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If we have a non-zero matrix $A\in\mathrm{Mat}_n(\mathbb{R})$ that is non-invertible. How do we prove that $A$ is both a left and right zero divisor?

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    $\begingroup$ Use the characteristic polynomial of A $\endgroup$
    – user261263
    Commented Sep 9, 2015 at 17:03
  • $\begingroup$ @Eugen: You should make it an answer. $\endgroup$
    – PseudoNeo
    Commented Sep 9, 2015 at 17:06
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    $\begingroup$ @PseudoNeo .. too lazy for that, please do it $\endgroup$
    – user261263
    Commented Sep 9, 2015 at 17:08

2 Answers 2

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If $A$ is non-invertible, then it sends at least one non-zero vector $v_1$ in $\mathbb{R}^n$ to $0$. Extend $v_1$ to a basis $\{v_1,\ldots,v_n\}$ for $\mathbb{R}^n$, and consider the projection operator $P_1(\sum_i \alpha_i v_i) = \alpha_1v_1$ as a nonzero matrix in $\mathbb{R}^n$. It follows that $AP_1v = 0$ for all $v$, and so $AP_1 = 0$ as a matrix.

Since $A$ is non-invertible if and only if its transpose $A^t$ is non-invertible, by the above argument you can find a nonzero matrix $P$ such that $A^tP = 0$. Then $P^tA = (A^tP)^t = 0^t = 0$.

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    $\begingroup$ Note to the OP: It would be a nice exercise to try and show that $A$ is a right divisor of $0$, not by using the (neat) transpose trick of Dan's, but to try to reproduce the construction he gave for the left divisor part. $\endgroup$
    – PseudoNeo
    Commented Sep 9, 2015 at 17:28
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    $\begingroup$ @PseudoNeo: I was thinking the same thing after posting this; I'll suggest it as an edit. :) Also, thanks for posting the complementary algebraic proof using the minimal polynomial of $A$. Cheers! $\endgroup$
    – Dan
    Commented Sep 9, 2015 at 18:28
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Even if I like a lot Dan's geometric take on the question, one can approach it in a more algebraic way.

Every matrix $A \in M_n(\mathbb R)$ has a minimal polynomial: that's a polynomial $\mu(X) = X^r + c_{r-1} X^{r-1} + \cdots + c_1 X + c_0$ such that $$\mu(A) = A^r + c_{r-1} A^{r-1} + \cdots + c_1 A + c_0 = 0$$ and whose degree is minimal among polynomials with this property.

Besides, the roots of $\mu$ are precisely the (complex) eigenvalues of $P$, so the noninvertibility of $A$ is equivalent to the vanishing of the constant coefficient $c_0$ of $\mu$.

So, if $A$ isn't invertible, the defining relation given by Cayley-Hamilton theorem takes the form $$A^r + c_{r-1} A^{r-1} + \cdots + c_1 A = 0,$$ which you can rewrite as $$A \left(A^{r-1} + c_{r-1} A^{r-2} + \cdots + c_1\right) = \left(A^{r-1} + c_{r-1} A^{r-2} + \cdots + c_1\right) A = 0.$$

Because of the minimality of $\mu$, the factor $A^{r-1} + c_{r-1} A^{r-2} + \cdots + c_1$ is non zero, so this expression proves that $A$ is both a left and a right divisor of $0$.

(Dispensable) comments on the relative merits of the geometric and algebraic proofs. Of course, using the projector as in Dan's answer is shorter, more low-tech and probably clearer. So why bother with this apparently contrived answer? Well, perhaps we shouldn't, but it's nice to notice that we have proven that not only $A$ divides $0$, which means that there is a non-zero $B$ such that $AB = 0$, but we have found such a $B$ who commutes with $A$ (so, in particular, $BA = 0$). Somehow our $B$ seems a bit more "natural" than the projector of Dan's answer (which depends moreover of the choice of basis ${v_1, \ldots, v_n}$).

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