Let M and N be two invertible square matrices | vedclass

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(C) Let the real matrices of order $2$ be $M = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ and $N = \begin{bmatrix} n_1 & 0 \ 0 & n_2 \end{bmatrix}

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For all diagonal matrices $M$

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(C) Let the real matrices of order $2$ be $M = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ and $N = \begin{bmatrix} n_1 & 0 \ 0 & n_2 \end{bmatrix}$.Since $M$ is invertible,$M^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.Then,$M N M^{-1} = \frac{1}{ad-bc} \begin{bmatrix} a & b \ c & d \end{bmatrix} \begin{bmatrix} n_1 & 0 \ 0 & n_2 \end{bmatrix} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.Calculating the product:$M N M^{-1} = \frac{1}{ad-bc} \begin{bmatrix} a & b \ c & d \end{bmatrix} \begin{bmatrix} n_1 d & -n_1 b \ -n_2 c & n_2 a \end{bmatrix} = \frac{1}{ad-bc} \begin{bmatrix} an_1d - bn_2c & -an_1b + bn_2a \ cn_1d - dn_2c & -cn_1b + dn_2a \end{bmatrix}$.For $M N M^{-1}$ to be diagonal,the off-diagonal elements must be zero:$ab(n_2 - n_1) = 0$ and $cd(n_1 - n_2) = 0$.If $M$ is a diagonal matrix,then $b = 0$ and $c = 0$,which satisfies these equations for any $n_1, n_2$.Thus,$M N M^{-1}$ is diagonal for all diagonal matrices $M$.

Let $M$ and $N$ be two invertible square matrices over $\mathbb{R}$ of order $2$ such that $N$ is diagonal. Then $M N M^{-1}$ is diagonal . . . . . .

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