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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 $(2)$?

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  • $\begingroup$ Do you know about the rank of a matrix? $\endgroup$ – Angina Seng Sep 4 '20 at 4:31
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    $\begingroup$ 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! $\endgroup$ – epsilon-emperor Sep 4 '20 at 4:34
  • $\begingroup$ it is false see yutsumura.com/if-the-matrix-product-ab0-then-is-ba0-as-well $\endgroup$ – Story123 Sep 4 '20 at 4:51
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    $\begingroup$ Are these two $B$ in 1 and 2 must be the same? If so, the claim is not true. $\endgroup$ – Zhanxiong Sep 4 '20 at 4:52
  • $\begingroup$ @Zhanxiong They are different. $\endgroup$ – epsilon-emperor Sep 4 '20 at 4:54
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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: $\{A : \det(A) = 0\}$

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  • $\begingroup$ Can you do this without invoking the concept of rank? $\endgroup$ – epsilon-emperor Sep 4 '20 at 5:19
  • $\begingroup$ 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. $\endgroup$ – Drew Brady Sep 4 '20 at 5:23
  • $\begingroup$ Just use the distributive property and write a matrix $B$ as a sum of matrices $B_j=\begin{bmatrix}0&\ldots &0&x_{1j}&0&\ldots&0\\0&\ldots&0&x_{2j}&0&\ldots &0\\\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\0&\ldots&0&x_{(n-1)j}&0&\ldots&0\\0&\ldots&0&x_{nj}&0&\ldots&0\end{bmatrix}$ $\endgroup$ – Invisible Jan 13 at 20:34
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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.

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  • $\begingroup$ The two B's should be different, I apologize. I've made edits accordingly, I hope they clarify it! $\endgroup$ – epsilon-emperor Sep 4 '20 at 4:54
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    $\begingroup$ @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 $B$. $\endgroup$ – Zhanxiong Sep 4 '20 at 5:04
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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.

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