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 just wanted to check that I am right about this.

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?

  • 1
    $\begingroup$ 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? $\endgroup$ Apr 13, 2014 at 14:22
  • $\begingroup$ 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! $\endgroup$
    – EHH
    Apr 13, 2014 at 14:27
  • $\begingroup$ I've removed that last bit now. $\endgroup$
    – EHH
    Apr 13, 2014 at 14:28

1 Answer 1


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 $0$.

  • $\begingroup$ I think I have misunderstood the notation, I have now added exactly how it appears on the question sheet at the end of the question. Does it look like anything you recognise? $\endgroup$
    – EHH
    Apr 13, 2014 at 14:42
  • $\begingroup$ 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? $\endgroup$
    – bradhd
    Apr 13, 2014 at 15:05
  • $\begingroup$ Yeah I thought that too but then the rest of the question (which I haven't posted) doesn't follow. I've emailed my lecturer to ask for clarification about the notation. Thanks for your help! $\endgroup$
    – EHH
    Apr 13, 2014 at 15:31

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