Question

Let V be the vector space of all 2 X 2 matrices over the field F. Prove that V has dimension 4 by exhibiting a basis for V which has four elements.

Answer 1

Answer:

See the proof below.

Step-by-step explanation:

We can define a basis of V with the following elements:

$X_1=\begin{matrix}1 & 0 \\0 & 0 \end{matrix}$

$X_2=\begin{matrix}0 & 1 \\0 & 0 \end{matrix}$

$X_3=\begin{matrix}0 & 0 \\1 & 0 \end{matrix}$

$X_4=\begin{matrix}0 & 0 \\0 & 1 \end{matrix}$

So then if we define the basis X as following:

$X = [X_1, X_2, X_3, X_4]$

$X =[\begin{pmatrix}1 & 0\\0 & 0\end{pmatrix},\begin{pmatrix}0 & 1\\0 & 0 \end{pmatrix},\begin{pmatrix}0 & 0\\1 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\0 & 1 \end{pmatrix}$

We see the the dimension for X is 4 $dim (V) = 4$ since the basis have a dimension of 4 $dim (X) =4$