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MATH 2030: ASSIGNMENT 6
Eigenvalues and Eigenvectors of n×n Matrices
Q.1: pg 309, q 2. For the given matrix,
A= 1 −9
1 −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 −9
= −(1−λ)(5+λ)+9
1 −5−λ
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 with
algebraic multiplicity 2.
(3) Computing the null space of the matrix A+2I = 3 −9 we find that a
1 −3
non-trivial solution to the homogeneous problem (A+2I)x = 0 will satisfy
x1 = −3x2. Thus the corresponding basis eigenvector for the eigenspace of
the eigenvalue λ = −2 of A is
x= 3
1
(4) The eigenvalue λ = −2 has algebraic multiplicity 2 and geometric multi-
plicity 1.
Q.2: pg 309, q 10. For the given matrix,
2 1 1 1
0 1 2 3
A=
0 0 3 3
0 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 matrix
is upper-triangular. The characteristic equation simply the product of the
diagonals
det(A−λI)=(2−λ)(1−λ)(3−λ)(2−λ).
(2) The eigenvalues of A are then λ = 2,1,3,2.
(3) ComputingthenullspacesofA−2I,A−I andA−3I wefindtheeigenspaces
are spanned by the following vectors
0 1 0
1 0 1
E =span , E =span , E =span .
1 2 3
0 0 1
0 0 0
(4) For λ = 1,2,3 have algebraic multiplicity and geometric multiplicity both
equal 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 is an eigenvalue of A−1 with corresponding eigenvector
λ
x.
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
−1 −1
x=λIx=A (λx)=λA x
Multiplying by 1/λ on both sides show that x is an eigenvector of A−1 with λ = 1
λ
since
A−1x= 1x.
λ
Q.4: pg 310, q 16. Suppose A is a 3×3 matrix with eigenvectors
1 1 1
v = 0 , v = 1 , v = 1
1 2 3
0 0 1
with corresponding eigenvalues λ = −1, λ = 1 and λ = 1 respectively. Find
1 3 2 3 3
2
20
A x, if x = 1 .
1
MATH 2030: ASSIGNMENT 6 3
A.4. We will give solutions for the vector given here and the vector given in the
2
text vb = 1 . It is easily shown that x = v1 + v3, while vb = 1v1 − 1v2 + 2v3.
2
Computing A20x is then
−20
1 20 3 +1
A20x = − v +(1)20v = 1
3 1 3
1
while the vector v yields
b
−20 −20
20 1 20 1 20 20 3 −3 +2 2
−20 −20
A vb = −3 v1 − 3 v2 +2(1) v3 − −3 2 +2 = −3 2 +2
Q.5: pg 310, q 17. With vi and λi and x as in the previous question, determine
Akx for arbitrary k.
A.5. Generalizing the result we find,
−k −k
k 1 k k (−1) 3 +1
A x= −3 v1+(1) v3= 1
1
while the vector v yields
b
−k −k
k 1 k 1 k k [(−1) −1]3 +2
−k
A vb = −3 v1 − 3 v2 +2(1) v3 − −3 2 +2
Q.6: pg 310, q 19.
• Show that for any square matrix A, At and A have the same characteristic
polynomial and hence the same eigenvalues.
• Give an example of a 2 × 2 matrix A for which At and A have different
eigenspaces.
A.6.
• Noting that det(At) = det(A) we examine the characteristic polynomial
of 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 its
transpose are the same, and hence they have the same eigenvalues.
• ConsiderthematrixA = 2 0 thishaseigenvaluesλ = 1,2witheigenspaces
2 1
spanned by
E =span 0 , E =span 1 .
1 1 2 2
The matrix At has the eigenspaces
E1 =span −2 , E2 = span 1 .
1 0
4 MATH 2030: ASSIGNMENT 6
Q.7: pg 310, q 22. If v is an eigenvector of A with corresponding eigenvalue λ
andcascalar, show that v is an eigenvector of A−cI with corresponding eigenvalue
λ−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 that
λ=0orλ=1aretheonly 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 2 :
5 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 the
following vectors respectively:
E =span −2 , E =span( 1
−2 5 5 1
• Using this result we see that the eigenvalues for A−1 are then λ = −1, 1
2 5
with eigenspaces
E−1 =span −2 , E1 = span( 1
2 5 5 1
• the eigenvalues and eigenspaces for A−2I are λ = −4,3 and
E =span −2 , E =span( 1
0 5 7 1
• the eigenvalues and eigenspaces for A+2I are λ = 0,7 and
E =span −2 , E =span( 1
−4 5 3 1
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