I know from a theorem that every hermitian and skew-hermitian matrix is similar to a diagonal matrix.

But, is this fact also true for symmetric and skew-symmetric matrices?

And, symmetric matrices have real eigenvalues, what about symmetric matrices that have complex entries?


2 Answers 2


This is just an "add-on" for the complex symmetric case.

No, complex symmetric matrices do not need to be diagonalizable. Consider $$ \pmatrix{1 & i\\ i & -1}, $$ which is symmetric but is not diagonalisable.

However, for any complex symmetric matrix $A$, there is a unitary matrix $U$ such that $A=UDU^T$, where $D$ is a nonnegative diagonal matrix (note that $^T$ stands here for the usual transposition, which is not same as the conjugate transpose usually seen in the context of complex matrices). This is referred to as the Takagi's factorization.


Every symmetric matrix is orthogonally diagonalizable. This is a standard theorem from linear algebra. So in particular, every symmetric matrix is diagonalizable (and if you want, you can make sure the corresponding change of basis matrix is orthogonal.)

For skew-symmetrix matrices, first consider $\begin{bmatrix}0&-1\\1&0\end{bmatrix}$. It's a rotation by 90 degrees in $\mathbb{R}^2$, so over $\mathbb{R}$, there is no eigenspace, and the matrix is not diagonalizable. It is of course, diagonalizable over $\mathbb{C}$ though.

See here for the corresponding statement about complex skew-symmetric matrices using unitary matrices instead of orthogonal ones. Note that the blocks in the matrix $\Sigma$ at this link are themselves diagonalizable, so $\Sigma$ is diagonalizable. Is $\Sigma$ diagonalizable with unitary matrices?

  • 2
    $\begingroup$ I like that you add the point that its diagonalizable over $\mathbb{C}$. I'm trying to compute $e^A$ in general for skew symmetric matrices, which was difficult because everyone says they're not diagonalizable, but they are. $\endgroup$
    – Kraigolas
    Commented Jul 26, 2019 at 20:51

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