Challenge Problems for Chapter 8

Suppose \(A\) is an \(n\times n\) matrix consisting entirely of real entries but \(a+ib\) is a complex eigenvalue having the eigenvector, \(\vec {x}+i\vec {y}\) Here \(\vec {x}\) and \(\vec {y}\) are real vectors. Show that then \(a-ib\) is also an eigenvalue with the eigenvector, \(\vec {x}-i\vec {y}\).
You should remember that the conjugate of a product of complex numbers equals the product of the conjugates. Here \(a+ib\) is a complex number whose conjugate equals \(a-ib.\)

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\(AX=\left ( a+ib\right )X\). Now take conjugates of both sides. Since \(A\) is real,
\[ A\overline {X}=\left ( a-ib\right ) \overline {X} \]
If \(A\) is \(2 \times 2\) and diagonalizable, show that \(C(A) = \{X \mid XA = AX\}\) has dimension \(2\) or \(4\).
If \(P^{-1}AP = D\), show that \(X\) is in \(C(A)\) if and only if \(P^{-1}XP\) is in \(C(D)\).
Let \(A\) be \(n \times n\) with \(n\) distinct real eigenvalues. If \(AC = CA\), show that \(C\) is diagonalizable.
Given a polynomial \(p(z) = r_{0} + r_{1}z + \dots + r_{n}z^{n}\) and a square matrix \(A\), the matrix \(p(A) = r_{0}I + r_{1}A + \dots + r_{n}A^{n}\) is called the evaluation of \(p(z)\) at \(A\). Let \(B = P^{-1}AP\). Show that \(p(B) = P^{-1}p(A)P\) for all polynomials \(p(x)\).
If \(A\) is diagonalizable and \(p(x)\) is a polynomial such that \(p(\lambda ) = 0\) for all eigenvalues \(\lambda \) of \(A\), show that \(p(A) = O\) (here, the final O is the zero matrix the same size as \(A\)).
Let \(P\) be an invertible \(n \times n\) matrix. If \(A\) is any \(n \times n\) matrix, write \(T_{P}(A) = P^{-1}AP\). Verify that:
1.
\(T_{P}(I) = I\)
2.
\(T_{P}(AB) = T_{P}(A)T_{P}(B)\)
3.
\(T_{P}(A + B) = T_{P}(A) + T_{P}(B)\)
4.
\(T_{P}(rA) = rT_{P}(A)\)
5.
\(T_{P}(A^{k}) = [T_{P}(A)]^{k}\) for \(k \geq 1\)
6.
If \(A\) is invertible, \(T_{P}(A^{-1}) = [T_{P}(A)]^{-1}\).
7.
If \(Q\) is invertible, \(T_{Q}[T_{P}(A)] = T_{PQ}(A)\).
Let \(A = \begin{bmatrix} 0 & a & b \\ a & 0 & c \\ b & c & 0 \end{bmatrix}\) and \(B = \begin{bmatrix} c & a & b \\ a & b & c \\ b & c & a \end{bmatrix}\).
1.
Show that \(x^{3} - (a^{2} + b^{2} + c^{2})x - 2abc\) has real roots by considering \(A\).
2.
Show that \(a^{2} + b^{2} + c^{2} \geq ab + ac + bc\) by considering \(B\).
Assume the \(2 \times 2\) matrix \(A\) is similar to an upper triangular matrix. If \(\mbox {tr} A = 0 = \mbox {tr} A^{2}\), show that \(A^{2}\) is equal to the zero matrix.
Show that \(A\) is similar to \(A^{T}\) for all \(2 \times 2\) matrices \(A\).
Let \(A =\begin{bmatrix} a & b \\ c & d \end{bmatrix}\). If \(c = 0\) treat the cases \(b = 0\) and \(b \neq 0\) separately. If \(c \neq 0\), reduce to the case \(c = 1\) using Exercise  4.
Suppose \(A\) is an \(n\times n\) matrix and let \(\vec {v}\) be an eigenvector such that \(A\vec {v}=\lambda \vec {v}\). Also suppose the characteristic polynomial of \(A\) is
\begin{equation*} \det \left ( z I-A\right ) =z ^{n}+a_{n-1} z ^{n-1}+\cdots +a_{1}z +a_{0} \end{equation*}
Explain why
\begin{equation*} \left ( A^{n}+a_{n-1}A^{n-1}+\cdots +a_{1}A+a_{0}I\right ) \vec {v}=0 \end{equation*}
Use this to prove that the Cayley-Hamilton theorem holds for any diagonalizable matrix \(A\). (The Cayley-Hamilton theorem says that \(A\) satisfies its characteristic equation, i.e.,
\begin{equation*} A^{n}+a_{n-1}A^{n-1}+\cdots +a_{1}A+a_{0}I=0 \end{equation*}
. (For a proof of the general case, see Theorem ??)
Suppose the characteristic polynomial of an \(n\times n\) matrix \(A\) is \(\lambda ^{n}-1\). Find \(A^{mn}\) where \(m\) is an integer.

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The eigenvalues are distinct because they are the \(n^{th}\) roots of \(1\). Hence if \(\vec {x}\) is a given vector with
\[ \vec {x}=\sum _{j=1}^{n}a_{j}\vec {v}_{j} \]
then
\[ A^{nm}\vec {x}=A^{nm}\sum _{j=1}^{n}a_{j}\vec {v}_{j}= \sum _{j=1}^{n}a_{j}A^{nm}\vec {v}_{j}=\sum _{j=1}^{n}a_{j}\vec {v}_{j}=\vec {x} \]
so \(A^{nm}=I\).

Bibliography

These problems come from Chapter 7 of Ken Kuttler’s A First Course in Linear Algebra. (CC-BY)

Ken Kuttler, A First Course in Linear Algebra, Lyryx 2017, Open Edition, pp. 359–401.