 | 40 −16
0 1 | −5 2
1 0 | −3 1
r1 →r1 +2r2
−−
−−−−→
.
0 1 | −5 2
Therefore,
−3
1
[b1 ]C = x =
and [b2 ]C = y =
.
−5
2
Then,
−3 1
P = [b1 ]C [b2 ]C =
.
−5 2
C←B
Note: From this example, we can see that
i
h
P .
c1 c2 | b1 b2 ∼ I | C←B
(b)
P =
B←C
=
P
−1
C←B
1
det P
C←B
−2 1
=
.
−5 3
31
2 −1
5 −3
(c)
[x]C = P [x]B
C←B
−3 1 2
=
−5 2 5
−1
=
.
0
Note: An alternate approach to finding P is to do as follows. Consider conversion to/from
C←B
standard coordinates. In other words, let x ∈ Rn . Then
PB [x]B = x
PC [x]C = x
[x]C = PC−1 x.
So,
[x]C = PC−1 x
= PC−1 PB [x]B .
Therefore,
P = PC−1 PB .
C←B
Example: In P2 find the change-of-coordinates matrix from the basis {1, t, t2 } to the basis
B = {1, 2t, 4t2 − 2}. Then, find [t]B .
Solution: To find the desired change-of-coordinates matrix, we need to find [1]B , [t]B , [t2 ]B . To
do this, we need to look at the following system.
1 = x1 + x2 (2t) + x3 (4t2 − 2) = 4x3 t2 + 2x2 t + (x1 − 2x3 )
t = y1 + y2 (2t) + y3 (4t2 − 2) = 4y3 t2 + 2y2 t + (y1 − 2y3 )
t2 = z1 + z2 (2t) + z3 (4t2 − 2) = 4z3 t2 + 2z2 t + (z1 − 2z3 ).
By equating coefficients, we could obtain a matrix system of equations [A|B], which we could
then solve. This does not look like fun.
An easier approach is to do the following. Since
1 = 1 · 1 + 0 · t + 0 · t2
2t = 0 · 1 + 2 · t + 0 · t2
4t2 − 2 = −2 · 1 + 0 · t + 4 · t2 ,
32
we obtain


1 0 −2
PB = 0 2 0  .
0 0 4
The desired change-of-coordinates matrix is PB−1 , which can easily be found.


1 0 21
PB−1 = 0 12 0  .
0 0 14
Then
 
0

[t] = 1 ,
0
so
   

0
0
1 0 21
1





1 = 21  .
[t]B = 0 2 0
0 0 14
0
0
33