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I have another interesting problem. I intuitively think this is false.

Problem: If matrix $A\in M_n(\mathbb{R})$ has $n$ distinct eiganvalues and $n$ orthogonal eigenvectors $q_1,q_2,q_3,..,q_n$. Then $A$ is symmetric matrix!

This is kind of inverse of spectral theorem. Is it enough to use spectral theorem or does inverse doesn't folow? I can't make up any matrix that has there properties but it's not symetric.

Thanks!

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It is true and relatively easy (compared to the converse). Let $P$ be the matrix whose columns are an orthonormal base of eigenvectors. Since they are orthogonal, $P^{-1}=P^t$ . Note that if $D$ is the diagonal matrix with the eigenvalues $q_i$ on the diagonal, then $P^{-1} A P = D$, hence $A = P D P^{-1} = P D P^{t}$, which is clearly symmetric.

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  • $\begingroup$ Thank you. So to sum it up. I can say, aplying spectral theorem it's correct! $\endgroup$ – techno Jun 20 at 15:55
  • $\begingroup$ You need to normalize your eigenvectors first. $\endgroup$ – Ted Shifrin Jun 20 at 15:55
  • $\begingroup$ @TedShifrin, you are right. I corrected it. $\endgroup$ – mlainz Jun 20 at 15:58
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    $\begingroup$ @techno Nothing to do with the spectral theorem. Just the change of basis formula. $\endgroup$ – Ted Shifrin Jun 20 at 16:02
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A matrix $A $ has the properties as in "Problem " if and only if $A $ is normal.

Example: $A= diag (i,-i)$ is normal, but not symmetric.

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    $\begingroup$ Why the downvote? Is the complex case forbidden? $\endgroup$ – Fred Jun 20 at 15:53
  • $\begingroup$ I didn't downvoted. Just edited, i forgot to put field mark R. Thank you for your reply. I will try your way. $\endgroup$ – techno Jun 20 at 15:54
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Symmetricity means that $\langle Ax,y\rangle = \langle x, Ay \rangle$ for all $x,y$.

Since $\{q_i\}$ form a basis, write $x$ and $y$ in terms of this basis:

$$x = \sum x_i q_i$$ $$y = \sum y_i q_i$$

Now, recalling that $\{q_i\}$ are orthogonal, compute

$$\langle Ax, y \rangle = \langle \sum x_i \lambda_iq_i, \sum y_i q_i \rangle = \sum x_i y_i \lambda_i$$

$$\langle x, Ay \rangle = \langle \sum x_i q_i, \sum y_i \lambda_iq_i \rangle = \sum x_i y_i \lambda_i$$

Thus, $A$ is indeed symmetric. Note that the argument works even if $\lambda_i$ are not distinct.

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  • $\begingroup$ Thank you for answering. Can i solve it using spectral theorem? It would be a lot easier for me because im new to linear algebra. $\endgroup$ – techno Jun 20 at 15:50

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