How do I find a matrix $B\neq 0$ so $AB = BA = 0$ Sorry if it's relatively easy, I just have no idea what to do here:
$A = \begin{pmatrix}
1 & 2 &3 \\ 
4 & 5 & 6\\ 
7 & 8 & 9
\end{pmatrix}$
how to find  $B\neq 0$ so $AB = BA = 0$
I only know if I need to find it, then A is linear dependant if that is useful for it somehow.
I probably can build a $3x3$ matrix with variables and solve a huge equation system but I don't think it's what I'm supposed to do here and there must be a smarter solution.
 A: Find non-zero vectors $v$ and $w$ such that 
$$v^TA=0=Aw$$
Then use $B=wv^T$.
[
This may seem like magic. Here is how I came up with it. 
I first thought of diagonalizing $A$. Write $A=SDS^{-1}$ where $D$ is a diagonal matrix. Then we find a diagonal $B_0$ such that $DB_0=B_0D=0$. Then we let $B=SB_0S^{-1}$.
From there, I saw this easier formulation, because the columns of $S$ are the right eigenvectors of $A$ and the rows of $S^{-1}$ are the left eigenvectors of $A$.
]
A: If $C_i$ is the i'th column of A, notice that 
$$-C_1 + 2C_2 = C_3$$
Meaning that
$$ \alpha C_1 - 2\alpha C_2 + \alpha C_3 = \vec{0} $$
Meaning that
$$ A\cdot \begin{pmatrix} \alpha \\ -2\alpha \\ \alpha \end{pmatrix} = \vec{0} $$
Meaning the columns of $B$ should be in this form. What about the rows?
$$-R_1 + 2R_2 = R_3 $$
So 
$$(\alpha , -2\alpha , \alpha) A = \vec{0}^T $$
Giving information about B's rows. 
We can conclude that
$$ B = \begin{pmatrix}
\alpha & -2\alpha & \alpha \\
-2\alpha & 4\alpha & -2\alpha\\
\alpha & -2\alpha & \alpha \end{pmatrix} $$
A: For $AB=O$ you need the three columns of $B$ to be (right)-eigenvectors of $A$ for the eigenvalue $0$. So [crunch, crunch] they are all multiples of $(1,-2,1)^T$. 
For $BA=O$ you need the three rows of $B$ to be (left)-eigenvectors of $A$ for the eigenvalue $0$. So [crunch, crunch] they are all multiples of  $(1,-2,1)$.
If we assume that $b_{11}=t$ we'll get:
$$
B=t
\left[
\begin{array}{ccc}
1 & -2 & 1\\
-2 & 4 & -2\\
1 & -2 & 1\\
\end{array}
\right]
$$
where we need to have $t\not=0$ to ensure $B\not=O$.
A: You may use the facts that $A$ has rank deficiency $1$ (so that $\operatorname{adj}(A)\ne0$) and $A\operatorname{adj}(A)=\operatorname{adj}(A)A=\det(A)I$.
