23
$\begingroup$

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?

$\endgroup$
35
$\begingroup$

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 $BA=AB=0$.

$\endgroup$
  • 2
    $\begingroup$ I was writing this but you were a little faster. $\endgroup$ – Jack Yoon Jul 15 '15 at 17:05
  • 3
    $\begingroup$ @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
14
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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 $AB = 0$.

$\endgroup$
2
$\begingroup$

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 $\mathbb K[A]$.

Proof:

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

$$AB=BA=p(A)=0.$$

$\endgroup$
1
$\begingroup$

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 $0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.