Let \(V\) be the vector space of all \(2 \times 2\) matrices over the field \(F\). Prove that \(V\) has dimension 4 by exhibiting a basis for \(V\) which has four elements.
Solution
Let \[ E_{11}=\begin{bmatrix}1&0\\0&0\end{bmatrix},\quad E_{12}=\begin{bmatrix}0&1\\0&0\end{bmatrix},\quad E_{21}=\begin{bmatrix}0&0\\1&0\end{bmatrix},\quad E_{22}=\begin{bmatrix}0&0\\0&1\end{bmatrix}. \] Every matrix \[ A=\begin{bmatrix}a&b\\c&d\end{bmatrix} \] has the unique expression \[ A=aE_{11}+bE_{12}+cE_{21}+dE_{22}. \] Thus the four matrix units span \(V\) and are linearly independent. They form a basis, so \[ \dim_F V=4. \]