Download Definition: A matrix transformation T : R n → Rm is said to be onto if

Definition: A matrix transformation T : Rn → Rm is said to be onto if evey vector
in Rm is the image of at least one vector in Rn .
Theorem 8.2.2: If T is a matrix transformation, T : Rn −→ Rn , then the following
are equivalent
(a) T is one-to-one
(b) T is onto
Example: Is the matrix transformation T : R2 → R3 , where T (x, y) = (x, y, x + y) is
onto?
Solution: T is onto if for any vector (a, b, c) ∈ R3 we can find a corresponding (x, y) ∈
R2 such that T (x, y) = (a, b, c). From here we get linear system
x = a
y = b
x+y = c




1 0 a
1 0
a
Row operations
0 1

b
T is onto if this sytem is consistent for all (a, b, c). 0 1 b 
→
1 1 c
0 0 c−b−a
So this system is consistent if c = a + b. Hence for (1, 2, 5) there is no (x, y) that is
mapped to (1, 2, 5) under T. So T is not onto.
Theorem 8.2.1 If T is a matrix transformation, T : Rn −→ Rm , then the following
are equivalent
(a) T is one-to-one
(b) nullspace of [T ] = {0}


1 0
Example: Consider the previous example we have [T ] = 0 1. Then null space
1 1
of [T ] is {0}. Hence T is one-to-one.
Chapter 5, Eigenvalues and Eigenvectors
Section 5.1 Eigenvalues and Eigenvectors
Definition: If A is an n × n matrix, the a nonzero vector x in Rn is called an eigenvector
of A if Ax is a scalar multiple of x, that is if Ax = λx for some scalar λ. The scalar λ is
called an eigenvalue of A, and x is said to be an eigenvector of A corresponding to λ.
• To find λ and then x:
Consider Ax = λx = λIx ⇒ (λI − A)x = 0. From here λ can be found by the
equation det (λI − A) = 0.
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Definition: det (λI − A) = 0 is called the characteristic equation of A, or characteristic polynomial of A.
Example:
equation of the matrix A and then find the eigen Find the characteristic

3
4 −1
values, A = −1 −2 1 
3
9
0
λ − 3 −4
λ − 3 −4 1 1
R +R λ + 2 −1 1→ 2 λ − 2 λ − 2 0 = (λ −
Solution: det (λI − A) = 1
−3
−3
−9
λ
−9 λ
2
2
2)(λ + λ − 6) = (λ − 2) (λ + 3) To find eigenvalues condisider det (λI − A) = 0 ⇒
λ = 2, λ = −3 are eigenvalues of A.
For a triangular matrix it is easier to find eigenvalues
Theorem 5.1.2: If A is an n×n triangular matrix (upper triangular, lower triangular,
or diagonal), then the eigenvalues of A are the entries on the main diagonal of A.
Idea: λI − A will also be triangular if A is triangular
Example:
2 1
λ − 2 −1
• A=
is upper triangular λI − A =
0 1
0
λ−1
det (λI − A) = (λ − 2)(λ − 1) ⇒ λ = 1, λ = 2




1 0 0
λ−1
0
0
0 
• A =  5 2 0 is lower triangular λI − A =  −5 λ − 2
−4 6 3
4
−6 λ − 3
det (λI − A) = (λ − 1)(λ − 2)(λ − 3) ⇒ λ = 1, λ = 2, λ = 3
Note: We can have eigenvalues that are complex numbers.
λ + 2
−2 −1
1
= λ2 − 4 + 5 = λ2 + 1
Example: A =
⇒ det (λI − A) = 5
2
−5 λ − 2
√
√
λ2 + 1 = 0 ⇒ λ = −1, λ = − −1.
Theorem 5.1.3 If A is an n × n matrix, the following statements are equivalent
(a) λ is an eigenvalue of A
(b) The system of equations (λI − A)x = 0 has nontrivial solutions.
(c) There is a nonzero vector x such that Ax = λx
(d) λ is a solution of the characteristic equation det (λI − A) = 0
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Definition: For λ an eigenvalue, the solution space of (λI − A)x = 0 is called the
eigenspace of A corresponding to λ. Nonzero vectors in the eigenspace of A corresponding
to λ are called the eigenvectors of A corresponding to λ.
−2 4
Example: Find eigenvalues of the matrix A =
. Also find eigenspaces of
3 2
the matrix and describe them geometrically.
Solution: characteristic equation: det λI − A = (λ − 4)(λ + 4) ⇒ λ = 4, λ = −4
are eigenvalues of A.
6 −4 x1
0
6 −4 0 1/2R1 +R2 6 −4 0
• Eigenspace forλ = 4 Solve
=
.
→
⇒
−3 2
x2
0
−3 2 0
0 0 0
x = t( 23 , 1).
Hence Eigenspace for λ = 4 = {x = t( 23 , 1)| − ∞ < t < ∞} is a line through origin
parallel to the vector ( 32 , 1)
−2 −4 x1
0
−2 −4 0 Row operations
=
.
→
• Eigenspace forλ = −4 Solve
−3 −6 x2
0
−3 −6 0
1 2 0
⇒ x = t(−2, 1).
0 0 0
Hence Eigenspace for λ = −4 = {x = t(−2, 1)| − ∞ < t < ∞} is a line through
origin parallel to the vector (−2, 1)
Eigenvalues of powers of a matrix
Theorem 5.1.4: If k is a positive integer, λ is an eigenvalue of a matrix A and x
is a corresponding eigenvector, then λk is an eigenvalue of Ak and x is a corresponding
eigenvector.
Idea: Ax = λx ⇒ A2 x = A(Ax) = A(λx) = λAx = λ2 x,
A3 x = A(A2 x) = A(λ2 x) = λ2 Ax = λ3 x . . .
Eigenvalues and invertibility
Theorem 5.1.5: A square matrix A is invertible if and only if λ = 0 is not an
eigenvalue of A.
Note: λ is an eigenvalue of A if and only if there is a nonzero vector x such that
Ax = λx. So λ = 0 is an eigenvalue of A if and only if there is a nonzero x such that
Ax = 0. But this is possible when A is not invertible.
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