Let \(f\) be the bilinear form defined in Exercise 6. Let \(W\) be the set of all matrices \(A\) in \(V\) such that \(f(A, B) = 0\) for all \(B\). Show that \(W\) is a subspace of \(V\). Describe \(W\) explicitly and find its dimension.
Solution
Recall \[ f(A,B) =n\operatorname{tr}(AB) -\operatorname{tr}(A)\operatorname{tr}(B). \]
1. The set \(W\) is a subspace of \(V\).
If \(A,C\in W\) and \(a,b\in\mathbb C\), then for every \(B\in V\), \[ f(aA+bC,B) =a f(A,B)+b f(C,B) =0. \] Thus \(aA+bC\in W\). Also \(0\in W\). Hence \(W\) is a vector subspace of \(V\).
2. Every scalar matrix lies in \(W\).
For \(A=cI\) and arbitrary \(B\), \[ \begin{aligned} f(cI,B) &=n\operatorname{tr}(cB) -\operatorname{tr}(cI)\operatorname{tr}(B)\\ &=nc\operatorname{tr}(B)-nc\operatorname{tr}(B)\\ &=0. \end{aligned} \] Therefore \[ \mathbb CI\subseteq W. \]
3. Every element of \(W\) is scalar.
Let \(A\in W\) and set \[ c=\frac{\operatorname{tr}(A)}{n}, \qquad C=A-cI. \] Then \(\operatorname{tr}(C)=0\). Since the scalar part \(cI\) belongs to \(W\) by step 2 and \(W\) is a subspace, \(C\in W\). Thus for every \(B\in V\), \[ 0=f(C,B)=n\operatorname{tr}(CB), \] so \[ \operatorname{tr}(CB)=0. \]
Take \(B=E_{ji}\). Then \[ \operatorname{tr}(CE_{ji})=C_{ij}. \] Hence every entry \(C_{ij}\) is zero. Thus \(C=0\) and \[ A=cI. \] Consequently \[ W=\mathbb CI. \]
4. Compute the dimension.
Since \(I\ne0\), the subspace \(\mathbb CI\) is one-dimensional. Therefore \[ \boxed{W=\mathbb CI, \qquad \dim W=1.} \]