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Recitation Notes Spring 16, 21-241: Matrices and Linear Transformations March 1, 2016

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Recitation Notes Spring 16, 21-241: Matrices and Linear Transformations March 1, 2016
Recitation Notes
Spring 16, 21-241: Matrices and Linear Transformations
March 1, 2016
Abstract
[TOPICS COVERED] Eigenvalues. Eigenvectors. Eigenspaces. Characteristic polynomial.
Fundamental Theorem of Invertible Matrices.
1
Administrative Matters
• No office hours today. Shifted to tomorrow 4.30 pm to 6.30 pm.
• Enjoy Spring Break! (:
2
Definitions / Notation
1. Eigenvalues
2. Eigenvectors
3. Eigenspaces
4. Characteristic polynomial
5. Subspace
3
Problems
1. Let A =
0 3
3 0
and v =
1
.
1
(a) Show that v is an eigenvector of A.
(b) Find the eigenvalues and associated eigenspaces of A.
(a) To show that v is an eigenvector of A, we want to show that Av = λv for some scalar
λ.
0 3
1
3
Av =
=
= 3v
3 0
1
3
Thus v is indeed an eigenvector of A.
1
The eigenvalues are the roots of the characteristic polynomial:
−λ 3
det(A − λI) = det
= λ2 − 9
3 −λ
Setting λ2 − 9 = (λ − 3)(λ + 3) = 0, we have λ = 3, −3.
λ = 3:
A − λI =
−3 3 0
3 −3 0
−3 3
3 −3
−−R−−
−−−−−→
7→R +R
−3 3 0
0 0 0
−−−−−−1−−→
1 −1 0
0 0 0
2
2
1
R1 7→− 3 R1
∴ E3 = span
1
1
λ = −3:
A − λI =
3 3 0
3 3 0
3 3
3 3
−−R−−
−−−−−→
7→R +R
3 3 0
0 0 0
−−−−−1−−→
1 1 0
0 0 0
2
2
1
R1 7→ 3 R1
∴ E−3 = span
1
−1


3 1 −1
2. Show that A =  1 1 1  has eigenvalue λ = 2. Find an eigenvector to this eigenvalue.
4 2 0
Solution. We need to show that 2 is a root of the characteristic polynomial det(A − λI),
i.e. det(A − 2I) = 0.
Page 2
Indeed,
det(A − 2I)


1 1 −1
= det  1 −1 1 
4 2 −2
= (1)(−1)(−2) + (1)(2)(−1) + (4)(1)(1) − (1)(2)(1) − (1)(1)(−2) − (4)(−1)(−1)
= 0
by Sarrus’ Rule. Thus 2 is an eigenvalue.
To find the eigenvectors corresponding to λ = 2,




1 1 −1 0
1 1 −1 0
−−−−−−−−−−−−−−−−→
 1 −1 1 0  −−
2 0 
R2 7→R2 −R1 , R3 7→R3 −4R1  0 −2
4 2 −2 0
0 −2 2 0


1 1 −1 0
−−R−−
−−−−−→  0 −2 2 0 
3 7→R3 −R2
0 0
0 0


1 1 −1 0
−−−−−−1−−→  0 1 −1 0 
R2 7→− 2 R2
0 0 0 0


1 0 0 0
−−R−−
−−−−−→  0 1 −1 0 
1 7→R1 −R2
0 0 0 0


  
0
0
Therefore a solution is v =  1 . Thus 2,  1  is an eigenvalue, eigenvector pair
1
1
of A.
3. Let a, b ∈ R. Find the eigenvalues and corresponding eigenspaces of A =
a b
−b a
over the
complex numbers.
Solution.
det(A − λI) = det
a−λ
b
−b a − λ
= (a − λ)2 + b2 = (a − λ + ib)(a − λ − ib)
Thus the eigenvalues are λ = a ± ib.
λ = a − ib:
A − λI
0
=
ib b 0
−b ib 0
Page 3
−−
−−−−−−−→
R 7→R −iR
2
2
1
ib b 0
0 0 0
−1
i
∴ Ea−ib = span
λ = a + ib:
A − λI
0
=
0
−ib b
−b −ib 0
−−
−−−−−−−→
R 7→R +iR
2
∴ Ea+ib = span
4. Let
A=
2 1
−1 0
2
1
i
1
−ib b 0
0 0 0
.
(a) Find the characteristic polynomial of A.
(b) Find the eigenvalues of A.
(c) Find the associated eigenspaces.
Solution.
(a) The characteristic polynomial is
2−λ 1
= −2λ + λ2 + 1 = (λ − 1)2 .
det(A − λI) = det
−1 −λ
(b) Since the roots of the characteristic polynomial are the eigenvalues, λ = 1 is the
eigenvalue.
(c)
A − λI
0
1
1 0 −−−−−−−−−→ 1 1 0
=
R2 7→R2 +R1
−1 −1 0
0 0 0
1
∴ E1 = span
−1
5. If v is an eigenvalue of A with corresponding eigenvalue λ, and c is a scalar, show that v is
an eigenvector of A − cI with corresponding eigenvalue λ − c.
Solution.
Av = λv =⇒ Av − cv = λv − cv =⇒ (A − cI)v = (λ − c)v
Thus (λ − c, v) is an eigenvalue, eigenvector pair of A − cI.
Page 4
6. Let A be an idempotent matrix, i.e. A2 = A. Show that λ = 0 and λ = 1 are the only
possible eigenvalues of A.
Solution. Let (λ, v) be an eigenvalue, eigenvector pair of A. Then we have
Av = λv
with v 6= 0.
λv = Av = A2 v = A(Av) = A(λv) = λ(Av) = λ(λv) = λ2 v
Thus
λv = λ2 v =⇒ (λ − λ2 )v = 0.
Since v 6= 0, it must be that
λ − λ2 = 0,
so
λ(1 − λ) = 0.
Indeed, the only possible values for λ are 0 and 1.
7. (a) Show that for square matrix A, A and AT have the same characteristic polynomial,
hence the same eigenvalues.
(b) Give an example where A and AT have different eigenspaces.
Solution.
(a)
det(A − λI) = det((A − λI)T ) = det(AT − λI T ) − det(AT − λI)
Thus the characteristic polynomials of A and AT are the same.
2 −1
T
. From part (a), we know λ = 1.
(b) Using A from problem 4, A =
1 0
T
1 −1
1 −1
A − λI =
Thus
AT
− λI
0
=
1 −1 0
1 −1 0
−−R−−
−−−−−→
7→R −R
∴ E1 = span
2
2
1
1
1
1 −1 0
0 0 0
This is a different eigenspace as in problem 4.
8. Let A be a 2 × 2 matrix with eigenvectors v 1 =
Page 5
1
−1
and v 2 =
1
1
corresponding to
eigenvalues λ1 =
1
2
and λ2 = 1. Find an expression for
Ak v
where v =
5
. What happens
1
when k → ∞?
Solution. v =
4
5
1
=2
1
−1
+3
1
1
= 2v 1 + 3v 2 Thus
Additional Notes
1. there are a few ways of beginning solutions on eigenvalue/vectors:
(a) (A − λI)v = 0,
(b) det(A − λI) = 0,
(c) Av = λv.
If you get stuck using one of these as a starting point in your proof, try the other two. They
may be surprisingly useful.
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