# 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)*

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## 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.

- The zero vector in $V$ is in $W$.
- For all $A, B\in W$, the sum $A+B \in W$.
- 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 $3$.

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