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How can I "show that the Hermitian Matrices form a real Vector Space"? I'm assuming this means the set of all Hermitian matrices.

I understand how a hermitian matrix containing complex numbers can be closed under scalar multiplication by multiplying it by i, but how can it be closed under addition? Wouldn't adding two complex matrices just produce another complex matrix in most instances?

Also, how can I find a basis for this space?

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    $\begingroup$ Hermitian matrices are not closed under multiplication by $i$, that's why they don't form a complex vector space, only a real one. $\endgroup$ – Michael Albanese Jan 28 '16 at 14:49
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Hint:

i give you the intuition for $2\times 2$ matrices that you can extend to the general case.

An Hermitian $2\times 2$ matrix has the form: $$ \begin{bmatrix} a&b\\ \bar b&c \end{bmatrix} $$ with: $a,c \in \mathbb{R}$ and $b\in \mathbb{C}$ ($\bar b $ is the complex conjugate of $b$).

Now you can see that, for two matrices of this form, we have:

$$ \begin{bmatrix} a&b\\ \bar b&c \end{bmatrix} + \begin{bmatrix} x&y\\ \bar y&z \end{bmatrix}= \begin{bmatrix} a+x&b+y\\ \bar{b}+\bar{y}&c+z \end{bmatrix} $$ and, since $\bar{b}+\bar{y}=\overline{b+y}$ the result is a matrix of the same form, i.e. an Hermitian matrix.

For the product we have: $$k \begin{bmatrix} a&b\\ \bar b&c \end{bmatrix}= \begin{bmatrix} ka&kb\\ k\bar b&kc \end{bmatrix} $$ so the result matrix is Hermitian only if $k$ is a real number. Finally we see that the null matrix is hermitian and the opposite of an Hermitian matrix is Hermitian, so the set of hermitian matrix is real vector space.

For the basis:

Note that an hermitian matrix can be expressed as a linear combination with real coefficients in the form: $$ \begin{bmatrix} a&b\\ \bar b&c \end{bmatrix}= a\begin{bmatrix} 1&0\\ 0&0 \end{bmatrix}+ \mbox{Re}(b) \begin{bmatrix} 0&1\\ 1&0 \end{bmatrix} +\mbox{Im}(b) \begin{bmatrix} 0&i\\ -i&0 \end{bmatrix} +c \begin{bmatrix} 0&0\\ 0&1 \end{bmatrix} $$ and ,since the four matrices at the right are linearly independent, these form a basis.

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  • $\begingroup$ I can't believe you answered this a year ago without an upvote. With regards, $\endgroup$ – user12802 Apr 21 '17 at 23:29

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