Spectral decomposition theorem

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).