# For a square matrix $A$, if $AB = 0$ for some nonzero $B$ then $CA = 0$ for a nonzero matrix $C$.

I've been asked to prove that the following statements are equivalent for any $$n$$x$$n$$ matrix $$A$$: (Note that A and B both are $$n$$x$$n$$)

1. $$AB = 0$$ for some non-zero $$B$$
2. $$CA = 0$$ for some non-zero $$C$$

How do I show this equivalence? I'm able to see that if $$(1)$$ holds, the columns of A are linearly dependent and the rows of B are linearly dependent. How do I go ahead?

Thanks!

P.S. The course I'm currently doing, however, has only covered concepts of matrix multiplication, elementary matrices, and system of equations - so it'd be great if you could provide a proof along those lines!

Also, I was wondering if there's a general closed-form or some way we can describe matrices A that satisfy $$(1)$$ and $$(2)$$?

• Do you know about the rank of a matrix? – Angina Seng Sep 4 '20 at 4:31
• Yes, I do! The course I'm currently doing, however, has only covered concepts of matrix multiplication, elementary matrices and system of equations - so it'd be great if you could provide a proof along those lines! – epsilon-emperor Sep 4 '20 at 4:34
• it is false see yutsumura.com/if-the-matrix-product-ab0-then-is-ba0-as-well – Story123 Sep 4 '20 at 4:51
• Are these two $B$ in 1 and 2 must be the same? If so, the claim is not true. – Zhanxiong Sep 4 '20 at 4:52
• @Zhanxiong They are different. – epsilon-emperor Sep 4 '20 at 4:54

If $$AB = 0$$ for some nonzero $$B$$, then $$Ax = 0$$ for some nonzero $$x$$ (in particular, one of the nonzero columns of $$B$$). But since this implies $$A$$ has linearly dependent columns, and $$A$$ is square, it means that $$A^T$$ also has linearly dependent columns, and so this implies that $$A^Ty = 0$$ for some nonzero $$y$$. Now consider $$C^T = [y \mid 0 \mid \cdots \mid 0]$$. Then $$A^TC^T = 0$$. Taking transposes, $$CA = 0$$, as required.

Note that this shows both directions: if $$CA = 0$$ for some nonzero $$C$$, then $$A^T B = 0$$ for $$B = C^T$$. Now apply the first direction again, which says that there exists $$D$$ nonzero such that $$DA^T = 0$$. Transpose this, $$AD^T = 0$$, which proves the other direction.

A closed form way to describe such matrices is as the set of singular $$n \times n$$ matrices: $$\{A : \det(A) = 0\}$$

• Can you do this without invoking the concept of rank? – epsilon-emperor Sep 4 '20 at 5:19
• Well, I guess the key idea is that if $A$ is square and has linearly dependent columns, then it necessarily has linearly dependent rows, which implies for example that $A^T$ also has linearly dependent columns. This is clear using the concept of rank, but you can also check it by hand. – Drew Brady Sep 4 '20 at 5:23
• Just use the distributive property and write a matrix $B$ as a sum of matrices $B_j=\begin{bmatrix}0&\ldots &0&x_{1j}&0&\ldots&0\\0&\ldots&0&x_{2j}&0&\ldots &0\\\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\0&\ldots&0&x_{(n-1)j}&0&\ldots&0\\0&\ldots&0&x_{nj}&0&\ldots&0\end{bmatrix}$ – Invisible Jan 13 at 20:34

If there exists $$B \neq 0$$ such that $$AB = 0$$, then this implies $$\mathrm{rank}(A) < n$$, for otherwise $$A$$ is nonsingular, forces $$AX = 0$$ only admits zero solution. Hence the columns of $$A$$ are linearly dependent, i.e., there exists a nonzero row vector $$c$$ such that $$cA = 0$$. Vertically expand $$c$$ (just adding zeros, for example) to an order $$n$$ matrix $$C$$ completes the proof.

• The two B's should be different, I apologize. I've made edits accordingly, I hope they clarify it! – epsilon-emperor Sep 4 '20 at 4:54
• @cogito_ai OK. I edited my answer. By the way, do not use $B'$ in your question to represent a different matrix from $B$, for in some textbook $B'$ means the transpose of $B$. – Zhanxiong Sep 4 '20 at 5:04

Let's not use rank, determinants or linear spaces.
If somehow, we can construct a $$C$$ such that $$CA=0$$, then we are done.

$$AB=0\implies$$ Columns of $$A$$ are linearly dependent.
Convert $$A$$ to Row reduced echelon form. Note that $$A$$ and $$A^T$$ will have same no. of pivots (leading entries). Let the no. of pivots =$$m\lt n$$

Therefore, columns of $$A^T$$ will also be linearly dependent. Hence, there exist nos. $$c_1,c_2,\cdots,c_n$$ (not all zeros) such that $$A^T\begin{bmatrix}c_1\\c_2\\...\\...\\c_n\end{bmatrix}=0$$
Let $$P$$ be a matrix whose one of the columns is $$\begin{bmatrix}c_1\\c_2\\...\\...\\c_n\end{bmatrix}$$ and rest of the columns are zero columns.
Hence, we have $$A^TP=0\implies (A^TP)^T=0^T\implies P^TA=0$$
Put $$P^T=C$$ and we are done.