Chapter 2: Problem 77
Suppose \(A\) is a complex matrix. Show that \(A A^{H}\) and \(A^{H} A\) are Hermitian.
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Chapter 2: Problem 77
Suppose \(A\) is a complex matrix. Show that \(A A^{H}\) and \(A^{H} A\) are Hermitian.
These are the key concepts you need to understand to accurately answer the question.
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Show (a) If \(A\) has a zero row, then \(A B\) has a zero row. (b) If \(B\) has a zero column, then \(A B\) has a zero column.
Let \(A=\left[\begin{array}{rr}1 & 2 \\ 4 & -3\end{array}\right]\), and let \(f(x)=2 x^{3}-4 x+5\) and \(g(x)=x^{2}+2 x+11\). Find (a) \(A^{2}\) (b) \(A^{3}\) (c) \(f(A)\), (d) \(g(A)\) (a) \(A^{2}=A A=\left[\begin{array}{rr}1 & 2 \\ 4 & -3\end{array}\right]\left[\begin{array}{rr}1 & 2 \\ 4 & -3\end{array}\right]=\left[\begin{array}{ll}1+8 & 2-6 \\ 4-12 & 8+9\end{array}\right]=\left[\begin{array}{rr}9 & -4 \\ -8 & 17\end{array}\right]\) (b) \(A^{3}=A A^{2}=\left[\begin{array}{rr}1 & 2 \\ 4 & -3\end{array}\right]\left[\begin{array}{rr}9 & -4 \\ -8 & 17\end{array}\right]=\left[\begin{array}{rr}9-16 & -4+34 \\ 36+24 & -16-51\end{array}\right]=\left[\begin{array}{rr}-7 & 30 \\ 60 & -67\end{array}\right]\) (c) First substitute \(A\) for \(x\) and \(5 I\) for the constant in \(f(x)\), obtaining $$ f(A)=2 A^{3}-4 A+5 I=2\left[\begin{array}{rr} -7 & 30 \\ 60 & -67 \end{array}\right]-4\left[\begin{array}{rr} 1 & 2 \\ 4 & -3 \end{array}\right]+5\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] $$ Now perform the scalar multiplication and then the matrix addition: $$ f(A)=\left[\begin{array}{rr} -14 & 60 \\ 120 & -134 \end{array}\right]+\left[\begin{array}{rr} -4 & -8 \\ -16 & 12 \end{array}\right]+\left[\begin{array}{ll} 5 & 0 \\ 0 & 5 \end{array}\right]=\left[\begin{array}{rr} -13 & 52 \\ 104 & -117 \end{array}\right] $$ (d) Substitute \(A\) for \(x\) and \(11 I\) for the constant in \(g(x)\), and then calculate as follows: $$ \begin{aligned} g(A) &=A^{2}+2 A-11 I=\left[\begin{array}{rr} 9 & -4 \\ -8 & 17 \end{array}\right]+2\left[\begin{array}{rr} 1 & 2 \\ 4 & -3 \end{array}\right]-11\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ &=\left[\begin{array}{rr} 9 & -4 \\ -8 & 17 \end{array}\right]+\left[\begin{array}{rr} 2 & 4 \\ 8 & -6 \end{array}\right]+\left[\begin{array}{rr} -11 & 0 \\ 0 & -11 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \end{aligned} $$ Because \(g(A)\) is the zero matrix, \(A\) is a root of the polynomial \(g(x)\).
Partition each of the following matrices so that it becomes a square block matrix with as many diagonal blocks as possible: $$ A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 3 \end{array}\right], \quad B=\left[\begin{array}{lllll} 1 & 2 & 0 & 0 & 0 \\ 3 & 0 & 0 & 0 & 0 \\ 0 & 0 & 4 & 0 & 0 \\ 0 & 0 & 5 & 0 & 0 \\ 0 & 0 & 0 & 0 & 6 \end{array}\right], \quad C=\left[\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 0 \end{array}\right] $$
Prove Theorem \(2.2(\text { iii })\) and (iv): (iii) \((B+C) A=B A+C A, \quad\) (iv) \(k(A B)=(k A) B=A(k B)\)
Let \(A=\operatorname{diag}(2,3,5)\) and \(B=\operatorname{diag}(7,0,-4) .\) Find (a) \(A B, A^{2}, B^{2}\) (b) \(f(A),\) where \(f(x)=x^{2}+3 x-2\) (c) \(A^{-1}\) and \(B^{-1}\)
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