# Matrix multiplication proof: $(A+B)^k$ for matrices such that $A^2=B^2=0$ and $AB=BA$

I have the following details about the matrices $A,B$:

$$A_{n\times n},B_{n\times n}$$ $$A^2 = B^2 = 0$$ $$AB=BA$$

I need to find $x \in \mathbb{N}$ so $(A+B)^x=0$

What implications can I make from the given details?

1. $A,B$ doesn't have to equal $0$ so $A^2 = B^2 = 0$
2. Can I imply from the 2nd and 3rd given details that $A=B$?
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Write out $(A+B)^x$ for $x=2$ and $x=3$ and simplify as far as you can. You should notice something nice. You can use the binomial theorem for these powers before simplification (because $AB=BA$). Then try $x=4$ using your result from $x=2$. – bgins Apr 6 '12 at 11:42
Concering 2.: $B$ might be a scaled version of $A=\begin{pmatrix}0& 1\\0 &0\end{pmatrix}$, so no. – draks ... Apr 6 '12 at 11:49

For all $x \ge 3$, The equation $(A+B)^x=0$ is true because for $x =3$, it is true.

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$$\mbox{ let } x > 2\mbox{ and } x\in \mathbb{N}$$ $$\textbf{(A+B)}^2 = \textbf{A}^2+\textbf{B}^2+2\textbf{AB} = \textbf{2AB}$$ $$\textbf{(A+B)}^x = \textbf{(A+B)}^2\textbf{(A+B)}^{x-2}$$ $$\Rightarrow \textbf{(A+B)}^x =2\textbf{AB}\textbf{(A+B)}^{x-2}$$ $$\Rightarrow \textbf{(A+B)}^x =2\textbf{AB}\textbf{(A+B)}\textbf{(A+B)}^{x-3}$$ $$\Rightarrow \textbf{(A+B)}^x =2(\textbf{ABA} + \textbf{ABB})\textbf{(A+B)}^{x-3}$$ $$\Rightarrow \textbf{(A+B)}^x =2(\textbf{AAB} + \textbf{ABB})\textbf{(A+B)}^{x-3}\quad(\because \textbf{BA = AB})$$ $$\Rightarrow \textbf{(A+B)}^x = \textbf{0} (\because\hspace{2pt}\textbf{A}^2 = \textbf{B}^2 = 0)$$ $$\text{If x = 2, I can not show }\textbf{(A+B)}^2 = 0$$ $$\text{ can anyone to prove that or give a counter example}$$

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I have added a counterexample showing that $(A+B)^2=0$ is not true in general. The counterexample uses $4\times4$ matrices. I have also tried to show that a counterexample cannot be found for smaller matrices. – Martin Sleziak Jan 30 '14 at 12:24

In another answer it was already shown that $(A+B)^3=0$ for any matrices fulfilling the given conditions. Let us give an example showing that in general the given conditions do not imply $(A+B)^2=0$.$\newcommand{\Ker}{\operatorname{Ker}}\newcommand{\inv}[1]{{#1}^{-1}}$

Consider the following $4\times 4$ matrices: $$A= \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \qquad B= \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$ We have $$AB=BA= \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$ and $A^2=B^2=0$.

Then we get $$(A+B)^2= \begin{pmatrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}^2= \begin{pmatrix} 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$ and $(A+B)^3=0$. (You may notice that $(A+B)^2=2AB=2BA$ which was derived in bsdshell's answer.)

This counterexample works for matrices over any field which does not have characteristic 2. If $\operatorname{char} F=2$ then we have $(A+B)^2=2AB=2BA=0$.

Now let me show how I found this counterexample. This argument also shows (if I have not missed something), that we cannot find a counterexample for $n<4$.

I will use row vectors. (If you are used to column vectors, you should modify some parts of the proof.) This means that to a $n\times n$ matrix $A$ over a field $F$ corresponds a linear map $f_A \colon F^n \to F^n$ given by $f_A(\vec x)=\vec xA$. The kernel of this map is $\Ker{f_A}=\{\vec x\in F^n; \vec xA=\vec 0\}$. The set $\Ker f_A$ is a subspace of $F^n$ and we can write the whole space as $F^n=S\oplus \Ker f_A$.

The condition $A^2=0$ means that $f_A\circ f_A=0$, i.e., $f_A(f_A(\vec x))=\vec 0$ for any vector $\vec x\in F^n$.

