If $A$ is a $n$ order matrix and $|A|=0$, then there always exists a non-zero matrix B of order $n$ such that $AB=0$. Is it true or false

Question

If $A$ is a $n$ order matrix and $|A|=0$, then there always exists a non-zero matrix B of order $n$ such that $AB=0$. Is it true or false. If true then prove it.

I am not getting any idea for how to solve this problem, any hints or suggestions?

where $|A|$ means determinant of A

Answer 1

$|A|=0$ implies there exists a vector $v$ such that $Av=0$. Let $B$ have $v$ as one of its columns, and let the rest of its columns be $0$.

Answer 2

I am saying the same thing as in the perfect answer by angryavian, but in different words.

Every matrix is a linear transformation, a special kind of function, $T\colon V\to W$ between vector spaces. The determinat is zero means some non-zero vector $v\in V$ in the domain of that function is sent to the zero vector.

Now define another linear transformation $T'$ (a function), on any vector space as its domain whose range contains only $v$ and its scalar multiples. This is easily arranged as below. Choose $U$ to be an arbitrary vector space and take a basis of $U$. Define $T\colon U\to V$ by sending every vector of the basis to that $v$.

Now matrix multiplication $AB$ corresponds to the function composition $T\circ T'$ and by way of our construction this composite function is the identically-zero function which translates to the product matrix $AB$ being the zero matrix.