Basis of a $2 \times 2$ matrix with trace $0$

I have a question that I do not understand and it goes like this:
Find a basis for the set $W$ of all matrices A in $M_{2\times2}$ with trace $0$: i.e. all matrices
$$
\begin{pmatrix}
a & b\\
c & d \
\end{pmatrix}
$$
with $a+d = 0$.
What is the dimension W?

Solutions Collecting From Web of "Basis of a $2 \times 2$ matrix with trace $0$"

So you really have the set of matrices of the form
$$
W = \{\pmatrix{a & b \\ c & -a}\}
$$
I claim that a basis is
$$
e_1 = \pmatrix{1 & 0 \\ 0 & -1}, \\
e_2 = \pmatrix{0 & 1 \\ 0 & 0},\\
e_3 = \pmatrix{0 & 0 \\ 1 & 0}.
$$
All you have to do is to prove that $e_1, e_2, e_3$ span all of $W$ and that they are linearly independent.

I will let you think about the spanning property and show you how to get started with showing that they are linearly independent. Assume that
$$
ae_1 + be_2 + ce_3 = 0.
$$
This means that
$$
\pmatrix{a & b \\ c & -a} = \pmatrix{0 & 0 \\ 0 & 0},
$$
and so $a = b = c = 0$. Hence we have linear independence.

Any required matrix of size 2×2 can be represented as a linear combination of

\begin{align}
\begin{bmatrix}
0&1\\0&0
\end{bmatrix}\\
\begin{bmatrix}
0&0\\1&0
\end{bmatrix}\\
\begin{bmatrix}
1&0\\0&-1
\end{bmatrix}\\
\end{align}
Dimension is 3.

Alternatively, $\operatorname{trace}\begin{pmatrix}a&b\\c&d\end{pmatrix}=0$ if and only if $(a,b,c,d)^T\perp(1,0,0,1)^T$ in $\mathbb{R}^4$.

Maybe it would help to forget the context and focus on the algebraic problem:

Find all solutions for $(a,b,c,d)$ to the linear system of one equation in four variables $a+d=0$. Write down a basis for the solution space. What is its dimension?

If you have trouble dealing with degeneracy (e.g. you think “but $b$ and $c$ aren’t in the equation!”), then as a temporary measure until you become more comfortable with it, it may help to write it as $a + 0b + 0c + d = 0$.