I'm reviewing the spectral decomposition theorem. If $Y$ is a symmetric matrix, then $Y$ can be decomposed as $Y=Q\Lambda Q'$, where the columns of $Q$ are the eigenvectors of $Y$ and $\Lambda$ is a diagonal matrix with the diagonal composed of the eigenvalues of $Y$.
From some part of this theorem can it be concluded that $Q$ is an orthogonal matrix? I know that the eigenvectors of $Y$ are orthogonal, so can it be concluded that the matrix $Q$ is orthogonal?
Best Answer
In your expression, $Y=Q\Lambda Q'$ and rows of $Q$ are the eigenvectors of $Y$.
In this, $Y=Q'\Lambda Q$ and columns of $Q$ are the eigenvectors of $Y$.
We know in a symmetric matrix, eigenvectors corresponding to distinct eigenvalues are always orthogonal. If $\bf x$ and $\bf y$ are eigenvectors corresponding to different eigenvalues $\lambda$ and $\alpha$ respectively, then $$\langle Y{\bf x},{\bf y}\rangle=\langle{\bf x},Y{\bf y}\rangle\implies\langle\lambda{\bf x},{\bf y}\rangle=\langle{\bf x},\alpha{\bf y}\rangle\implies(\lambda-\alpha)\langle{\bf x},{\bf y}\rangle=0\implies\langle{\bf x},{\bf y}\rangle=0$$ since $\lambda-\alpha\neq 0$.
Now find an orthonormal basis for each eigenspace that make the set of all eigenvectors of $Y$ an orthonormal set in $\mathbb{R}^n$ (as different eigenspaces are orthogonal to each other).