Is the set of all matrices $A$ that have $Tr(BAB)=0$ ($B$ is any fixed matrix) a vector space?

Question

I'm really struggling to write the $tr(BAB)$ with the formal definition (sigma and products). I would appreciate it if someone can help me with how I approach questions like these with traces and products of matrices. Thanks

Answer 1

To show that something is a vector space you need it to be closed under addition an scalar multiplication. So assume that $\operatorname{tr}(BAB)=\operatorname{tr}(BA'B)=0$. Let us use properties of traces (but not the definition): $$\operatorname{tr}(B(A+A')B)=\operatorname{tr}(BAB+BA'B)=\operatorname{tr}(ABA)+\operatorname{tr}(BA'B)=0+0=0.$$

Similarly, if $c$ is a scalar, we have $$\operatorname{tr}(B(cA)B)=\operatorname{tr}(c(BAB))=c\operatorname{tr}(BAB)=c\cdot 0 = 0.$$

The first calculation tells us that our space is closed under sums. The second tells us that our space is closed under scalar multiplication.

Answer 2

To complement Aaron's nice answer, one could also notice that the map

$$f \colon \mathrm{Mat}(n \times n, k) \to k, \ \ \ \ \ A \mapsto \mathrm{tr}(BAB)$$

is linear. (Here, $k$ is any field and $\mathrm{Mat}(n \times n, k)$ is the space of $n \times n$-matrices with entries in $k$). Of course, one proves this with similar arguments as in Aaron's answer.

Then your desired space is just the kernel of $f$, hence a vector subspace of $\mathrm{Mat}(n \times n, k)$.