Description

Dane Johnson
1
Let A be an n × n matrix and J = {f(t) ∈ K[t] | f(A) = 0}. To show that J is an ideal, we check that the zero polynomial is in J, that if f,g ∈ J then f + g ∈ J, and that if f ∈ J and g ∈ K[t] then gf ∈ J.
Let O denote the zero polynomial, then O(A) = 0, so O ∈ J.
Suppose f,g ∈ J. Then f(A) = 0 and g(A) = 0, so it follows that (f + g)(A) = f(A) + g(A) = 0. Thus, f + g ∈ J.
Let g ∈ K[t] and f ∈ J. Then (gf)(A) = g(A)f(A) = g(A)0 = 0.
Let p ∈ K[t] denote the characteristic polynomial of A. Then p is of degree n and by the Cayley Hamilton Theorem, p(A) = 0. Therefore p ∈ J and since J is an ideal, we also have that p2 ∈ J. Then p2 is a polynomial of degree n2 and it is the case that p2(A) = p(A)p(A) = 0 as well.
2
Let ad(A) denote the classical adjoint of the n × n matrix A, where ad(A) =(co(A))T. That is, the adjoint is the transpose of the cofactor matrix of A. Let (Aad(A))ij denote the element in the ith row and jth column of the product Aad(A), aij the element of A in row i and column j, and bij the element of ad(A) in row i and column j. Then considering the jth column,
n n
(Aad(A))ij = Xaikbki = X(−1)kdetaikdet(Ajk) .
k=1 k=1
When i = j this computes the determinant of A but when i 6= j, this computes the determinant of a matrix with a repeated row, which must have a 0 determinant. So we conclude that (Aad(A))ij =det(A) when i = j and (Aad(A))ij = 0 when i 6= j. Therefore Aad(A) =det(A)I.
we apply the reasoning above to see that (ad(A)A)T = AT(co(AT))T = det(AT)I = det(A)I. But since (ad(A)A)T = det(A)I, which is a diagonal matrix, we conclude that ad(A)A = det(A)I as well.
3
Let A be an upper triangular n × n matrix.
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If we let Aij denote the element of A in the ith row and jth column, then Aij = 0 for i > j. First we prove that the product of two upper triangular matrices is upper triangular.
Let B be an n × n upper triangular matrix as well. We define Bij and (AB)ij similarly to Aij. Note that if i = 1, then 1 = i > j cannot occur for any column number j and since the definition of upper triangular requires only that (AB)ij = 0 for i > j, we assume i > 1 in what follows.
The element in the ith row and jth column of AB is found by:
n i−1 n
(AB)ij = XAikBkj = XAikBkj + XAikBkj.
k=1 k=1 k=i
Inspecting the summation from k = 1 to k = i − 1, we see that since k < i, Aik = 0 in each term so that = 0. This means that
.
Using the result above, we see that since A2 is the product of two upper triangular n×n matrices, A2 is also upper triangular. Then since A and A2 are upper triangular, A3 = A2A is upper triangular. Applying the reasoning inductively we conclude that Ak = Ak−1A is upper triangular for all positive powers k.
We cannot prove that Ak is upper triangular for k < 0 without further assuming that A is invertible. We do not know whether A is invertible here. •
Since A is an upper triangular matrix, its eigenvalues λ1,λ2,…,λn are the n elements on the diagonal of A (not necessarily distinct). Since f is a polynomial, f(A) is the result of exponentiating A, multiplying by elements k ∈ K and adding matrices. That is, if f is a degree m polynomial then f(A) = k0I + k1A1 + … + kmAm. Since Ap is upper triangular for all nonnegative integers p, the result of multiplying an upper triangular matrix by a scalar is upper triangular, and the sum of two upper triangular matrices is upper triangular, we conclude that f(A) must be an upper triangular matrix. Next we note that if λ is an eigenvalue of A with corresponding eigenvalue v, then since Av = λv we have
kApv = kλAp−1v = kλAp−2Av = kλ2Ap−2v = … = kλpv k ∈ K,p ∈ N .
This shows that kλp is an eigenvalue of kAp. Then for f(A) = k0I + k1A + … + kmAm, if λ is an eigenvalue of A with corresponding eigenvalue v, then
f(A)v = k0I+k1A+…+kmAm = k0λ0v+k1λv+…+kmλmv = (k0+k1λ+kmλm)v .
This implies that the eigenvalues of f(A) are f(λi), where each λi is one of the n diagonal elements of A.
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Let A be a nonsingular matrix and let λ be an eigenvalue of A. Then for an eigenvector v corresponding to the eigenvalue λ we have Av = λv. Since A is invertible, we have λ 6= 0 and

Therefore, if λ is an eigenvalue of A then λ−1 is an eigenvalue of A−1.
•
Suppose A is a 3 × 3 upper triangular matrix with eigenvalues −1,0,1. Since 0 is an eigenvalue of A, A is not invertible. Consider however, the matrix A3 − 3A2 + I.
Using our results above, the eigenvalues of A3 − 3A2 + I are
13 − 3(1)2 + 1 = 1 − 3 + 1 = −1
03 − 3(0)2 + 1 = 1
(−1)3 − 3(−1)2 + 1 = −3 .
Note that since none of the eigenvalues of A3 − 3A2 + I are 0, then the matrix A3 − 3A2 + I is invertible.
Therefore the eigenvalues of (A3 − 3A2 + I)−1 are −1,1 and −1/3.
4
Let A be an n×n matrix with eigenvalues 1,2, and 3 with corresponding eigenvalues v1,v2, and v3. In general for an n × n matrix with eigenvalue λ and corresponding eigenvector v,
Av = λv =⇒ A100v = λA99v = λ2A98v = … = λ100v .
Therefore λ100 is an eigenvalue of A100.
So we conclude that 1,2100, and 3100 are eigenvalues of A100. Let T : Rn → Rn be the transformation given by T(x) = Ax. Then there exists a basis B such that [T]B→B is an upper triangular matrix. The eigenvalues of the transformation T are the same as the matrix A since A is one matrix representation of the transformation T and the eigenvalues of T do not depend on the choice of basis used in the representation of T. By the previous exercise, the eigenvalues of the matrix ([T]B→B)100 are found by applying the polynomial f(λ) = λ100 to each eigenvalue of [T]B→B. Thus the eigenvalues of ([T]B→B)100 are just 1,2100, and 3100. Since these must be the same set of eigenvalues as the matrix A100, we conclude that A100 does not have any eigenvalues aside from 1,2100, and 3100.