Is every noninvertible matrix over a field a zero divisor?

Related to this: What are sufficient conditions for a matrix to be a zero divisor over a noncommutative ring?


New and improved: If $A$ is singular we can get $AB=BA=0$ with no more work.

Original Yes. If $A$ is a singular square matrix then there exists a non-zero vector $v$ with $Av=0$. So if $B$ is the square matrix that has every column equal to $v$ then $AB=0$.

Better There is a non-zero "column vector" $v$ with $Av=0$. There is also a non-zero "row vector" $w$ with $wA=0$. Let $B=(b_{jk})$, where $b_{jk}=v_jw_k$. Then every row of $B$ is a multiple of $w$ and every column of $B$ is a multiple of $v$, so $BA=AB=0$.

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    $\begingroup$ I was writing this but you were a little faster. $\endgroup$ – Jack Yoon Jul 15 '15 at 17:05
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    $\begingroup$ @JackYoon Yeah, you gotta be quick. If I see a question's already been answered I won't post an answer unless I feel I have something significant to add that hasn't been said. But if new answers appear while I'm typing my criteria are a little looser - I'll go ahead and post if I just like the way I said it better than the way they said it. Those guys have no business posting while I'm typing, heh-heh... $\endgroup$ – David C. Ullrich Jul 15 '15 at 17:11
  • $\begingroup$ And it is also easy to see that a left-factor $C$ with $CA=O$ exists exactly if this is the case. $\endgroup$ – Jeppe Stig Nielsen Jul 16 '15 at 8:44

Let $p(X) = a_n X^n + \cdots + a_1 X + a_0$ denote the minimal polynomial of $A$. The constant coefficient $a_0$ is, up to sign, the product of the distinct eigenvalues of $A$, and thus vanishes. Hence $p(A) = A(a_n A^{n-1} + \cdots + a_1) = 0$, and the polynomial $a_n A^{n-1} + \cdots + a_1\not = 0$ by the fact that $p$ is minimal.


If $A$ isn't invertible, its rows are linearly dependent, so you can find a column vector $v$ such that $Av=0$. You fill out $v$ into a square matrix (with zeros even, if you like) and you've shown $A$ is a zero divisor.

It's pretty trivial to see, then, that a matrix over $A$ any ring $R$ is a zero divisor iff there exists a nonzero vector $v$ over $R$ (with the right length) such that $Av=0$.

There are probably not many things you can say about the general case that are better than this.

  • $\begingroup$ By “trivial” do you mean a zerodivisor? $\endgroup$ – Dap Dec 21 '17 at 4:16
  • $\begingroup$ @Dap yes, corrected ! Thanks $\endgroup$ – rschwieb Dec 21 '17 at 11:27

Yes. If $A$ is any (non-zero) singular matrix, then there will be some (non-trivial) linear subspace $S$ so that $Av = 0$ for any $v \in S$. There will be a (non-zero) projection $P$ onto this sub-space (given an inner product, you can use the orthogonal projection), that is a matrix $B$ so that $Bv \in S$ for all $v$. Then the matrix $AB = 0$.


I want to reproof the statement of @DavidC.Ullrich with the method of @anomaly and add a supplement.

Theorem: Let $A$ be a square matrix with entries in a field $\mathbb K$ and $\det(A)=0$. Then there exists a square non-zero matrix $B$ with $AB=0=BA$.

Supplement: $B$ can be taking from $\mathbb K[A]$.


Let $p(X)$ be the minimal polynomial of $A$. Then we can write $p(X)$ as $q(X)X$ because the constant coefficient of $p$ vanishes since it is $\det(A)$.

Taking $B:=q(A)$ (which is non-zero because $p$ is minimal) gives us



Yes this is true for another reason not already mentioned. If $A$ is a singular matrix of rank $1\le r < n$, then up to changes of basis in both domain and codomain, we have $$ A=\begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix}, $$ Where $I_r$ is the $r\times r$ identity matrix. Put $$ B=\begin{pmatrix} 0 & 0 \\ 0 & I_{n-r} \end{pmatrix}. $$ Then $AB = BA = 0$ but neither $A$ nor $B$ is $0$.


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