Show that the matrix $$A = \begin{bmatrix}a&h\\h&b\end{bmatrix} ,\quad a \ne b$$ is transformed to diagonal matrix $D = P^{-1}AP$, where $$P = \begin{bmatrix}\cos x& -\sin x\\\sin x& \cos x\end{bmatrix}$$ and $$\tan2x=\frac{2h}{(a-b)}$$
I understand that a $n \times n$ matrix $A$ is diagonalizable if there is a diagonal matrix $D$ such that $A$ is similar to $D$, that is, if there is an invertible matrix $P$ such that $P^{-1}AP= D$. Also, columns of $P$ are $n $ linearly independent eigenvectors of $A$ and the diagonal entries of $D$ are the eigenvalues of $A$ corresponding to the eigenvectors in $P$ in the same order. I am unable to proceed solving the above problem with these leads.
Best Answer
You seem to know the theory behind the problem. So there is no reason you cannot find the eigenvalues of $A$ and the corresponding eigenvectors required for the diagonalisation. Note that since $A$ is symmetric, it is orthogonally diagonalisable.
To do this in a slightly different setting but essentially the same method, consider the quadratic form
$$Q(X,Y)=(X\quad Y)\begin{pmatrix}a&h\\h&b\end{pmatrix}\begin{pmatrix}X\\Y\end{pmatrix}=aX^2+2hXY+bY^2\quad,\,\,a\ne b$$
Let $x$ be the angle by which the coordinate axes $X,Y$ should be rotated about the origin so that $Q$ is transformed to another quadratic form in which the product term is absent. The vanishing of the product term is equivalent to diagonalising the matrix $A$ or the quadratic form $Q$.
Suppose $u,v$ is our set of coordinate axes.
The required transformation is given by
$$\begin{pmatrix}u\\v\end{pmatrix}=\begin{pmatrix}\cos x&-\sin x\\\sin x&\cos x\end{pmatrix}\begin{pmatrix}X\\Y\end{pmatrix}$$
Or, $$\begin{pmatrix}X\\Y\end{pmatrix}=\begin{pmatrix}\cos x&\sin x\\-\sin x&\cos x\end{pmatrix}\begin{pmatrix}u\\v\end{pmatrix}$$
So, $$Q(X,Y)=Q'(u,v)=Au^2+2Huv+Bv^2$$
, where
$$A=a\cos^2x+2h\sin x\cos x+b\sin^2 x,$$
$$B=a\sin^2x-2h\sin x\cos x+b\cos^2 x,$$
$$H=(b-a)\sin x\cos x+h(\cos^2x-\sin^2 x)$$
The product term $uv$ in $Q'$ vanishes iff
\begin{align} H=0&\implies(b-a)\sin x\cos x+h(\cos^2 x-\sin^2 x)=0 \\&\implies (b-a)\sin 2x+2h\cos 2x=0 \\&\implies \tan 2x=\frac{2h}{a-b} \end{align}
Thus using the rotation matrix $P$, we have transformed the matrix $A$ to a diagonal matrix $D$ which is nothing but the matrix associated with the quadratic form $Au^2+Bv^2$.