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I come across this question in a linear algebra reference book. I understand that if $A$ is diagonalizable and its eigenvectors of distinct eigenvalues are orthogonal, $A$ will be symmetric as it will be orthogonally diagonalizable. But what if the matrix itself is not diagonalizable? will the statement remains true or not? Or is it impossible for this to happen? Thanks. (Working on real space $\mathbb{R}$)

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  • $\begingroup$ If the matrix is not diagonalizable, then it is not symmetric (Spectral Theorem). $\endgroup$ – User8128 Jun 22 '17 at 2:31
  • $\begingroup$ Suppose you start with any matrix you want as long as you can form a basic using some of its eigenvectors $\{v_1,...,v_n\}$. Then define the scalar product by defining it to be $v_i\cdot v_j=\delta_{i,j}$, where $\delta_{i,j}$ is the Kronecker's Delta and extend the definition by linearity. $\endgroup$ – OR. Jun 22 '17 at 2:38
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You have two cases to deal with: eigenvectors spanning, and eigenvectors not spanning the space.

It is easy to give an example where the eigenvectors do not span: $$ \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $$ has one eigenvector, $(1,0)$, with eigenvalue $0$, and this is clearly orthogonal to all of the other eigenvectors, vacuously. This matrix is also clearly not diagonalisable.

So suppose we have a matrix $A$, of which the eigenvectors with distinct eigenvalues orthogonal, and eigenvectors of $A$ span the space. Then we can produce an orthonormal basis $\{ e_i \}$ of the space by using Gram–Schmidt on each eigenspace. But then we can write the matrix as $$ A = \sum_i \lambda_i e_i e_i^T $$ (where the $\lambda_i$ are the eigenvalues, not necessarily distinct), which follows by expanding $v$ in the orthonormal basis and using the linearity of $A$. But then $ A^T = \sum_i \lambda_i (e_i e_i^T)^T = \sum_i \lambda_i e_i e_i^T = A $, so $A$ is symmetric.

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