Is there a matrix $A$ such that for every other matrix $B$, we have $\mbox{tr}(AB) = 0$? I'm struggling to prove/disprove this notion. I've figured if such matrix exists, it has to be nilpotent and non-invertible, and the sum of its eigenvalues is 0. Can anyone chip in?
Edit : Oh yeah, $A \neq 0$
 A: Another sensible approach here: let $E_{ij}$ denote the matrix whose $i,j$ entry is $1$ and whose other entries are zeros.  Show that
$$
\operatorname{tr}(AE_{ji}) = a_{ij}
$$
This can only be zero for all $i,j$ if $A$ is the zero matrix.

A nifty trick in proving the above is to note that $E_{ij} = e_ie_j^T$, where $e_i$ denotes the $i$th standard basis vector.  In particular, we have
$$
\operatorname{tr}(AE_{ji}) = \operatorname{tr}(Ae_je_i^T) = \operatorname{tr}(e_i^TAe_j) = e_i^TAe_j = a_{ij}
$$
A: Other answers have presented proofs that this must be the zero matrix, but here are some general avenues of attack for analyzing a set S of matrices, applied to the case where S is the set of matrices with your property.


*

*Is it closed under addition? $tr(M_1+M_2)=tr(M_1)+tr(M_2)$, hence $tr((A_1+A_2)B)=tr(A_1B)+tr(A_2B)$. With a bit more work, it can be shown that S is a group under addition.

*Is it closed under multiplication?
Given $A$ in S, we have $tr(ACB)$ = $tr(A(CB))$, which, since $A$ is in S, is zero. But it can also be written as $tr((AC)B)$.  So given any $A$ in S and arbitrary $C$, $AC$ is in S. Thus, S is an ideal. (Technically, I've shown that it's a right ideal, but not that it's a left ideal, but that follows from $tr(AB) = tr(BA)$). 

*What entries have to be non-zero? Since S is an ideal, we can permute any element of S and the result will be an element of S. So if an element of S has any non-zero entry, we can permute that entry to be $a_{11}$. We can also scale it to be 1. So if there are any nontrivial elements of S, then there is an element for which $a_{11}$ is 1

*Is $E_{ij}$ in S (where $E_{ij}$ is the matrix with $a_{ij} = 1$ and zero elsewhere)? If we have an element $A$ of S whose $a_{11}$ entry is 1, then $E_{11}A= E_{11}$ ( where $E_{ij}$ is the matrix with $a_{ij} = 1$ and zero elsewhere). So if there are any nontrivial elements of S, then $E_{11}$ is in S. A similar proof applies to any $E_{ij}$.


At this point, we've reached an absurdity, as $E_{11}$ is clearly not in S.
A: Let $B=A^t$.  Then $\operatorname{trace}(AA^t)$ is the sum of squares of all coefficients of the real matrix $A$.  If this is zero then $A=0$.  So the zero matrix is the only possibility.
Somebody should have gotten around to pointing out that the underlying fundamental fact is that the Hilbert-Schmidt operator provides a norm on the space of matrices, also known as the Frobenius norm.
A: Perhaps the interest of the question appears more clearly when introducing the $K$-vector space $M_n(K)$ of $n\times n$ matrices with coefficients in $K$ (where $K$ is any field) endowed with the map $(A, B)\to <A,B>:= tr(AB^t)$. This map is a $K$-bilinear symmetric form : $K$-bilinearity is routine, and the symmetry comes from the well known identities $tr(AB)=tr(BA)$ and $tr(C^t)=tr(C)$. The OP question can be reformulated as : is the form <.,.> non degenerate, in other words is it a scalar product (but note that this last terminology is not a universal standard) ?
Now $M_n(K)$ has dimension $n^2$, with a canonical basis consisting of the elementary matrices $E_{ij}$ whose entries are $0$ except the $(i,j)$ entry which is $1$. It is an easy calculation (or a "nifty trick" in Omnomnomnom's words) to show that $tr(E_{ij}E_{hk})$ is the $(kh)$ entry of $E_{ij}$, so that $<E_{ij},E_{kh}>=1$ if $(i,j)=(k,h)$, $0$ otherwise. This means that the $ E_{ij}$'s form an orthonormal basis of $M_n(K)$. If we rewrite this basis as  $\epsilon_1 ,...,\epsilon_{n^2}$ (the order is arbitrary), and denote the coordinates of $A,B$ in this basis by $(\alpha_i,\beta_j)$, then $<A,B>=\alpha_1\beta_1 +...+\alpha_{n^2}\beta_{n^2}$, a well known expression. Note that although the bilinear form <.,.> is non degenerate, the associated quadratic form defined by $Q(A)=<A,A>$ is not necessarily definite, i.e. it can admit non zero isotropic vectors since the solutions of the equation $Q(A)=0$ depend on the base field.
A: @Omnomnomnom's answer works over any unital ring. What if the ring is non-unital? 
Well then, there are beasts like the quotient of $2\Bbb Z$ by the ideal $4\Bbb Z$. It has two elements, $\bar 0$ and  $\bar 2$, of which all possible products, that is $\ \bar 0\times \bar 0$, $\ \bar 0\times \bar 2,\ $$\bar 2\times \bar 2$, are equal to $\bar 0$. It follows, since the entries of a product matrix are made of (sums of) products of entries, that any matrix $A\in \mathcal{M}_{n, n}(2\Bbb Z/4\Bbb Z)$, $A\neq 0$,  satisfies the required property that for all $B\in \mathcal{M}_{n, n}(2\Bbb Z/4\Bbb Z)$, $\operatorname{Tr}(AB)=0$.
A: No. For $A\not=0$ we have $\mathrm{tr}(AA^\top)>0$. 
This is because $AA^\top$ is positive semi-definite (see Gramian matrix), hence its trace (as the sum of all the non-negative eigenvalues) is positive unless $A$ was zero.
