Dr. Marques Sophie
Office 519
Linear algebra
Spring Semester 2015
marques@cims.nyu.edu
Problem Set # 10
The following equations are considered over the reals numbers. All the answers should
be justified unless mentioned differently.
Problem 1 :
Let λ be an eigenvalue of an invertible matrix A. Show that λ´1 is an eigenvalue of
A´1 .
Solution : Let λ be an eigenvalue of an invertible matrix A. Then there is a non zero
vector x such that
Ax “ λx
So that, multiplying the equality by A´1 ,
x “ A´1 λx “ λA´1 x
And multiplying the equality by λ´1 , we get
A´1 x “ λ´1 x
with x ‰ 0.
That means that λ´1 is a eigenvalue for A´1 .
Problem 2 :
Show that if A2 is the zero matrix, then the only eigenvalue of A is 0.
Solution : If A2 is the zero matrix, then detpA2 q “ detpAqdetpAq “ 0, so that detpAq “ 0.
By the invertible matrix theorem, we have that that 0 is an eigenvalue for A.
Now we prove that 0 is the only eigenvalue. Indeed, suppose that lambda is an eigenvalue
for A, then there is a nonzero vector x such that
Ax “ λx
Then
A2 x “ λAx
So that
A2 x “ λ2 x
Since A2 “ 0, then λ2 x “ 0. And since x ‰ 0, this implies that λ2 “ 0, thus λ “ 0.
Finally, we have proven that λ “ 0 is the only eigenvalue.
Problem 3 :
Show that λ is an eigenvalue of A if and only if λ is an eigenvalue of AT .
1
Solution : We know that λ is an eigenvalue for A if and only if λ is a root of the
characteristic polynomial.
We will prove that the characteristic polynomial of A equals the one of AT that will
prove that A and AT have the same eigenvalues and this will solve the problem. We
have prove in class that detpMq “ detpMT q for any matrix M, thus
detpA ´ λIq “ detppA ´ λIqT q “ detpAT ´ λIq
since pλIqT “ λI.
So, we have proven that the characteristic polynomial of A detpA ´ λIq is equal to the
characteristic polynomial of AT which is detpAT ´ λIq.
Problemˆ4 :
˙
ˆ
˙
ˆ
˙
0.6 0.3
3{7
0.5
Let A “
, v1 “
and x0 “
.
0.4 0.7
4{7
0.5
1. Find a basis for R2 , consisting of v1 and another eigenvector v2 of A.
2. Verify that x0 may be written in the form x0 “ v1 ` cv2 .
3. For k “ 1, 2, ¨ ¨ ¨ , define xk “ Ak x0 . Compute x1 and x2 , and write a formula for
xk . Then show that xk ÞÑ v1 as k increases.
Solution :
1. Since A is a 2 ˆ 2 matrix, the eigenvalues are easy to find, and factorizing the
characteristic polynomial is easy when one of the two factors is known.
ˆ
˙
0.6 ´ λ
0.3
detp
q “ p0.6´λqp0.7´λq´0.3¨0.4 “ λ2 ´1.3λ`0.3 “ pλ´1qpλ´3q
0.4
0.7 ´ λ
The eigenvalues are 1 and 0.3. For the eigenvalue 0.3, solve pA ´ 0.3Iqx “ 0 :
ˆ
˙ ˆ
˙
0.6 ´ 0.3
0.3
0
0.3 0.3 0
“
0.4
0.7 ´ 0.3 0
0.4 0.4 0
Here, x1 ` x2 “ 0, with xˆ
The general solution is not needed. Set x2 “ 1 to
2 free. ˙
´1
find an eigenvector v2 “
. A suitable basis for R2 is tv1 , v2 u.
1
ˆ
˙ ˆ
˙
ˆ
˙
1{2
3{7
´1
2. Write x0 “ v1 ` 2v2 :
“
`c
.
1{2
3{7
1
By inspection, c is ´1{14. (The value of c depends on how v2 is scaled.)
3. For k “ 1, 2, ¨ ¨ ¨ , define xk`1 “ Ak x0 . Then x1 “ Apv1 ` cv2 q “ Av1 ` cAv2 “
v1 ` cp0.3qv1 , because v1 and v2 are eigenvector. Again,
x2 “ Ax1 “ Apv1 ` cp0.3qv2 q “ Av1 ` cAv2 “ v1 ` cp0.3qv2
Continuing, the general pattern is xk “ v1 ` cp0.3qk v2 , As k increase, the second
term tends to 0 and so xk tends to v1 .
Problem 5 :
Diagonalize the matrices, if possible.
2
1.
¨
˛
2 0 ´2
A“˝ 1 3 2 ‚
0 0 3
2.
