# A problem with eigenvectors of a Hermitian(?) Matrix

I have this matrix:

$$A=\begin{bmatrix} 1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{bmatrix}$$

Eigenvalues are $$\lambda_i=0,1,3$$

Corresponding eigenvectors:

$$v_{\lambda=0}$$$$\begin{bmatrix} 1 \\1\\1 \end{bmatrix}$$

$$v_{\lambda=1}$$$$\begin{bmatrix} 1 \\0\\-1 \end{bmatrix}$$ $$v_{\lambda=3}$$$$\begin{bmatrix} 1 \\-2\\1 \end{bmatrix}$$

Now, $$A=XDX^{-1}$$ as it should (where D is the eigenvalues matrix). But my notes say the following:

And in a sense $$A$$ is hermitian (because it is symmetric), so we should have that $$A=XDX^T$$ but $$XDX^T=\begin{bmatrix} 4 & -6 & 2 \\ -6 & 12 & -6 \\ 1 & -2 & 1 \end{bmatrix}\neq A$$

What am I doing wrong here?

EDIT: After using Jose's hint everything works:

• There must be some problem with your computation; if $D$ is diagonal, then $XDX^\top$ should definitely be (at least) symmetric. – Theo Bendit May 12 at 12:45
• You need to make sure you normalize $X$ in the right way. As Jose wrote. – mathreadler May 12 at 12:49
• Please replace the screenshot by a text quotation of it. – Christoph May 12 at 12:52

Your matrix $$X$$ is not orthogonal: the columns do not have norm $$1$$. Divide each column by its norm, and all will be fine then. Not being orthogonal means that $$X^{-1}\neq X^T$$.
• I think you mean $X$ is not orthogonal. $A$ is equal to its conjugate transpose, hence it is Hermitian. – Theo Bendit May 12 at 12:43