# Vector Space of 2 by 2 Traceless Matrices

## Problem 601

Let $V$ be the vector space of all $2\times 2$ matrices whose entries are real numbers.
Let
$W=\left\{\, A\in V \quad \middle | \quad A=\begin{bmatrix} a & b\\ c& -a \end{bmatrix} \text{ for any } a, b, c\in \R \,\right\}.$

(a) Show that $W$ is a subspace of $V$.

(b) Find a basis of $W$.

(c) Find the dimension of $W$.

(The Ohio State University, Linear Algebra Midterm)

## Solution.

### (a) Show that $W$ is a subspace of $V$.

To show that $W$ is a subspace of $V$, we verify the following subspace criteria.

1. The zero vector in $V$ is in $W$.
2. For all $A, B\in W$, the sum $A+B \in W$.
3. For all $A\in W$ and $r\in \R$, the scalar multiplication $rA\in W$.

Note that the zero vector in $V$ is the zero matrix $\begin{bmatrix} 0& 0 \\ 0& 0 \end{bmatrix}$.
The zero matrix corresponds to the matrix in $W$ with $a=b=c=0$, and hence the zero vector of $V$ is in $W$.
Condition 1 is met.

To verify condition 2, let
$A=\begin{bmatrix} a & b\\ c& -a \end{bmatrix} \text{ and } A’=\begin{bmatrix} a’ & b’\\ c’& -a’ \end{bmatrix}$ be arbitrary elements in $W$, where $a, b, c, d, a’, b’, c’, d’\in \R$.
Then we have
$A+A’=\begin{bmatrix} a & b\\ c& -a \end{bmatrix} +\begin{bmatrix} a’ & b’\\ c’& -a’ \end{bmatrix}=\begin{bmatrix} a+a’ & b+b’\\ c+c’& -(a+a)’ \end{bmatrix}$ and this is of the form of elements of $W$. Hence $A+A’\in W$ and condition 2 is met.

Finally, let us check condition 3.
Let $\begin{bmatrix} a & b\\ c& -a \end{bmatrix}$ be an arbitrary element in $W$ and let $r\in \R$ be any real number.

Then we have
\begin{align*}
rA=r\cdot \begin{bmatrix}
a & b\\
c& -a
\end{bmatrix}=\begin{bmatrix}
ra & rb\\
rc& -(ra)
\end{bmatrix}
\end{align*}
and this is of the form of the elements of $W$. Thus $rA\in W$ and condition 3 is met.

Therefore, by the subspace criteria, we conclude that $W$ is a subspace of $V$.

### (b) Find a basis of $W$.

Any vector $A=\begin{bmatrix} a & b\\ c& -a \end{bmatrix}$ in the subspace $W$ can be written as
$\begin{bmatrix} a & b\\ c& -a \end{bmatrix}=a\begin{bmatrix} 1 & 0\\ 0& -1 \end{bmatrix}+b\begin{bmatrix} 0 & 1\\ 0& 0 \end{bmatrix}+c\begin{bmatrix} 0 & 0\\ 1& 0 \end{bmatrix}.$ It follows that the matrices
$\begin{bmatrix} 1 & 0\\ 0& -1 \end{bmatrix}, \begin{bmatrix} 0 & 1\\ 0& 0 \end{bmatrix}, \begin{bmatrix} 0 & 0\\ 1& 0 \end{bmatrix}$ span the subspace $W$.

If the above linear combination is the zero matrix, then $a=b=c=0$.
This implies that these matrices are linearly independent.

Thus,
$\left\{\, \begin{bmatrix} 1 & 0\\ 0& -1 \end{bmatrix}, \begin{bmatrix} 0 & 1\\ 0& 0 \end{bmatrix}, \begin{bmatrix} 0 & 0\\ 1& 0 \end{bmatrix}\,\right\}$ is a basis for $W$.

### (c) Find the dimension of $W$.

As the basis we obtained contains three vectors, the dimension of $W$ is $3$.

Close