Let A be a square matrix. Prove that there exists a nonzero square matrix B such that AB = 0 if and only if det(A) = 0 Let A be a square matrix. Prove that there exists a nonzero square matrix B such that AB = 0
if and only if det(A) = 0.
Hi all, need help to prove this question for my homework. I am new to matrices and proving. I understand to prove a if and only if, we need to prove left to right side and right to left side.
For right to left, I thought of doing so by contradiction: ie, Suppose det(A) != 0, then A is invertible, so IB = 0, B = 0 which contradicts the statement. Hence det(A) has to be 0.
However, I am stuck from this onwards.
Is my approach correct and if it is wrong, what are the steps that I need to take?
Your reply will be appreciated.
 A: Your approach was somewhat unclear but was, I think, correct. I've rewritten it below.
Suppose $AB=0$, with $B\ne0$.
If $|A|\ne 0$ then $A$ is invertible, let its inverse be $C$ and then $$0=C(AB)=(CA)B=B, $$ a contradiction.
Hence$|A|= 0$.
Suppose $|A|=0$.
Then there is a non-zero vector $v$ such that $Av=0$. Let $B$ be the square matrix with every column equal to $v$.
Then $AB=0$.
A: Suppose $A$ is of order $m\times m, m\ge 1$.
If $AB=[0]$ then $Ab_j=0$ for every column vector $b_j$ of B.
As $B$ is non zero, atleast one of $b_j$'s does not consist of all zeroes .
It follows that there exists a $J: 1\le J\le m$ and  $b_J\ne 0$ is in null (A). Since $A0=0=Ab_J$, it follows that dimension of null space of $A$ is $>1$, which means that $A$ is not full rank (note that rank $A<m$). Hence $A$ is not invertible. That is, rows of $A$ are linearly dependent whence it follows that $|A|=0$.
Conversely, if $|A|=0$ then $A$ is singular and there exists a non zero vector $v\in \mathbb R^m$ such that $Av=0$. Set $B=[v, 0, 0,...,0]_{m\times m}$ and note that $AB=0$.
