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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:enter image description here

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:

enter image description here

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  • $\begingroup$ There must be some problem with your computation; if $D$ is diagonal, then $XDX^\top$ should definitely be (at least) symmetric. $\endgroup$ – Theo Bendit May 12 at 12:45
  • $\begingroup$ You need to make sure you normalize $X$ in the right way. As Jose wrote. $\endgroup$ – mathreadler May 12 at 12:49
  • $\begingroup$ Please replace the screenshot by a text quotation of it. $\endgroup$ – Christoph May 12 at 12:52
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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 $X^{-1}\neq X^T$.

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  • $\begingroup$ I think you mean $X$ is not orthogonal. $A$ is equal to its conjugate transpose, hence it is Hermitian. $\endgroup$ – Theo Bendit May 12 at 12:43
  • $\begingroup$ I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos May 12 at 12:45
  • $\begingroup$ Done. And I've added another sentence too. $\endgroup$ – José Carlos Santos May 12 at 12:50
  • $\begingroup$ Thanks, it works now ! $\endgroup$ – Scavenger23 May 12 at 12:55

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