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How might I find matrix $M\in M_2(\mathbb C)$ such that $M^t A =M^{-1}$ where $A=\left[ \begin{array}{cc} a & a \\ a & a+1 \\ \end{array} \right]\in M_2(\mathbb N)$ without using the brute force method of writing $M=\left[ \begin{array}{cc} a & b \\ c & d \\ \end{array} \right]$? (I don't quite know how to solve the resulting system even in the brute force case anyway.)

Thanks in advance.

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Would it help to write your system as $\mathbf M^\top \mathbf A\mathbf M=\mathbf I$? –  J. M. Nov 30 '11 at 12:13

3 Answers 3

We can rewrite your equation as $$MM^T = A^{-1},$$ where $$A^{-1} = \left[\begin{array}{cc} 1 + \frac{1}{a} & -1 \\ -1 & 1\end{array}\right]$$ is real, symmetric positive-definite. The Cholesky decomposition of $A^{-1}$ thus supplies such an $M$: $$M = \left[\begin{array}{cc}\sqrt{\frac{a+1}{a}} & 0\\-\sqrt{\frac{a}{a+1}} & \sqrt{\frac{1}{a+1}}\end{array}\right].$$

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You can also rephrase the problem as find a matrix $M$ such that $M^TAM = I$. It is relatively simple once you see that eliminating the matrix entries are easy to handle with $A$. $$ \pmatrix{1&0\\-1&1} \pmatrix{a & a \\a & a+1}\pmatrix{1&-1\\0&1} = \pmatrix{a&0\\0&1}$$ Seeing that we have an extra $a$ term left we remove it by an extra step (assuming $a\neq 0$) $$ \pmatrix{\frac{1}{\sqrt{a}}&0\\0&1} \pmatrix{a &0 \\0 & 1} \pmatrix{\frac{1}{\sqrt{a}}&0\\0&1}= \pmatrix{1&0\\0&1} $$ Combining the two steps into one, gives

$$ \underbrace{\pmatrix{\frac{1}{\sqrt{a}}&0\\0&1}\pmatrix{1&0\\-1&1}}_{M^T} \pmatrix{a & a \\a & a+1} \underbrace{\pmatrix{1&-1\\0&1}\pmatrix{\frac{1}{\sqrt{a}}&0\\0&1}}_M = \pmatrix{1&0\\0&1} $$

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How did you see that the matrix could be "diagonalized" by transpose matrices? –  charlie Nov 30 '11 at 21:15
    
No brains but just a bit of experience with a nose for what can eliminate what. My work involves a lot of these congruence and similarity transformations. I have started with Woodbury Matrix Identity and your question is a special case. You can start from there too :) –  user13838 Nov 30 '11 at 21:47
    
Thanks, percusse :-) –  charlie Dec 1 '11 at 8:34

Clearly $a$ is nonzero, or else $M^\top A$ is non-invertible and it cannot be equal to the invertible matrix $M^{-1}$. Now, $$ \begin{align} A=\begin{pmatrix}a&a\\a&a+1\end{pmatrix} &=\begin{pmatrix}a&a\\a&a\end{pmatrix}+\begin{pmatrix}0&0\\0&1\end{pmatrix}\\ &=\begin{pmatrix}\sqrt{a}\\ \sqrt{a}\end{pmatrix}\begin{pmatrix}\sqrt{a}&\sqrt{a}\end{pmatrix}+\begin{pmatrix}0\\1\end{pmatrix}\begin{pmatrix}0&1\end{pmatrix}\\ &=\begin{pmatrix}\sqrt{a}&0\\ \sqrt{a}&1\end{pmatrix} \begin{pmatrix}\sqrt{a}&\sqrt{a}\\0&1\end{pmatrix} = (M^{-1})^\top M^{-1}. \end{align} $$ So you may set $M=\begin{pmatrix}\sqrt{a}&\sqrt{a}\\0&1\end{pmatrix}^{-1}=\begin{pmatrix}\frac{1}{\sqrt{a}}&-1\\0&1\end{pmatrix}$.

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