MATH 2030: ASSIGNMENT 6

Eigenvalues and Eigenvectors of n× n Matrices

Q.1: pg 309, q 2. For the given matrix,

A =

[1 −91 −5

]calculate

(1) The characteristic polynomial of A.(2) The eigenvalues of A.(3) A basis for each eigenspace of A(4) the algebraic and geometric multiplicity of each value

A.1.

(1) The characteristic polynomial of A will be det(A− λI):∣∣∣∣1 − λ −91 −5 − λ

∣∣∣∣ = −(1 − λ)(5 + λ) + 9

expanding this we find the polynomial

λ2 + 4λ+ 4 = (λ+ 2)2.

(2) Equating this polynomial to zero, we find that the roots will be λ = −2,−2;this is the only value to satisfy det(A − λI) = 0, -2 is an eigenvalue withalgebraic multiplicity 2.

(3) Computing the null space of the matrix A+ 2I =

[3 −91 −3

]we find that a

non-trivial solution to the homogeneous problem (A+ 2I)x = 0 will satisfyx1 = −3x2. Thus the corresponding basis eigenvector for the eigenspace ofthe eigenvalue λ = −2 of A is

x =

[31

](4) The eigenvalue λ = −2 has algebraic multiplicity 2 and geometric multi-

plicity 1.

Q.2: pg 309, q 10. For the given matrix,

A =

2 1 1 10 1 2 30 0 3 30 0 0 2

calculate

(1) The characteristic polynomial of A.(2) The eigenvalues of A.

1

2 MATH 2030: ASSIGNMENT 6

(3) A basis for each eigenspace of A(4) the algebraic and geometric multiplicity of each value

A.2.

(1) Taking the determinant of the matrix A− λI is easily done as this matrixis upper-triangular. The characteristic equation simply the product of thediagonals

det(A− λI) = (2 − λ)(1 − λ)(3 − λ)(2 − λ).

(2) The eigenvalues of A are then λ = 2, 1, 3, 2.(3) Computing the null spaces of A−2I, A−I and A−3I we find the eigenspaces

are spanned by the following vectors

E1 = span

0100

, E2 = span

1000

, E3 = span

0110

.

(4) For λ = 1, 2, 3 have algebraic multiplicity and geometric multiplicity bothequal to 1 for each eigenvalue respectively.

Q.3: pg 310, q 13. Prove that if A is invertible with eigenvalue λ and correspond-ing eigenvector x, then 1

λ is an eigenvalue of A−1 with corresponding eigenvectorx.

A.3. If x is an eigenvalue of A, with eigenvalue λ then Ax = λx. As A is invertible,we may apply its inverse to both sides to get

x = λIx = A−1(λx) = λA−1x

Multiplying by 1/λ on both sides show that x is an eigenvector of A−1 with λ = 1λ

since

A−1x =1

λx.

Q.4: pg 310, q 16. Suppose A is a 3 × 3 matrix with eigenvectors

v1 =

100

, v2 =

110

, v3 =

111

with corresponding eigenvalues λ1 = − 1

3 , λ2 = 13 and λ3 = 1 respectively. Find

A20x, if x =

211

.

MATH 2030: ASSIGNMENT 6 3

A.4. We will give solutions for the vector given here and the vector given in the

text vb =

212

. It is easily shown that x = v1 + v3, while vb = 1v1 − 1v2 + 2v3.

Computing A20x is then

A20x =

(−1

3

)20

v1 + (1)20v3 =

3−20 + 111

while the vector vb yields

A20vb =

(−1

3

)20

v1 −(

1

3

)20

v2 + 2(1)20v3 −

3−20 − 3−20 + 2−3−20 + 2

2

=

2−3−20 + 2

2

Q.5: pg 310, q 17. With vi and λi and x as in the previous question, determineAkx for arbitrary k.

