Direct sum related to trace is zero space

Question

Let $W$ be the space of $n\times n$ matrices whose trace is zero. Find a subspace $W'$ so that $\mathbb{R}^{n\times n}=W\oplus W'$ (Here $\oplus$ is direct sum notation.)

So, I have to find $W'$ such that $W\cap W'=0$. But no clue how to find such set $W'$. It is certainly NOT matrices whose trace is non-zero (then we can not get $W\cap W'=0$).

Any help appreciated.

Answer 1

Hint The subspace $W$ is precisely the kernel of the nontrivial linear map $\operatorname{tr} : \Bbb R^{n \times n} \to \Bbb R$, so it has codimension $1$.

So, any $1$-dimensional subspace $W'$ not in $W$ (so, the line $\langle A \rangle$ for any $A \not\in W$) will do. On the other hand, there is a canonical choice for $A$, namely, the identity matrix $I$, so that $W'$ is precisely the space of scalar matrices, $\lambda I$, $\lambda \in \Bbb R$. By construction the trace map $\operatorname{tr}$ is essentially the projection onto this factor in the decomposition $\Bbb R^{n \times n} = W \oplus \langle I \rangle$: More precisely, the projection is $$A \mapsto \tfrac{1}{n} \operatorname{tr} A .$$

Answer 2

Define $f:\Bbb R^{n\times n}\to\Bbb R$ as $f(M)=Tr(M)$. This function is linear and $W=\ker f$. So $\dim W=n^2-1$. Then $W'$ can be the $1$-dimensional space generated by any matrix not in $W$.

Answer 3

$W' = \operatorname{Span}(I)$ is the subspace you are looking for. Alternatively, for any $i\leq n$, the subspace $W' = \operatorname{Span}(E_{ii})$ also works, where $E_{ii}$ is the matrix that is all $0$ except for position $(i,i)$, where it is $1$. Or mix the two matrices before you take the span. Whatever you like.