Vector Space of 2 by 2 Traceless Matrices

Ohio State University exam problems and solutions in mathematics

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.

  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 $3$.


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