Notes on Homework ?
Some calculations omitted; ask if you don’t know how to do a part.
1. Let A =
4 −1
.
6 −1
(a) Its characteristic polynomial is
p A ( z) = det( A − zI2 )
4 −1
z 0
= det(
−
)
6 −1
0 z
4−z
−1
= det(
)
6
−1 − z
= z2 − 3z + 2.
(b) The eigenvalues of A are the roots of p A ( z), namely, 1 and 2.
(c) We calculate the associated eigenspaces:
• We have v ∈ ε A (1) if and only if Av = v; so, let v =
x
y
, and solve the matrix
equation
4 −1
6 −1
x
x
=
,
y
y
which is equivalent to the system of equations
( 4 − 1) x − y = 0
6 + (−1 − 1) y = 0
This system has one free variable,
and thus a one-parameter family of solutions,
1
spanned by, for example,
.
3
1
).
Explicitly, ε A (1) = span (
3
• Similarly, ε A (2) consists of those v such that Av = 2v; solve this, and find that
1
ε A (2) = span(
).
2
(d) Since p A ( z) factors as z2 − 3z + 2 = ( z − 1)1 ( z − 2)1 , we have α A (1) = α A (2) = 1. (For
any other number λ, α A (λ) = 0.)
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008
(e) We have γ A (1) = dim ε A (1) = 1, and similarly γ A (2) = dim ε A (2) = 1; and for any
other λ, γ A (λ) = 0.
(f) Since for all λ, γ A (λ) = α A (λ), the matrix is diagonalizable.
(g) Since 0 is not an eigenvalue of A, there is no nontrivial solution to Av = 0 · v, and A is
injective; and an injective map from a vector space to itself is invertible.
2. With even less annotation:
(a)
pC ( z) = det(C − zI2 )
1−z
1
= det(
)
−1 −1 − z
= z2
(b) The eigenvalues are the roots of pC ( z) = z2 ; 0 is the only eigenvalue.
(c) We need to solve Cv = 0v; the solution space is (calculation omitted) εC (0) = span (
(d)
(e)
(f)
(g)
1
−1
αC (0) = 2, since pC ( z) = ( z − 0)2 .
γC (0) = 1, since dim εC (0) = 1.
The matrix C is not diagonalizable, since there’s an eigenvalue, 0, with γC (0) < al phaC (0).
Since 0 is an eigenvalue, the nullspace of C has positive dimension (in fact, dimension
one), and thus C is not invertible.
3. Let A = [T ]E ←E .
(a) As above, compute the characteristic polynomial
p A ( z) = det( A − z id )
= z2 − z − 6
= ( z − 3)( z + 2).
So the eigenvalues of A are 3 and −2.
1
(b) For λ ∈ {−2, 3}, solve the equation Av = λv. One finds that v1 =
is an eigen1
2
is an eigenvalue for −2.
value for 3, and v2 =
1
(c) Let B be the basis
Then
T (v1 ) = 3v1 = 3v1 + 0v2 , so
{v1, v2 } with v1 and v2 as above.
0
3
. Similarly, [T (v2 )]B =
. Therefore,
that [T (v1 )]B =
0
−2
[T ]B←B = [T (v1 )]B [T (v2 )]B
3 0
=
.
0 −2
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008
).
(d) Just compute:
−1 −7 10
1
−5 8
1
−1 2
−7 10
1
=
1
1 −1
−5 8
3 0
=
0 −2
[id]B←E [T ]E id E ←B =
1 2
1 1
2
1
2
1
= [T ]B←B .
4. Since λ is an eigenvalue of A, there is a nonzero element v ∈ Fn with Av = λv.
(a) Compute:
(α A)v = α · Av
= α · λv
= (αλ)v.
Therefore, v is an eigenvector of α A with eigenvalue αλ.
(b) If k = 0, then we define A0 = id, and the statement follows immediately.
Otherwise, one could prove this by induction on k, but: Let v be as above. Then
Ak v = Ak−1 Av
= A k− 1 λ v
= λ A k− 1 v
= λ Ak−2 Av
= λ 2 A k− 2 v
· · · = λk v.
To prove this by induction on k: The claim is clearly true for k = 1. Otherwise, suppose it is
known that Ak−1 v = λk−1 v. Then Ak v = AAk−1 v = Aλk−1 v = λk−1 Av = λk−1 λv = λk v.
(c) Let p( z) = ∑kN=0 α k zk be a polynomial with coefficients in F. Let’s figure out the effect
of p( A) (a sum of multiples of powers of A on v. Using the first two parts, we have
N
p( A ) v = ( ∑ α k A k ) v
k= 0
N
=
∑ αk Ak v (distribute)
k= 0
N
=
∑ αk λk v
k= 0
= p(λ)v.
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008
5. Let p A ( z) be the characteristic polynomial of A. We evaluate p A (0) in two different ways.
On one hand, since
p A ( z) = det( A − zI )
if we substitute z = 0 we have
p A (0) = det( A − 0I ) = det A.
On the other hand, if p A ( z) = an zn + · · · + a1 z + a0 , then substituting z = 0 yields p A (0) =
a0 , the constant term of p A ( z).
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008