Chapter 6: Problem 8
Let \(A\) be an Hermitian matrix and let \(B=i A .\) Show that \(B\) is skew Hermitian.
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Chapter 6: Problem 8
Let \(A\) be an Hermitian matrix and let \(B=i A .\) Show that \(B\) is skew Hermitian.
These are the key concepts you need to understand to accurately answer the question.
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Let \(A\) be a symmetric \(n \times n\) matrix. Show that \(e^{A}\) is symmetric and positive definite.
Let \(A\) and \(B\) be \(n \times n\) matrices. Show that if none of the eigenvalues of \(A\) are equal to \(1,\) then the matrix equation \\[ X A+B=X \\] will have a unique solution.
Show that if \(A\) is skew Hermitian and \(\lambda\) is an eigenvalue of \(A\) then \(\lambda\) is purely imaginary (i.e., \(\lambda=b i\) where \(b\) is real).
The transition matrix in Example 5 has the property that both its rows and its columns add up to 1 In general, a matrix \(A\) is said to be doubly stochastic if both \(A\) and \(A^{T}\) are stochastic. Let \(A\) be an \(n \times n\) doubly stochastic matrix whose eigenvalues satisfy \\[ \lambda_{1}=1 \quad \text { and } \quad\left|\lambda_{j}\right|<1 \text { for } j=2,3, \ldots, n \\] Show that if \(\mathbf{e}\) is the vector in \(\mathbb{R}^{n}\) whose entries are all equal to \(1,\) then the Markov chain will converge to the steady-state vector \(\mathbf{x}=\frac{1}{n} \mathbf{e}\) for any starting vector \(\mathbf{x}_{0} .\) Thus, for a doubly stochastic transition matrix, the steady-state vector will assign equal probabilities to all possible outcomes.
In each of the following, factor the matrix \(A\) into a product \(X D X^{-1},\) where \(D\) is diagonal: (a) \(A=\left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right)\) (b) \(A=\left(\begin{array}{rr}5 & 6 \\ -2 & -2\end{array}\right)\) (c) \(A=\left(\begin{array}{cc}2 & -8 \\ 1 & -4\end{array}\right)\) (d) \(A=\left(\begin{array}{rrr}2 & 2 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & -1\end{array}\right)\) (e) \(A=\left(\begin{array}{rrr}1 & 0 & 0 \\ -2 & 1 & 3 \\ 1 & 1 & -1\end{array}\right)\) (f) \(A=\left[\begin{array}{lll}1 & 2 & -1 \\ 2 & 4 & -2 \\ 3 & 6 & -3\end{array} |\right.\)
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