Eigenvalues and Eigenvectors of n Matricesmcnuttd/LinAlg-Assignme… · · 2012-07-04non-trivial...
Transcript of Eigenvalues and Eigenvectors of n Matricesmcnuttd/LinAlg-Assignme… · · 2012-07-04non-trivial...
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).