# Finding a basis for all $2\times2$ matrices A such that…

Finding a basis for all $2\times2$ matrices A such that

$$\left[ \begin{matrix} 1 \hspace{5pt} 2 \\ 0 \hspace{5pt} 3 \end{matrix} \right]A=\left[ \begin{matrix} 0 \hspace{5pt} 0\\ 0 \hspace{5pt} 0 \end{matrix} \right]$$

My process:

Since $A$ is all $2\times2$ matrices, I computed $\left[ \begin{matrix} 1 \hspace{5pt} 2 \\ 0 \hspace{5pt} 3 \end{matrix} \right]\left[ \begin{matrix} a \hspace{5pt} b \\ c \hspace{5pt} d \end{matrix} \right]$. This gave me $\left[ \begin{matrix} a+2c \hspace{5pt} b+2d \\ 3c \hspace{5pt} 3d \end{matrix} \right]$.

So since each slot in the matrix must equal $0$ I got this system of equations:

$a+2c = 0 \\b+2d = 0 \\3c=0\\3d=0$ $\hspace{5pt}$ $\rightarrow$ $\hspace{5pt}$ $a=-2c\\b=-2d\\c=0\\d=0$ $\hspace{5pt}$ $\rightarrow$ $\hspace{5pt}$ $a=0\\b=0\\c=0\\d=0$

So Iis the only solution for $A$ in $\left[ \begin{matrix} 1 \hspace{5pt} 2 \\ 0 \hspace{5pt} 3 \end{matrix} \right]A=\left[ \begin{matrix} 0 \hspace{5pt} 0\\ 0 \hspace{5pt} 0 \end{matrix} \right]$ being $A=\left[ \begin{matrix} 0 \hspace{5pt} 0\\ 0 \hspace{5pt} 0 \end{matrix} \right]$ meaning there's no basis? Since $\left[ \begin{matrix} 0 \hspace{5pt} 0\\ 0 \hspace{5pt} 0 \end{matrix} \right]$ isn't linearly independent?

Am I facing the problem correctly? Thanks for all the help in advance.

By Sylvester's rank inequality we have that: $$rank\left(\left[ \begin{matrix} 1 \hspace{5pt} 2 \\ 0 \hspace{5pt} 3 \end{matrix} \right] A\right) \ge rank\left(\left[ \begin{matrix} 1 \hspace{5pt} 2 \\ 0 \hspace{5pt} 3 \end{matrix} \right]\right) + rank(A) - 2 = rank(A)$$

On the other hand we have that $rank(\left[ \begin{matrix} 0 \hspace{5pt} 0\\ 0 \hspace{5pt} 0 \end{matrix} \right]) = 0$.

Therefore, $rank(A) = 0$, so $A = \left[ \begin{matrix} 0 \hspace{5pt} 0\\ 0 \hspace{5pt} 0 \end{matrix} \right]$.

So you are correct that there is no basis.

• Thanks a lot for the help! – Kenshin Mar 11 '15 at 2:23

In my opinion, you solved the problem correctly. Since the matrix $B=\begin{bmatrix} 1 & 2\\0 & 3\end{bmatrix}$ is not singular and its inverse matrix is denoted as $B^{-1}$, from the equation($BA={\bf 0}$) \begin{gather} \begin{bmatrix} 1 & 2\\0 & 3 \end{bmatrix}A= \begin{bmatrix} 0 & 0\\0 & 0 \end{bmatrix} \end{gather} then we can get $A=B^{-1}\times {\bf 0}={\bf 0}$, i.e., $$A=B^{-1}\times\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$$ Hope this can help you.

Determinant of $\pmatrix{1 &2\cr 0&3\cr}$ is $3$, hence it is non-singular; call this matrix $B$. So we have $A$ is $B^{-1}$ times the zero matirx hence $A$ is the zero matrix.