# short question regarding convention - symmetric matrices and transpose

I have a short question because wikipedia is extremly vague on this subject.

Suppose I have the matrix $A=\begin{pmatrix} i & 1 \\ 1 & -i\end{pmatrix}$.

Is it symmetric? I mean, in the complex field transpose is defined not like in the real field. I'd like someone to help clarify the difference between symmetric, hermatian, and how transposition is defined over R and C. Is this matrix symmetric? is it hermatian?

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Symmetric yes, Hermitian no. –  Amzoti Dec 26 '13 at 0:25

A square matrix $A$ is called symmetric if $A=A^T$ , and Hermitian if $A=A^H=\bar A^T$. So we don't care whether elements of matrix are real or complex. For real matrices, symmetric is equivalent to Hermitian. But symmetric matrix in complex may not be Hermitian (say $A$ in your example).

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And a Hermitian matrix need not be symmetric. –  Alex Becker Dec 26 '13 at 1:02
In such a case it is very common to say complex symmetric, since just "symmetric" often implies "real". –  Vedran Šego Dec 26 '13 at 1:25

If you ask: Is $A^t=A$ then the answer is yes.

But for complex matrices, transposing has no real useful meaning. What you want is conjugate transpose and clearly $A^H \neq A$. Nothing more can be said of the matrix.

Why do you need to transpose AND take the complex conjugates? The way I like to explain is that a complex number $u + i \,v$ can be mapped to a real matrix $$u + i\,v \Leftrightarrow \begin{bmatrix}u & v\\-v &u\end{bmatrix}$$ Notice that transposing the matrix results in taking complex conjugate.

So... when dealing with complex matrices, transposition should always be accompanied by taking complex conjugates. The two go together!

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I asked this because i was asked "Is any symmetric matrix over an algebraically closed field diagonlizable" i looked at this matrix. I think it's symmetric, it's not diagonlizable. I just wanted to clarify that this is indeed a symmetric matrix over the complex numbers. –  Oria Gruber Dec 26 '13 at 0:34
Complex symmetric matrices do have their applications and are researched. Check, for example, Takagi decomposition. –  Vedran Šego Dec 26 '13 at 1:25
Good point. I don't like over generalizations and I realize I did so in my final comment. Mea culpa –  user44197 Dec 26 '13 at 1:31

Yes it is symmetric, as $A^T=A$. What this matrix is not, is Hermitian. Recall, that a matrix $A$ is called Hermitian if $\bar A^T=A$, and in the case of matrices with complex entries we usually we are not wondering if they a symmetric (or anti-symmetric) but hermitian (or skew-hermitian).

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