A.5. Generalizing the result we find,

Akx =

(−1

3

)kv1 + (1)kv3 =

(−1)−k3−k + 111

while the vector vb yields

Akvb =

(−1

3

)kv1 −

(1

3

)kv2 + 2(1)kv3 −

[(−1)−k − 1]3−k + 2−3−k + 2

2

Q.6: pg 310, q 19.

• Show that for any square matrix A, At and A have the same characteristicpolynomial and hence the same eigenvalues.

• Give an example of a 2 × 2 matrix A for which At and A have differenteigenspaces.

A.6.

• Noting that det(At) = det(A) we examine the characteristic polynomialof A and use this fact, det(A − λI) = det([A − λI]t) = det(At − λIt) =det(At − λI). This shows the characteristic polynomials for A and itstranspose are the same, and hence they have the same eigenvalues.

• Consider the matrixA =

[2 02 1

]this has eigenvalues λ = 1, 2 with eigenspaces

spanned by

E1 = span

([01

]), E2 = span

([12

]).

The matrix At has the eigenspaces

E1 = span

([−21

]), E2 = span

([10

]).

4 MATH 2030: ASSIGNMENT 6

Q.7: pg 310, q 22. If v is an eigenvector of A with corresponding eigenvalue λand c a scalar, show that v is an eigenvector of A−cI with corresponding eigenvalueλ− c.

A.7. Make the matrix A− cI and contract with the vector x, one finds

(A− cI)x = Ax− cx = λx− cx = (λ− c)x.

This proves that the vector x corresponding to λ the eigenvalue of A is an eigen-vector corresponding to λ− c for the matrix A− cI.

Q.8: pg 311, q 21. Let A be an idempotent matrix, meaning A2 = A. Show thatλ = 0 or λ = 1 are the only possible eigenvalues of A.

A.8. Suppose λ is any eigenvalue of A with corresponding eigenvector x, then λ2

will be an eigenvalue of the matrix A2 with corresponding eigenvector x. However,A2 = A and so λ2 = λ for the eigenvector x. This can only occur if λ = 0 or 1.

Q.9: pg 310, q 23. For the matrix,

A =

[3 25 0

]:

• Find the eigenvalues and eigenspaces of this matrix.• Using the appropriate theorem, and the previous example determine the

eigenvalues and eigenspaces of A−1, A− 2I and A+ 2I.

A.9.

• This matrix has eigenvalues λ = −2, 5 with eigenspaces spanned by thefollowing vectors respectively:

E−2 = span

([−25

]), E5 = span(

([11

])• Using this result we see that the eigenvalues for A−1 are then λ = − 1

2 ,15

with eigenspaces

E− 12

= span

([−25

]), E 1

5= span(

([11

])• the eigenvalues and eigenspaces for A− 2I are λ = −4, 3 and

E0 = span

([−25

]), E7 = span(

([11

])• the eigenvalues and eigenspaces for A+ 2I are λ = 0, 7 and

E−4 = span

([−25

]), E3 = span(

([11

])

MATH 2030: ASSIGNMENT 6 5

Q.10: pg 311, q 39. Use the helpful fact:

Proposition 0.1. any square matrix A that may be partitioned as A =

[P QO S

]where P and S are square matrices and O is the zero matrix, then detA = (detP )(detS).

to prove that if a square matrix A =

[P QO S

]partitioned so that P, S are square

matrices then the characteristic polynomial of A is

cA(λ) = cP (λ)cS(λ).

A.10. Computing A − λI we may partition this new matrix using In−p and Ipwhere p is the size of the matrix P,

A− λI =

[P − λIp Q

O S − λIn−p

]Taking the determinant of this matrix we find det(A− λI) = det(P − λIp)det(S −λIn−p), noting that cA(λ) = det(A− λI) we find this last identity is exactly whatwas needed

cA(λ) = cP (λ)cS(λ).

References

[1] D. Poole, Linear Algebra: A modern introduction - 3rd Edition, Brooks/Cole (2012).