No, not in general. The matrix
$$
\begin{bmatrix}1&1\\0&1\end{bmatrix}
$$
is famously not similar to a diagonal matrix (a property commonly known as diagonalizable). Symmetric matrices, on the other hand, are always diagonalizable. In fact, if $S$ is a symmetric matrix, we can even require that $S = PDP^{-1}$ where $D$ is diagonal and $P$ is orthogonal. (This is assuming that the entries in the matrices here are real. If they are complex we use slightly different words to take into account that ne never really transpose matrices, but instead transpose them and simultaneously take the complex conjugate, but the results stay the same.)
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In general, no. Take $\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$, for instance.
No, not in general. The matrix $$ \begin{bmatrix}1&1\\0&1\end{bmatrix} $$ is famously not similar to a diagonal matrix (a property commonly known as diagonalizable). Symmetric matrices, on the other hand, are always diagonalizable. In fact, if $S$ is a symmetric matrix, we can even require that $S = PDP^{-1}$ where $D$ is diagonal and $P$ is orthogonal. (This is assuming that the entries in the matrices here are real. If they are complex we use slightly different words to take into account that ne never really transpose matrices, but instead transpose them and simultaneously take the complex conjugate, but the results stay the same.)