Finding dimension of space

Question

Let $V = M_n(\mathbb{C})$ and define

$$ f(A,B) = n\cdot\mathrm{tr}(AB) - \mathrm{tr}(A)\mathrm{tr}(B)\ . $$

Find $\mathrm{dim} (S)$ where

$$ S = \left\{ v \in V\ | \ \text{for all}\ u \in V, \ f(v,u) = 0\right\}\ . $$

I first tried to check some base cases: for $n=1$, $f(A,B) = 0$ so $\mathrm{dim}(S) = n = 1$, but for $n>1$ it's not easy to find $\mathrm{dim}(S)$.

So I proved that $f(A,B)$ is symmetric bilinear form over $V$. Then I tried to prove that $f$ is an inner product, but I didn't manage to prove it and I don't know if this is the right way.

I try to find a basis and count when faced with such a problem. — James S. Cook, Aug 30 '13 at 10:01
Your form $f$ is complex bilinear and symmetric; there is no such thing as being positive definite for such forms. That notion requires a Hermitian (sesquilinear) form instead. — Marc van Leeuwen, Aug 30 '13 at 10:05
@user91015 you might be correct, I'll post my calculation in a minute or two. — James S. Cook, Aug 30 '13 at 10:46

score 3 · Answer 1 · answered Aug 30 '13 at 13:32

You are looking for the set$~S$ of all matrices $\def\tr{\operatorname{tr}}A\in M_n(\Bbb C)$ such that $$ n\tr(AB)=\tr(A)\tr(B)\qquad\text{for all $B\in M_n(\Bbb C)$,} $$ which as you remarked is a (complex) subspace of $M_n(\Bbb C)$. Note that since $\tr(I_n)=n$, one has $I_n\in S$. So the dimension is at least$~1$.

Now $(A,B)\mapsto \tr(AB)$ is a non-degenerate symmetric bilinear form, as one easily checks (only the zero matrix gives$~0$ with all elementary matrices). If we let $B$ run through the $n^2-1$ dimensional subspace$~T$ of $M_n(\Bbb C)$ of matrices with trace$~0$, we see that $A$ has to be in the orthogonal complement of$~T$ for this form. So $1\leq\dim S\leq n^2-(n^2-1)=1$ and $\dim S=1$.

@user91015 this is the clever answer which avoids the computation given in mine. Very pretty. — James S. Cook, Aug 30 '13 at 13:38

score 1 · Answer 2 · answered Aug 30 '13 at 10:51

Consider $E_{ij}$ the matrix with zero in every entry except where is takes value $1$ in the $(i,j)$-th component. We can write these as $E_{ij} = e_ie_j^T$ the product of $n \times 1$ and $1 \times n$ standard basis vectors and their transposes. Clearly, $e_i^T e_j = e_i \cdot e_j = \delta_{ij} = \begin{cases} 0 & i \neq j \\ 1 & i=j \end{cases}$. Consider then, $$ E_{ij}E_{kl} = (e_ie_j^T)(e_ke_l^T) = e_i \delta_{jk}e_l^T = \delta_{jk}E_{il} $$ Any matrix may be written $A = \sum_{ij} A_{ij}E_{ij}$ for appropriate complex scalars $A_{ij}$. Likewise for $B$. Now plug these into the defining condition for $S$ and see what condition this forces on the components $A_{ij}$. Note: $$ AB = \sum_{ijk} A_{ik}B_{kj}E_{ij} $$ But, $\text{trace}(C) = \sum_i C_{ii}$ hence $$ \text{trace}(AB) = \sum_{ik} A_{ik}B_{ki} $$ On the other hand, $$ \text{trace}(A)\text{trace}(B) = \sum_{i} A_{ii} \sum_k B_{kk} = \sum_{ik}A_{ii}B_{kk} $$ Therefore, in general, $f(A,B)=0$ yields the condition that follows as $n\text{trace}(AB)-\text{trace}(A)\text{trace}(B) = \text{trace}(nAB)-\text{trace}(A)\text{trace}(B)$ $$ 0 = \sum_{ik} (nA_{ik}B_{ki} -A_{ii} B_{kk}) \qquad \star $$ We want to characterize all $A$ such that $f(A,B)=0$ for all $B$. Therefore, we can choose particular $B$ as to simplify the condition above. For example $B=E_{pq}$ has $\text{trace}(B)=\delta_{pq}$ and $B_{ij} = \delta_{pi}\delta_{qj}$ and it follows that $\star$ reduces to: $$ 0 = \sum_{ik} (nA_{ik}\delta_{pk}\delta_{qi} -A_{ii}\delta_{pk}\delta_{qk} )$$ Therefore, keep in mind $p,q$ are free indices here, $$ 0 = nA_{qp} -\text{trace}(A)\delta_{pq} $$ hence, swapping $p$ with $i$ and $q$ with $j$ we find $$ A_{ij} = \frac{1}{n}\text{trace}(A)\delta_{ij}$$ Thus, if $f(A,B)=0$ for all $B$ then $$ A = \sum_{ij} A_{ij}E_{ij} = \sum_{ij}\frac{1}{n}\text{trace}(A)\delta_{ij}E_{ij} =\frac{1}{n}\text{trace}(A) \sum_i E_{ii} = \frac{\text{trace}(A)}{n}I.$$ Thus $S = span(I)$.

So, I bet you're correct, I suspect you could find this from a more clever line of reasoning. Truth is that I happen to enjoy this sort of calculation.

score 1 · Answer 3 · answered Sep 04 '13 at 10:37

Here is an alternative proof. Consider the inner product $\langle X,Y\rangle=\operatorname{tr}(Y^\ast X)$. If $A\in S$, the given condition says that $$ \langle A,B\rangle = \frac1n\langle A,I\rangle\langle I,B\rangle = \left\langle \frac1n\langle A,I\rangle I,\ B\right\rangle $$ for all $B\in M_n(\mathbb{C})$. Hence $A=\frac1n\langle A,I\rangle I=\frac1n\operatorname{tr}(A)I$, i.e. $A$ is a scalar multiple of $I$. Therefore $\dim S=1$.

Finding dimension of space

3 Answers3