I know by theorem that every hermitian and skew-hermitian matices is similar to diagonal


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?

  • $\begingroup$ Is the following matrix diagonalizable: $\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}$? That's an actual question by me (is that matrix considered diagonalized?). $\endgroup$ – Jared Nov 19 '14 at 6:17
  • 1
    $\begingroup$ @Jared Yes, it is already diagonal. $\endgroup$ – alex.jordan Nov 19 '14 at 6:49

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?


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