¨
˛
1 2 ´3
B “ ˝ 2 5 ´2 ‚
1 3 1
Solution :
1.
˛
2´λ
0
´2
3´λ
2 ‚
A ´ λI “ ˝ 1
0
0
3´λ
¨
The characteristic equation is
ˇ
ˇ 2´λ
0
´2
ˇ
ˇ
3´λ
2
detpA´λIq “ ˇ 1
ˇ 0
0
3´λ
So, the eigenvalues ¨
of A
0
˝
1
For λ “ 2, A´2I “
0
ˇ
ˇ
ˇ
ˇ
ˇ
0
ˇ “ p3´λq ˇ 2 ´ λ
ˇ
ˇ 1
3
´
λ
ˇ
ˇ
ˇ
ˇ “ p3´λqp2´λqp3´λq “ p3´λq2 p2´
ˇ
given
˛
¨
˛ to be 2 and 3.
1 1 0 0
´2
2 ‚and row reducing rA´2I, 0s yields ˝ 0 0 1 0 ‚.
0 0 0 0
1
˛
¨
˛
´1
´1
The general solution is x2 ˝ 1 ‚ and a basis for the eigenspace is v1 “ ˝ 1 ‚.
0
0
˛
¨
2´λ
0
´2
˝
1
3´λ
2 ‚ and row reducing rA ´ 3I, 0s yields
For λ “ 3, A ´ 3λ “
0
0
3´λ
¨
˛
1 0 2 0
˝ 0 0 0 0 ‚.
0 0 0 0
¨ ˛
¨
˛
0
´2
The general solution is x2 ˝ 1 ‚ ` x3 ˝ 0 ‚, and nice for the eigenspace is
0
1
¨ ˛ ¨
˛
0
´2
tv1 , v2 u “ t˝ 1 ‚, ˝ 0 ‚u. From v1 , v2 and v3 , construct
0
1
¨
˛
´1 0 ´2
P “ rv1 , v2 , v3 s “ ˝ 1 1 0 ‚
0 0 1
are
0
1
0
¨
3
Then set
¨
˛
2 0 0
D“˝ 0 3 0 ‚
0 0 3
where the eigenvalues in D correspond to v1 , v2 and v3 , respectively.
2.
˛
1´λ
2
´3
5 ´ λ ´2 ‚
B ´ λI “ ˝ 2
1
3
1´λ
¨
Then the characteristic equation
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ 5 ´ λ ´2 ˇ
ˇ 2 ´2 ˇ
ˇ 2 5´λ ˇ
ˇ ´ 2ˇ
ˇ
ˇ
ˇ
detpB ´ λIq “
p1 ´ λq ˇˇ
ˇ 1 1 ´ λ ˇ ´ 3ˇ 1
3
1´λ ˇ
3 ˇ
“
p1 ´ λqpp5 ´ λqp1 ´ λq ` 6q ´ 2p2p1 ´ λq ` 2q ´ 3p6 ´ p5 ´ λqq
“
6 ´ 6λ ` p1 ´ λqp5 ´ 6λ ` λ2 q ´ 4 ´ 4 ` 2λ ´ 18 ` 15 ´ 9λ
“ 6 ´ 6λ ` 5 ´ 6λ ` λ2 ´ 5λ ` 6λ2 ´ λ3 ´ 4 ´ 4 ` 2λ ´ 18 ` 15 ´ 9λ
“
7λ2 ´ λ3 ´ 24λ “ λp´λ2 ` 7λ ´ 24q
∆ “ 49 ´ 4 ¨ 24 ă 0
2
So the polynomial ´λ ` 7λ ´ 24 does not factorize into linear coefficient over R.
Thus we know that B is not diagonalizable.
Problem 6 :
˛
¨
pp´1q
Define T : P2 Ñ R3 by Tppq “ ˝ pp0q ‚
pp1q
1. Find the image under T of pptq “ 5 ` 3t.
2. Show that T is a linear transformation.
3. Find the matrix for T relative to the basis t1, t, t2 u for P2 and the standard basis
for R3 .
