Let \(V\) be the vector space of all \(2 \times 2\) matrices over the field \(F\). Let \(W_1\) be the set of matrices of the form \[ \begin{bmatrix} x & -x \\ y & z \end{bmatrix} \] and let \(W_2\) be the set of matrices of the form \[ \begin{bmatrix} a & b \\ -a & c \end{bmatrix}. \]
Prove that \(W_1\) and \(W_2\) are subspaces of \(V\).
Find the dimensions of \(W_1\), \(W_2\), \(W_1 + W_2\), and \(W_1 \cap W_2\).
Solution
1. \(W_1\) and \(W_2\) are subspaces of \(V\).
One has \[ W_1=\left\{ x\begin{bmatrix}1&-1\\0&0\end{bmatrix} +y\begin{bmatrix}0&0\\1&0\end{bmatrix} +z\begin{bmatrix}0&0\\0&1\end{bmatrix}:x,y,z\in F \right\}, \] so \(W_1\) is a span. Likewise \[ W_2=\left\{ a\begin{bmatrix}1&0\\-1&0\end{bmatrix} +b\begin{bmatrix}0&1\\0&0\end{bmatrix} +c\begin{bmatrix}0&0\\0&1\end{bmatrix}:a,b,c\in F \right\}. \] Hence both are subspaces.
2. \(\dim W_1=\dim W_2=3\).
The three displayed spanning matrices for each space have disjoint independent coordinate parameters, hence are linearly independent.
3. The intersection is\[ W_1\cap W_2 =\left\{ \begin{bmatrix}x&-x\\-x&z\end{bmatrix}:x,z\in F \right\}, \]so \(\dim(W_1\cap W_2)=2\).
A matrix lies in both spaces exactly when \[ \begin{bmatrix}x&-x\\y&z\end{bmatrix} =\begin{bmatrix}a&b\\-a&c\end{bmatrix}. \] Comparing entries gives \(a=x\), \(b=-x\), \(y=-x\), and \(c=z\), with \(x,z\) free.
4. \(W_1+W_2=V\), so \(\dim(W_1+W_2)=4\).
The dimension formula gives \[ \dim(W_1+W_2)=3+3-2=4. \] Since \(V=M_2(F)\) has dimension \(4\), the subspace \(W_1+W_2\) must equal \(V\).