# Show the Hermitian matrices, with trace(g*g1,1)=0 form a vector space.

This is a question from an example sheet that I think may have a mistake in it.

Show that the set of Hermitian matrices $A \in H_2 (\mathbb{C})$ with Trace$(A\cdot A_{(1,1)})=0$ is a real three dimensional vector space.

So Hermitian $2\times2$ matrices have reals on the diagonal and a complex number and it's conjugate on it's off diagonal. Then the condition Trace$(A\cdot A_{(1,1)})=0$ mean that,

$$(A_{(1,1)})^2 + A_{(2,2)}A_{(1,1)} =0$$

now if $A_{(1,1)} \neq0$ we have matrices of the form;

$$\begin{pmatrix} a & b \\ \bar{b} & -a \end{pmatrix} ~~~~~s.t.~~~~ a \in \mathbb{R}, b\in \mathbb{C}$$

But if $A_{(1,1)}= 0$ then our conditions are also all satisfied and we get matrices of the form;

$$\begin{pmatrix} 0 & b \\ \bar{b} & d \end{pmatrix} ~~~~~s.t.~~~~ d \in \mathbb{R}, b\in \mathbb{C}$$

but summing a matrix of each type we get,

$$\phi=\begin{pmatrix} a & b \\ \bar{b} & d-a \end{pmatrix} ~~~~~s.t.~~~~ a \in \mathbb{R}, b\in \mathbb{C}$$

But then Trace$(\phi \cdot \phi _{(1,1)}) = da \neq 0$ if $a$ and $d$ aren't. So this can't be a vector space since it isn't closed under addition.

I think I may have misunderstood the notation though, the exact way it is written on the question sheet is $$Trace(AA_{1,1})$$

does this have a usual meaning different to the one I gave?

• I don't really understand your notations, but if you do not allow for the zero matrix, how is your set going to be a vector space? – Giuseppe Negro Apr 13 '14 at 14:22
• My notation is that A(i,j) is the jth entry in the ith row of the matrix. You're right though I can't make that restriction, thanks! – EHH Apr 13 '14 at 14:27
• I've removed that last bit now. – EHH Apr 13 '14 at 14:28

If I understand your notation correctly, you are right: this isn't a vector space, for the reasons you gave. For example, $A = \left(\begin{array}{cc}1&0\\0&-1\end{array}\right)$ and $B = \left(\begin{array}{cc}0&0\\0&1\end{array}\right)$ are both in that set, but $A+B = \left(\begin{array}{cc}1&0\\0&0\end{array}\right)$ is not in that set, as the trace of $1(A+B)$ is $1$, not $0$.
• No, it doesn't. I'd have interpreted it the same way you did. By the way, the set of $2\times 2$ Hermitian matrices $A$ with $\operatorname{Trace}(A)=0$ is a 3-dimensional real vector space. Perhaps that's what the question is meant to say? – bradhd Apr 13 '14 at 15:05