Find $2\times 2$ matrices $A$ and $B$ such that $AB=0$ but $BA$ does not equals to $0$ Find $2\times 2$ matrices $A$ and $B$ such that $AB=0$ but $BA$ does not equals to $0$ (please show working and the concepts used) Thanks :)
 A: There is not a "canonical answer" to this question: by that I mean that there are lots of pairs $A,B$ which will answer the question, and even two experts in the field might not arrive at the same answer.  Let me explain how to get started.
In theory we could write $A = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]$ and $B = \left[ \begin{array}{Cc} e & f \\ g & h \end{array} \right]$, set $AB = 0$, and then try to choose values for $a,b,c,d,e,f,g,h$ which make $BA$ something nonzero.  If you actually try to do this though you find that the equations are rather unwieldy.  E.g. that $AB = 0$ means
$ae + bg = af + bh = ce + dg = cf + dh = 0$.
This is a system of equations, but not linear equations: quadratic equations.  Linear algebra does not teach us how to solve such equations in a systematic way (rather, solving systems of quadratic equations in several variables is as hard as solving polynomial equations of any degree, which is a deep subject, algebraic geometry).  
To proceed, we should probably set some of the variables equal to $0$: enough to make the equations simple but not so many so that $BA = 0$.  Here is a suggestion for how to do this: take advantage of the fact that if $A$ has a row of zeros, then the same row of $AB$ must be zero, but $BA$ is not forced to have zero entries.  Similarly, if $B$ has a column of zeros, then the same column of $AB$ must be zero, but $BA$ is not forced to have zero entries.  This sugests that $A$ should have at least one row (and clearly not two rows, so maybe one further zero entry?) and $B$ should have at least one column (and....).  From this point there is still trial and error, but it is much easier and motivated by some ideas that can be used for other problems as well.
A: It is more or less "trial and error", but you know that at least one of the matrices A or B must have an eigenvalue equals to 0, so you could try your luck with
$$\begin{bmatrix}0&1\\0&0\end{bmatrix},\begin{bmatrix}1&1\\0&0\end{bmatrix},\begin{bmatrix}0&1\\0&1\end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix},\dots$$
For example you could take $A=\begin{bmatrix}0&0\\0&1\end{bmatrix}$ and $B=\begin{bmatrix}0&1\\0&0\end{bmatrix}$
Then  $AB= 0$ but $BA\neq0$.
A: I'll outline an approach using "rank-1 matrices". Getting comfortable working with such matrices might take some effort, but in the long run it would also be quite worthwhile.  With practise, all of the steps below can become part of your internal thought process, removing the need for any calculation.  

Let's narrow down the possibilities:


*

*We can't have $A$ invertible or else $BA = A^{-1} (AB) A = 0$, contrary to assumption. For essentially the same reason, we can't have $B$ invertible. I see this immediately because I know that $AB$ and $BA$ are "similar" when one of them is invertible.

*We can't have $A = 0$ or $B=0$ since, in both cases, we get $BA=0$, contrary to assumption.


To paraphrase the above, $rank(A) \neq 0$ or $2$. Likewise, $rank(B) \neq 0$ or $2$. The only remaining possibility is
$$rank(A) = rank(B) = 1.$$
But, rank 1 matrices are pretty easy to describe: they just look like $xy^T$ where $x$ and $y$ are (nonzero) $2 \times 1$ column vectors. For fun, you might want to check that the null space of $xy^T$ is the orthocomplement of $y$. The column space of $xy^T$ is the line through $x$. 
Anyway, let's say
\begin{align*}
 A = x y^T  && B = z w^T
\end{align*}
Then, 
$$ AB = x y^T z w^T.$$
But,  $y^Tz$ is really just the number $y \cdot z$, the dot product of $y$ and $z$. So, 
$$AB = x (y \cdot z) w^T = (y \cdot z)  \underbrace{x w^T}_\text{rank-1}.$$
Similarly, 
$$BA = (x \cdot w) \underbrace{z y^T}_\text{rank-1}$$
So, pretty much, we just need to make $y \cdot z = 0$, and $x \cdot w \neq 0$ and we'll be fine.  A simple way to procceed here is to take
\begin{align*} x = y = w =\begin{bmatrix} 1 \\ 0 \end{bmatrix} && z = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \end{align*}
This leads to 
\begin{align*}
A = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} && B = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \end{align*}
