Download Part 1 - UBC Math

Math 221, Section 2.5 and 2.6, Part 1
(1) Subspaces (2) Basis and dimension
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Subspace
Recall from Sec 1.3: for example in R3 ,
h1i
I Take u = 2 then
3
h1i
Span{u} = {s 2 |where s is a scalar}
3
I
is a 1-dimensional subspace (a line).
h1i
h1i
Take u = 2 and v = 0 , then u, v are linearly independent
2
3
Span{u, v} = {s
I
h1i
2
3
+t
h1i
0
2
|where s, t are scalars}
is a 2-dimensional subspace (a plane).
h3i
Take w = 2 , then
7
[uvw] =
h1 1 3i
202
327
→ ··· →
h1 0 1i
012
0 0 0 RREF
,
implies that w = 1u + 2v, and Span{u, v, w} = Span{u, v}.
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Defining properties of subspace
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A subspace is a subset H of Rn that has three properties.
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The zero vector 0 is in H.
If u, v are in H, then the sum u + v is in H.
If u is in H and c is a scalar, then the scalar multiple cu is in
H.
In words, a subspace is closed under addition and scalar
multiplication
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Verify that: H = Span{v1 , v2 } is a subspace of Rn .
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0 is in H, because 0 = 0v1 + 0v2 is a linear combination of
v1 , v2 .
If u = av1 + bv2 and v = cv1 + dv2 , then
u + v = (a + c)v1 + (b + d)v2
which shows that u + v is a linear combination of v1 , v2 .
If u = av1 + bv2 , then
cu = c(av1 + bv2 ) = cav1 + cbv2
which shows that cu is a linear combination of v1 , v2 .
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Examples
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Rn itself is a subspace because it has the three properties
required for a subspace.
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The zero subspace {0}, consisting of only the zero vector in
Rn , is a subspace.
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The subspace spanned by v1 , . . . , vr ,
Span{v1 , . . . , vr } = {all linear combinations of v1 , . . . , vr }
is a subspace of Rn .
(The verification of this statement is similar to the argument given on the
previous page.)
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Non-examples
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A line L not passing through the origin is not a subspace,
because it does not contain the origin, as required.
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Fig. 2 shows that L is not closed under addition or scalar
multiplication.
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Non-examples
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In general, a translation of a subspace away from the origin by
a non-zero vector p, which is
p + Span{v1 , . . . , vr }
is not a subspace of Rn .
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Basis: the smallest spanning subset
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Because a subspace contains an infinite number of vectors,
some problems involving a subspace are handled best by
working with a small finite subset of vectors that span the
subspace.
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The fewer the spanning vectors, the better.
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The smallest possible spanning set must be linearly
independent.
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(Recall from Sec 1.7, if we put the vector together to form a matrix A,
then all columns of REF (A) contains a pivot.)
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h1i
h1i
h3i
Example: again take u = 2 , v = 0 , w = 2 .
3
2
7
We checked that w = u + 2v, so
H = Span{u, v, w} = Span{u, v}
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Basis and dimension
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A basis for a subspace H of Rn is a linearly independent
subset in H that spans H.
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Note: Given a subspace, there are more than one choices of
bases.
The dimension of a subspace H is the number of vectors in
any basis of H.
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Although the choices of bases are different, but they all have
the same number, which must be the dimension.
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Basis and dimension
Example: again take the vectors in Sec 1.3.
h3i
h1i
h1i
I Take u = 2 , v = 0 , w = 2 .
3
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2
7
We checked that w = u + 2v, so
H = Span{u, v, w} = Span{u, v}
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The subset {u, v, w} is linearly dependent, so this subset is
not a basis of H.
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The subset {u, v} is linearly independent, so this subset is a
basis of H, and dim(H) = 2.
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Basis and dimension
Other examples
I The columns of an n × n identity matrix are the standard
vectors denoted by e1 , . . . , en .
h1i
h0i
h0i
I
E.g. in R3 , e1 =
I
e1 , e2 , e3 are linearly independent.
Hence they form a basis for the subspace
h1i
h0i
h0i
Span{e1 , e2 , e3 } = {x 0 + y 1 + z 0 |where x, y , z are scalars}
0
0
1
hx i
= { y |where x, y , z are scalars}
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0
0
, e2 =
1
0
, e3 =
0
1
.
z
It is the whole R3 .
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The set e1 , . . . , en is called the standard basis for Rn .
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Basis and dimension
Other subset of n vectors in Rn form a basis, as long as they are
linearly independent.
h1i
h1i
h1i
I Take u = 2 , v = 0 , w = 2 .
3
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2
5
We checked that u, v, w are linearly independent. So
H = Span{u, v, w}
is a subspace in R3 .
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dim(H) is equal to the number of vectors which is 3, so it
must be the whole R3 .
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Invertible matrices and basis
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Suppose you have r linearly independent vectors {v1 , . . . , vr }
in Rn .
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If r ≤ n, then they form a basis of Span{v1 , . . . , vr }, which is
an r -dimensional subspace in Rn
If r = n, then
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Span{v1 , . . . , vn } = Rn .
Combining with the Invertible Matrices Theorem (a.⇔ e.),
if A = [v1 . . . vn ], then
for {v1 , . . . , vn } forming a basis of Rn ,
A must be invertible.
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Invertible matrices and basis
Examples
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Previous examples

I
I
I
1
 2
 3
1
 2
3
form

1 1
0 2  is invertible, so the columns form a basis in R3 .
2 5 
1 3
0 2  is non-invertible/singular, so the columns do not
2 7
a basis in R3 .
(Sec 1.8, Ex 15)
Are the columns of
The REF has
4
−2
16
−3
form a basis in R2 ?
pivots, so the columns
a basis for R2 .
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Invertible matrices and basis
Examples
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(Sec 1.8, Ex 17)

0

0
Are the columns of
−2

5 6
0 3  form a basis in R3 ?
4 2
The REF has
pivots, so the
I (Sec 1.8, Ex 18)

1
3
Are the columns of  1 −1
−3 2
columns
a basis for R3 .

5
1  form a basis in R3 ?
−4
The REF has
pivots, so the columns
a basis for R3 .
I (Sec 1.8, Ex 19)


3
6
Are the columns of  −8 2  form a basis in R3 ?
1 −5
No, because
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Quiz 4
The bold numbers indicate similar types of questions that may
appear in Quiz 4.
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Sec 2.2: Practice problem 2, Exercises 1,3,5,8,9a-d,10,13, 17,
29, 31 (use the algorithm I showed in class), 35
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Sec 2.3: Practice problems 1,2,3, Exercises 1,3,5, 11, 12, 13,
14,
The following questions require the big Invertible Matrix Theorem
and some thinking. They will not be in the quiz, but some of them
could be in the next midterm and the final. You are strongly
suggested to think about them.
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Sec 2.2: 21, 22, 23, 24,
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Sec 2.3: 15, 17, 18, 19, 20, 21, 24, 35, 36, 39
You may check the solution at:
http://www.slader.com/textbook/9780321385178-linear-algebraand-its-applications-4th-edition/
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