# What is a basis for the space of $n\times n$ Hermitian matrices?

So I was working on a specific problem related to Hermitian matrices. If we let $H_n$ denote the set of n x n Hermitian matrices. We're told that $H_n$ is a real vector space under matrix addition and scalar multiplication by a real number. I don't understand why though, because what would prevent you from adding two imaginary numbers?

Then we're told to write a basis for $H_2$. Now for a Hermitian matrix, it's equal to its conjugate transpose, so a basis would be:

$$\left(\begin{array}{ccc} 0 & 1 \\ 1 & 0\\ \end{array}\right) , \left(\begin{array}{ccc} 0 & 0 \\ 0 & 1\\ \end{array}\right), \left(\begin{array}{ccc} 1 & 0 \\ 0 & 0\\ \end{array}\right), \left(\begin{array}{ccc} 0 & i \\ -i & 0\\ \end{array}\right)$$

Which should be the same as the basis for a symmetric matrix, correct?

From there though, we have $sH_n$ represent skew-Hermitian matrices. I believe the basis for $sH_n$ is:

Edit: updated my basis here.

$$\left(\begin{array}{ccc} 0 & -1 \\ 1 & 0\\ \end{array}\right), \left(\begin{array}{ccc} 0 & i \\ i & 0\\ \end{array}\right)$$

because the diagonals would have to be zero right, to make the conjugate transpose of A be equal to -A?

From there though, they ask if $sH_n$ is a real vector space, and I'm not sure what to answer. It's real if we only use reals?

I'm asked the same thing for $U_n$, where $U_n$ represents n x n unitary matrices. I'm not even sure how to come up with a basis for this set??

Edit: Technically, the question says "Is $U_n$ a real vector space? Write a basis if possible."

I suppose I couldn't write a basis because $U_n$ is not a real vector space then. How would I show that though?

• The basis for $SH_2$ you have is now correct with the edit, but you're missing a sign in the imaginary matrix you added to the first basis. It can't be both Hermitian and skew-Hermitian (cause it's not $0$)! – anon Dec 14 '11 at 22:52
• @anon - yep, just noticed that! I realized that if they're the same sign, then they'll be skew-Hermitian, so I changed that to fix it. Thanks for all your help! I guess my last question is: How do I show that unitary matrices aren't closed? Would a counter example suffice, or is there an elegant proof? – John Doe Dec 14 '11 at 22:54
• Both $I_n$ and $-I_n$ are unitary matrices. What happens when you add them together? – anon Dec 14 '11 at 22:59
• Ah, that does it. How did you come up with that example? Just past experience or is there a good system to derive such counter examples? – John Doe Dec 14 '11 at 23:31
• If you pick two points on a unit circle, almost surely they will not be exactly a distance of $1$ apart. I intuited that the same principle applies with adding unitary matrices: pick any two and almost surely they will not add to another unitary matrix. So I decided to make the first one a very simple choice of $I_n$ and saw eventually that the second one could also be a simple choice, namely $-I_n$. I guess you could also say that $A+A\not\in U_n$ for any $A\in U_n$ as well - whatever works. – anon Dec 14 '11 at 23:36

• I take it your specific question in the first instance is: I don't understand why it has to be a real vector space because we can add together complex numbers just as well, can't we? Yes, but don't forget the little detail of complex conjugates. Observe that $$i\begin{pmatrix}0&i\\-i&0\end{pmatrix}+\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}0&0\\2&0\end{pmatrix}$$ is a $\mathbb{C}$-linear combination of Hermitian matrices, and the result is not Hermitian.

• With entries strictly in $\mathbb{R}$, Hermitian matrices are just symmetric matrices so your basis is correct and is indeed the very one for symmetric matrices. However, the problem probably wants the matrices here to include components from $\mathbb{C}$, so what you have is incomplete. By linearity, try treating the real and imaginary parts of the elements of $H_n$ separately.

• As for the basis for what you call $SH_2$, what about the following matrix, which isn't represented by what you have? And do you know what "diagonals" refers to in a matrix? $$\begin{pmatrix}0&1\\-1&0\end{pmatrix}$$ Of course, this one probably wants complex hermitian matrices as well.

• Careful. The addition of two unitary matrices is generally not unitary, so try to understand what the question means...

• For the $sH_2$ basis, I wrote it incorrectly. And yep, I meant "diagonal" not the plural, because I want to say that the diagonal must be zero for skew hermitian and the other two must be equivalent, but with opposite sign. In fact, I believe what you've given is the basis for skew-symmetric matrices correct? I updated my basis for both Hermitian and skew-Hermitian matrices. – John Doe Dec 14 '11 at 22:43
• @John: Yes, it is the basis for skew-symmetric $2\times2$ matrices, even over $\mathbb{C}$. But not the one for $SH_2\subset M_{2,2}(\mathbb{C})$, because there is one matrix with imaginary numbers that is needed to complete the basis then.. Keep in mind the matrices themselves might have complex number entries but the field we're working over (the field of scalars with which we multiply the matrices by) is $\mathbb{R}$. – anon Dec 14 '11 at 22:48

a basis for $H_2$ over $\mathbb{R}$ is $$\left( \begin{array}{cc} 1&0\\ 0&0\\ \end{array} \right), \left( \begin{array}{cc} 0&0\\ 0&1\\ \end{array} \right), \left( \begin{array}{cc} 0&i\\ -i&0\\ \end{array} \right), \left( \begin{array}{cc} 0&1\\ 1&0\\ \end{array} \right)$$ being a real vector space doesnt mean that the entries have to be real, just that it is closed under multiplication by real scalars

In the general case, a standard basis is given (up to normalisation) by $$E_{ii},\qquad E_{ij}+E_{ji}, \qquad i(E_{ij}-E_{ji}),$$ for all $$i,j=1,...,n$$ with $$i, and where I used the notation $$E_{ij}\equiv e_i e_j^*$$, where $$e_i$$ is the unit vector that is nonzero only in its $$i$$-th component: $$(e_i)_j =\delta_{ij}$$ and $$(E_{ij})_{k\ell}=\delta_{ik}\delta_{j\ell}$$.

Clearly all these matrices are Hermitian. To verify that form an orthogonal basis, we just need to check that $$\mathrm{Tr}(E_{ii} (E_{jk}+E_{kj})) =\mathrm{Tr}(E_{ii}(E_{jk}-E_{kj}))) =\mathrm{Tr}(E_{ii}E_{jk})) = \delta_{ij}\delta_{ik} = 0, \qquad (\text{as } j