If A is a $2\times2$ symmetric matrix ($A^T = A$) where $b$ does not equal zero ($a$'s are on the diagonal, $b$'s occupy the other $2$ spaces), find a matrix $X$ such that $X^T AX$ is diagonal.
What is the simple way to solve this problem (using orthogonal diagonalization intro linear algebra)?
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$\begingroup$You can also just go with the computations ;)
You easily get the two eigenvalues : $\lambda_1=a-b$ and $\lambda_2=a+b$ and the corresponding eigenvectors:
$V_1=\begin{pmatrix} -1 \\ 1 \end{pmatrix}$ and $V_2=\begin{pmatrix} 1 \\ 1 \end{pmatrix}$
you can also make them have norm 1. Then:
$V_1=\frac{1}{\sqrt{2}}\begin{pmatrix} -1 \\ 1 \end{pmatrix}$ and $V_2=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$
And thus you have your two matrices $P$ and $D$ such that $A=PDP^{-1}$:
$D=\begin{pmatrix} a-b & 0 \\0 & a+b \end{pmatrix}$ and $P=\frac{1}{\sqrt{2}}\begin{pmatrix} -1 & 1 \\1 & 1 \end{pmatrix}$
your matrix $P$ is clearly symmetric, and $P^{-1}=P^T$. And there you have it:
$A=PDP^{-1}=PDP^T$ which gives $D=P^TAP$
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