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Prove/ Disprove: Let $n$ be a positive integer. Let $A$, $B$ be two $n \times n$ square matrices over the complex numbers. If $AB = BA$ and $\ker A = \ker A^2$ and $\ker B = \ker B^2$ then $\ker AB = \ker A + \ker B$.

(Recall that $\ker A$ is the set of all vectors $v$ such that $Av = 0$.)

Background: I am teaching linear algebra this semester. I did not like the standard proof of the Jordan canonical form I found in the textbooks, and thought I could prove it differently, directly from the axioms for a vector space, without using either the determinant, or the classification theorem for finite abelian groups. If the statement above is true, I believe I have a proof for the Jordan canonical form for $T$ by setting $A = (T-\lambda_1I)^{n_1}$ and $B=(T-\lambda_2I)^{n_2}$ for appropriate $n_1$ and $n_2$.

Note 1: If $A = B = \left( \begin{array}{cc} 0 & 1\\\0 & 0 \end{array} \right)$ then $AB = BA$ but $\ker AB \neq \ker A + \ker B$.

Note 2: It is easy to find $A, B$ such that $\ker A = \ker A^2$ and $\ker B = \ker B^2$ and $B$ maps a vector outside $\ker A + \ker B$ to $\ker A$, so that $\ker AB \neq \ker A + \ker B$.

Hence, both conditions are necessary.

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A neat proof of the Jordan canonical form can be found here: matheplanet.com/matheplanet/nuke/html/article.php?sid=1028 (in german). In particular, Satz 1 might interest you. –  Martin Brandenburg Nov 5 '11 at 10:07
    
Thanks, Martin. Satz 1 would certainly give me the kind of proof I am looking for. If I'm not mistaken, it says that: Claim: If g,h are polynomials in one variable whose gcd is 1, then for every endomorphism $\alpha$, the kernel $\ker (gh)(\alpha)$ is a direct sum of $\ker g(\alpha)$ and $\ker h(\alpha)$. Proof: 1. If there is a vector $v$ in the intersection of the kernels of $g(\alpha)$ and $h(\alpha)$, then its annihilator (in the polynomial ring) is trivial, so $v=0$. 2. Proves the easy direction of the inclusion: $\ker g(\alpha) \subseteq \ker (gh) (\alpha)$. 3. $\alpha$ commutes wi –  Manoj Nov 5 '11 at 11:50
    
3. $\alpha$ commutes with $W=\ker (gh)(\alpha)$, so $W$ is invariant under $\alpha$. Let $\gamma$ be the restriction of $\alpha$ to $W$. We have already shown that the two other kernels of interest live inside $W$. 4. Restricted to $W$, we know that $(gh)(\gamma)$ is identically zero. Therefore the image of $h(\gamma)$ must be contained in the kernel of $g(\gamma)$. [The proof seems to get this backwards. Perhaps they are using left multiplication?] 5. But NOW! $dim(\ker h(\gamma)) + dim(im h(\gamma)) = dim W$ by the rank-nullity theorem! 6. Therefore, $dim(\ker h(\gamma)) + dim(\ker g(\ga –  Manoj Nov 5 '11 at 11:51
    
6. Therefore, $dim(\ker h(\gamma)) + dim(\ker g(\gamma)) >= dim W$. But hold on, these kernels intersect trivially, as we already showed. So we're done! –  Manoj Nov 5 '11 at 11:51
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I believe you might be interested in the book Linear Algebra Done Right by Sheldon Axler. He has a similar dislike of determinants. –  Loop Space Nov 5 '11 at 22:34

3 Answers 3

up vote 26 down vote accepted

Since $\ker A = \ker A^2$, the map $\bar{A} : V/\ker A \to V/\ker A$ is injective. Since $V/\ker A$ is finite dimensional, this map is surjective. So for any $x \in V$ we can find $y \in V$ and $z \in \ker A$ such that $x = Ay + z$.

Now suppose $ABx = 0$ and let $x = Ay + z$ as above. Then $0 = ABx = ABAy + ABz = A^2By + BAz = A^2By$ because $AB = BA$ and $Az = 0$.

So $By \in \ker A^2 = \ker A$ whence $ABy = 0$. So $B(Ay) = 0$ and $Ay \in \ker B$.

Hence $x = Ay + z \in \ker B + \ker A$.

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1+. More generally, for an artinian object $V$ in an abelian category and two commuting endomorphisms $A,B$ of $V$ such that $\ker(A^2)=\ker(A)$ and $\ker(B^2)=\ker(B)$ (as subobjects of $V$), then $\ker(AB)=\ker(A) + \ker(B)$ (as subobjects of $V$). –  Martin Brandenburg Nov 5 '11 at 15:10
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+1. Very neat proof. –  Todd Trimble Nov 5 '11 at 15:50
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+1. Why don't all books approach the JCF theorem the way suggested in this post? –  Bill Johnson Nov 5 '11 at 18:00
    
Thanks, Konstantin. It is a very neat proof! I notice that you did not make use of the assumption that $\ker B^2 =\ker B$. –  Manoj Nov 6 '11 at 3:11
    
So you seem to have proved something stronger: If $AB=BA$ and either $\ker A = \ker A^2$ or $\ker B=\ker B^2$ then $\ker AB=\ker A+\ker B$. –  Manoj Nov 6 '11 at 3:13

Here is a proof. First notice that the space of all matrices of size $n$ is finite dimensional. Therefore matrices $A,...,A^{m}$ are linearly dependent for a sufficiently large $m$ (this is instead of Hamilton-Cayley). Hence there exists a polynomial $f(x)$ such that $f(A)=0$, $f(0)=0$. Represent $f(x)$ as $x^kg(x)$ where $g(0)\ne 0$, $k\ge 1$. Dividing by $g(0)$ we can assume that $g(0)=1$. We have $A^kg(A)=0$. Hence $Ag(A)=0$ (since $Ker A= Ker A^2$). Now let $ABx=0$. We have that $(g(A)-1)x=A*h(A)x$ (for some $h$), so $(g(A)-1)x$ is in $Ker B$. Since $g(A)x$ is in the kernel of $A$, we have the decomposition $x=g(A)x-(g(A)-1)x$ where the first summand in $Ker A$, the second in $Ker B$.

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Thanks, Mark, quite a nice proof. –  Manoj Nov 6 '11 at 3:21

Here's another statement of more or less the same result. The ideas in the proof are from this proof in German that Martin Brandenburg linked to in his comment.

Claim: Let $n$ be a non-negative integer. Let $A$, $B$ be two $n×n$ square matrices over the complex numbers. If $AB=BA$ and $\ker A \cap \ker B = \{0\}$ then $\ker AB=\ker A \bigoplus \ker B$.

Note: Assuming that $\ker A\cap\ker B=\{0\}$ is not a big restriction, since we can always quotient out to eventually reduce to this case.

Proof: Since $A,B$ commute, it is clear that $\ker A \bigoplus \ker B\subseteq \ker AB$. Further, $\ker AB$ is invariant under $A, B$. Since all the action is taking place within $\ker AB$, we may assume without loss of generality that $\ker AB$ is the entire space, of dimension $n$. This implies $\operatorname{im} B\subseteq \ker A$.

By the rank-nullity theorem, $\dim\ker B + \dim\operatorname{im} B = n$. Hence, $\dim\ker A + \dim\ker B\geq n$. Since these spaces intersect trivially by assumption, we are done.

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