Find a matrix that is related to another one via multiplication

Question

Given

\begin{bmatrix}a&b\\c&d\end{bmatrix} which multiplies \begin{pmatrix}I_1\\I_2\end{pmatrix} to give \begin{pmatrix}I'_1\\I'_2\end{pmatrix} Can another square matrix M express in terms of a,b,c,d such that M times \begin{pmatrix}(I_1+I_2)/2\\(I_1-I_2)/2\end{pmatrix} gives \begin{pmatrix}(I'_1+I'_2)/2\\(I'_1-I'_2)/2\end{pmatrix}

Answer 1

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{pmatrix}I_1\\I_2\end{pmatrix}=\begin{pmatrix}I'_1\\I'_2\end{pmatrix}$$ And $$\begin{bmatrix}1/2&1/2\\1/2&-1/2\end{bmatrix}\begin{pmatrix}I'_1\\I'_2\end{pmatrix}=\begin{pmatrix}(I'_1+I'_2)/2\\(I'_1-I'_2)/2\end{pmatrix} \\ \implies \begin{bmatrix}1/2&1/2\\1/2&-1/2\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{pmatrix}I_1\\I_2\end{pmatrix}=\begin{pmatrix}(I'_1+I'_2)/2\\(I'_1-I'_2)/2\end{pmatrix}$$

Note that $$\begin{bmatrix}1/2&1/2\\1/2&-1/2\end{bmatrix}\begin{pmatrix}I_1\\I_2\end{pmatrix}=\begin{pmatrix}(I_1+I_2)/2\\(I_1-I_2)/2\end{pmatrix} \\ \implies \begin{pmatrix}I_1\\I_2\end{pmatrix}=\begin{bmatrix}1 & 1\\ 1 & -1 \end{bmatrix}\begin{pmatrix}(I_1+I_2)/2\\(I_1-I_2)/2\end{pmatrix}$$ So the matrix $M$ is given by $$M = \begin{bmatrix}1/2&1/2\\1/2&-1/2\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}1 & 1\\ 1 & -1 \end{bmatrix}=0.5\begin{bmatrix}a+b+c+d & a-b+c-d\\ a+b-c-d & a-b-c+d \end{bmatrix}$$

Answer 2

Think about this: what matrix multiplication takes $[I_1, I_2]^T$ to $[(I_1+I_2), (I_1-I_2)]^T$? And what does the contrary for $I'_1, I'_2$ (actually the symbols used here are irrelevant)? And how can you associate them with the $abcd$ matrix to attain the desired matrix?

Answer 3

Notice $\pmatrix{1/2&1/2\\1/2&-1/2}\pmatrix{I_1\\I_2}=\pmatrix{(I_1+I_2)/2\\(I_1-I_2)/2}$. So we are trying to find $A'$ such that $$ A'\pmatrix{1/2&1/2\\1/2&-1/2}\pmatrix{I_1\\I_2}=\pmatrix{1/2&1/2\\1/2&-1/2}\pmatrix{I_1'\\I_2'}, $$ that is, $$ \pmatrix{1&1\\1&-1}A'\pmatrix{1/2&1/2\\1/2&-1/2}\pmatrix{I_1\\I_2}=\pmatrix{I_1'\\I_2'}. $$ Thus we can take $\pmatrix{1&1\\1&-1}A'\pmatrix{1/2&1/2\\1/2&-1/2}\pmatrix{I_1\\I_2}=A$, i.e. $$ A'=\pmatrix{1/2&1/2\\1/2&-1/2}A\pmatrix{1&1\\1&-1}. $$

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Hope this helps.