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# (a) If$A = \left[ \begin{array} { l l } { 1 } & { 1 } \\ { 0 } & { 1 } \end{array} \right]$, show that$U ^ { - 1 } A U$is not diagonal for any invertible complex matrix U. (b) If$A = \left[ \begin{array} { l l } { 0 } & { 1 } \\ { - 1 } & { 0 } \end{array} \right]$, show that$U ^ { - 1 } A U$is not upper triangular for any real invertible matrix U.

Solution

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$\pmb{(a)}$ Let

$U=\begin{bmatrix}a+ib & c+id\\e+if & g+ih\end{bmatrix}$

be an invertible complex matrix, and let's assume that

$U^{-1}AU=D=\begin{bmatrix}x & 0\\0 & y\end{bmatrix}$

is a diagonal matrix. Therefore, we get the following.

\begin{align*} U^{-1}AU=D&\implies AU=UD\\\\ &\implies \begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}\begin{bmatrix}a+ib & c+id\\e+if & g+ih\end{bmatrix}=\begin{bmatrix}a+ib & c+id\\e+if & g+ih\end{bmatrix}\begin{bmatrix}x & 0\\0 & y\end{bmatrix}\\\\ \pmb{(*)}&\implies \begin{bmatrix}(a+e)+(b+f)i & (c+g)+(d+h)i\\e+if & g+ih\end{bmatrix}=\begin{bmatrix}(a+ib)x & (c+id)y\\(e+if)x & (g+ih)y\end{bmatrix}\\\\ &\implies e+if=(e+if)x\quad\land\quad g+ih=(g+ih)y\\\\ &\implies x=1\quad\land\quad y=1\\\\ &\stackrel{\pmb{(*)}}{\implies}(a+e)+(b+f)i=a+ib\quad\land\quad (c+g)+(d+h)i=c+id\\\\ &\implies e+if=0\quad\land\quad g+ih=0\\\\ &\implies U=\begin{bmatrix}a+ib & c+id\\0 & 0\end{bmatrix}\\\\ &\implies \det(U)=0\\\\ &\implies\pmb{\text{U is not an invertible matrix}} \end{align*}

This is a contradiction, and therefore there is no invertible complex matrix $U$ such that $U^{-1}AU$ is a diagonal matrix.

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