Solution :
1. The image of p under T is
¨
˛ ¨ ˛
pp´1q
2
Tppq “ ˝ pp0q ‚ “ ˝ 5 ‚
pp1q
8
2. Let p and q be polynomials in P2 and let c be any scalar. Then,
¨
˛ ¨
˛ ¨
˛ ¨
˛
pp ` qqp´1q
pp´1q ` qp´1q
pp´1q
qp´1q
Tpp`qq “ ˝ pp ` qqp0q ‚ “ ˝ pp0q ` qp0q ‚ “ ˝ pp0q ‚`˝ qp0q ‚ “ Tppq`Tpqq
pp ` qqp1q
pp1q ` qp1q
pp1q
qp1q
¨
˛ ¨
˛
¨
˛
pcpqp´1q
c ¨ pp´1q
pp´1q
Tpcpq “ ˝ pcppq0q ‚ “ ˝ c ¨ pp0q ‚ “ c ˝ pp0q ‚ “ c ¨ Tppq
pcpqp1q
c ¨ pp1q
pp1q
4
3
3. Let B “ t1, t, t2 u and
˛ te1 , e2 , e3 u be the standard basis¨for R ˛. Since rTpb1 qs “
¨ “
´1
1
Tpb1 q “ Tp1q “ ˝ 1 ‚, rTpb2 qs “ Tpb2 q “ Tptq “ ˝ 0 ‚ and rTpb3 qs “
1,
1,
˛
¨
1
Tpb3 q “ Tpt2 q “ ˝ 0 ‚, the matrix for T relative to B and is
1,
˛
¨
1 ´1 1
rrTpb1 qs , rTpb2 qs , rTpb3 qs s “ ˝ 1 0 0 ‚
1 1 1
Problem 7 :
Find the B-matrix for the transformation x ÞÑ Ax where B “ tb1 , b2 u with
ˆ
˙
ˆ
˙
ˆ
˙
´4 ´1
´1
´1
A“
, b1 “
, b2 “
6
1
2
1
ˆ
˙
´1 ´1
Solution : If P “ rb1 , b2 s “
, then the B- matrix is
2
1
ˆ
˙ˆ
˙ˆ
˙ ˆ
˙
1
1
´4 ´1
´1 ´1
´2 ´2
´1
P AP “
“
´2 ´1
6
1
2
1
0 ´1
Problem 8 :
Let A ˆbe 2˙ˆ 2 matrix with
eigenvalues
3 and 1{3 and corresponding eigenvectors
ˆ
˙
1
´1
v1 “
and v2 “
. Let txk u be a solution of the difference equation
1
1
ˆ ˙
9
xk`1 “ Axk , x0 “
.
1
1. Compute x1 “ Ax0 .
2. Find a formula for xk involving k and the eigenvectors v1 and v2 .
Solution :
1. To find the action of A on x0 , express x0 in terms of v1 and v2 . That is, find c1 and
c2 such that x0 “ c1 v1 ` c2 v2 . This is certainly possible because the eigenvectors v1
and v2 are linearly independent (by inspection and also because they correspond
to distinct eigenvalues) and hence form a basis for R2 . (Two linearly independent
vectors in R2 automatically span R2 .) . The row reduction
ˆ
˙ ˆ
˙
1 ´1 9
1 0 5
rv1 , v2 , v3 s “
„
1 1 1
0 1 ´4
shows that x0 “ 5v1 ´ 4v2 . Since v1 and v2 are eigenvectors (for eigenvalues 3 and
1{3).
ˆ
˙
49{3
x1 “ Ax0 “ 5Av1 ´ 4Av2 “ 15v1 ´ 4{3v2 “
41{3
5
2. Each time A acts on linear combination of v1 and v2 , the v1 term is multiplied
by the eigenvalue 3 and the v2 term is multiplied by the eigenvalue 3 and the v2
term is multiplied by the eigenvalue 1{3 :
x2 “ Ax1 “ Ar15v1 ´ 4{3v2 s “ 5p3q2 v1 ´ 4p1{3q2 v2
In general, xk “ 5p3qk v1 ´ 4p1{3qk v2 ,for k ě 0.
Problem 9 :
Classify the origin as an attractor, repeller, or saddle point of the dynamical system
xk`1 “ Axk . Find the directions of greatest attraction and/or repulsion.
ˆ
˙
1.7 ´0.3
A“
´1.2 0.8
Solution :
detpA ´ λIq “ λ2 ´ 2.5λ ` 1 “ 0
Then
?
2.52 ´ 4
λ“
“ 2 and 0.5
2
The origin is a saddle point because one eigenvalue is greater than 1 and the other
eigenvalue is less than 1 in magnitude. The direction of greatest repulsion is through
the origin and the eigenvector v1 found below. Solve pA ´ 2Iqx “ 0
ˆ
˙ ˆ
˙
´0.3 ´0.3 0
1 1 0
„
´1.2 ´1.2 0
0 0 0
ˆ
˙
´1
so x1 “ ´x2 and x2 is free. Take v1 “
. The direction of greatest attraction is
1
through the origin and the eigenvector v2 below.
Solve pA ´ 0.5Iqx “ 0
ˆ
˙ ˆ
˙
1.2 ´0.3 0
1 ´0.25 0
„
´1.2 0.3 0
0
0
0
ˆ ˙
1
So x1 “ 0.25x2 and x2 is free. Take v2 “
.
4
2.5 ˘
6