Finding the basis of $U=\{ A\in M_{2\times 3}(\mathbb R) |a_{11}+a_{21}+a_{22}+a_{23}=0 \}$

Question

Find the basis of $U=\{ A\in M_{2\times 3}(\mathbb R) |a_{11}+a_{21}+a_{22}+a_{23}=0 \}$

So we set $a_{11}=-a_{21}-a_{22}-a_{23}$

Then why is this a basis ?

I get that its dimension should be 5 but I don't understand how to they got to this basis.

Answer 1

The condition for our set $U$ is satisfied, as you note, if and only if $a_{11}=-a_{21}-a_{22}-a_{23}$. That means that we can write any such matrix as $$ \begin{pmatrix}-a_{21}-a_{22}-a_{23} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\end{pmatrix} $$ for any choice of the $a_{ij}$.

Now, by linearity, we can write this matrix as $$ \begin{pmatrix}0 & a_{12} & 0\\0 & 0 & 0\end{pmatrix}+\begin{pmatrix}0 & 0 & a_{13}\\0 & 0 & 0\end{pmatrix}+\begin{pmatrix}-a_{21} & 0 & 0\\a_{21} & 0 & 0\end{pmatrix}+\begin{pmatrix}-a_{22} & 0 & 0\\0 & a_{22} & 0\end{pmatrix}+\begin{pmatrix}-a_{23} & 0 & 0\\0 & 0 & a_{23}\end{pmatrix}, $$ separating out the contributions by each (independent!) variable.

From here, we note that we can factor out scalars; this leaves us with writing any such matrix as $$ a_{12}\begin{pmatrix}0 & 1 & 0\\0 & 0 & 0\end{pmatrix}+a_{13}\begin{pmatrix}0 & 0 & 1\\0 & 0 & 0\end{pmatrix}+a_{21}\begin{pmatrix}-1 & 0 & 0\\1 & 0 & 0\end{pmatrix}+a_{22}\begin{pmatrix}-1 & 0 & 0\\0 & 1 & 0\end{pmatrix}+a_{23}\begin{pmatrix}-1 & 0 & 0\\0 & 0 & 1\end{pmatrix} $$ So, this shows you that a matrix is in your set if and only if it is a linear combination of these five constant matrices, which also happen to be linearly independent. Thus, they form a basis.

But wait -- that's different than the answer you were given! Well, that's alright, as bases aren't by any means unique. You can get to the answer you were given by solving $a_{11}+a_{21}+a_{22}+a_{23}=0$ for $a_{23}$ instead of $a_{11}$, then following the same process.