# Basis for a subspace of matrix $2\times2$with sum 0 entries.

hamel-basismatricesvector-spaces

Consider the subspace of matrices of the form $$\left(\begin{matrix} a & b\\ c & d \end{matrix}\right)$$ such that $$a+b+c+d=0$$ I have to find a basis. I was thinking for example of letting $$3$$ variables free and then the fourth will be the negative linear combination.

$$\left\lbrace\left(\begin{matrix} 0 & b\\ c & d \end{matrix}\right),\left(\begin{matrix} -b-c-d & 0\\ 0 & 0 \end{matrix}\right):b,c,d\in\mathbb{R}\right\rbrace$$
Is that enough? In addition for here the dimension is $$2$$ right?

I have another idea of putting
$$\left\lbrace\left(\begin{matrix} 0 & b\\ 0 & 0 \end{matrix}\right), \left(\begin{matrix} 0 & 0\\ c & 0 \end{matrix}\right), \left(\begin{matrix} 0 & 0\\ 0 & d \end{matrix}\right), \left(\begin{matrix} -b-c-d & 0\\ 0 & 0 \end{matrix}\right):b,c,d\in\mathbb{R}\right\rbrace$$ all of these matrices are linearly independent so the dimension is $$4$$? I have a confusion here

Here's a basis:$$\left\{\begin{bmatrix}1&0\\0&-1\end{bmatrix},\begin{bmatrix}0&1\\0&-1\end{bmatrix},\begin{bmatrix}0&0\\1&-1\end{bmatrix}\right\}.\tag1$$It is clear thet each of these matrices belong to your space and that they span the whole space: if $$a+b+c+d=0$$, then $$d=-a-b-c$$, and therefore\begin{align}\begin{bmatrix}a&b\\c&d\end{bmatrix}&=\begin{bmatrix}a&b\\c&-a-b-c\end{bmatrix}\\&=a\begin{bmatrix}1&0\\0&-1\end{bmatrix}+b\begin{bmatrix}0&1\\0&-1\end{bmatrix}+c\begin{bmatrix}0&0\\1&-1\end{bmatrix}.\end{align}Finally, the set $$(1)$$ is linearly independent. So, it's a basis. Since it has $$3$$ elements, your space is $$3$$-dimensional.