Let $\vec x_1,\dots,\vec x_l$ be any basis of the subspace $S$. Then vectors $f_A(\vec x_1),\dots,f_A(\vec x_l)$ belong to $\Ker f_A$. Moreover, they are linearly independent. (Indeed, $c_1f_A(\vec x_1)+\dots+c_lf_A(\vec x_l)=\vec0$ implies $f_A(c_1\vec x_1+\dots+c_l\vec x_l)=\vec0$ and thus we get $c_1\vec x_1+\dots+c_l\vec x_l\in\Ker f_A\cap S=\{\vec 0\}$. This implies $c_1=\dots=c_l=0$.) This implies that $\dim\Ker f_A\ge l$ and we can choose a basis for $\Ker f_A$ of the form $\vec x_{l+1}=f_A(\vec x_1),\dots,\vec x_{2l}=f_A(\vec x_l),\vec x_{2l+1},\dots,\vec x_n$.

Notice that the matrix of the map $f_A$ with respect to this basis is $$A_1= \begin{pmatrix} 0 & I & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}$$ where the 4 blocks in the upper left corner are all $l\times l$ and the rightmost bottom block is $(n-2l)\times(n-2l)$.

Since the matrix $A_1$ is similar to the matrix $A$, we have $A_1=PA\inv P$ for some invertible matrix $P$. If we choose $B_1=PB\inv P$, then we have $A_1B_1=PAB\inv P=PBA\inv P=B_1A_1$ and $A_1^2=PA^2\inv P=0=PB^2\inv P=B_1^2$. We also have $A+B=\inv P(A_1+B_1)P$ and $(A+B)^2=\inv P(A_1+B_1)^2P$. So the question whether we can find the matrices fulfilling given conditions such that $(A+B)^2\ne0$ is equivalent to the question whether we can find such matrices where $A$ already has a special form $A=A_1= \begin{pmatrix} 0 & I & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}$; i.e., we can work with $A_1$, $B_1$ instead of $A$, $B$.

Now we can divide the matrix $B_1$ into blocks of the same sizes. $$B_1= \begin{pmatrix} B_{11} & B_{12} & B_{13} \\ B_{21} & B_{22} & B_{23} \\ B_{31} & B_{32} & B_{33} \end{pmatrix}$$

Then we get $$A_1B_1= \begin{pmatrix} B_{21} & B_{22} & B_{23} \\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}= \begin{pmatrix} 0 & B_{11} & 0\\ 0 & B_{21} & 0\\ 0 & B_{31} & 0 \end{pmatrix} =B_1A_1$$ which implies that
$$B_{11}=B_{22}\qquad\text{ and }\qquad B_{21}=B_{23}=B_{31}=0,$$
so we have $$B_1= \begin{pmatrix} B_{11} & B_{12} & B_{13} \\ 0 & B_{11} & 0 \\ 0 & B_{32} & B_{33} \end{pmatrix}.$$

We also know that $B_1^2=0$ which implies $B_{11}^2=0$.

If we are working with $2\times2$ or $3\times3$ matrices, then $B_{11}$ must be $1\times1$ and $B_{11}^2=0$ already implies $B_{11}=0$. This, in turn implies $(A_1+B_1)^2=A_1^2+2B_1A_1+B_1^2=2B_1A_1= 2\begin{pmatrix} 0 & B_{11} & 0\\ 0 & B_{21} & 0\\ 0 & B_{31} & 0 \end{pmatrix} =0$. So no counterexample for $n\le3$ can be found.

The counterexample for $n=4$ was constructed simply by taking $B_{22}=B_{11}$ equal to some non-zero $2\times2$ matrix with $B_{11}^2=0$; and all other blocks were chosen to be zero. (In this case $l=2$ and $n=2l=4$.)

NOTE: Originally I had here more complicated proof which only worked for $2\times2$ matrices and used Jordan normal form.

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What I have written above is solution only for $2\times2$ matrices. I guess that similar approach might work for $n\times n$ matrices, but it seems that the proof is becoming somewhat messy. – Martin Sleziak Jan 29 '14 at 13:31
Nice try, Looking for a simple proof. – bsdshell Jan 30 '14 at 1:58
Note: The above comments refer to my first attempt. (I have rewritten my post completely since then.) – Martin Sleziak Jan 30 '14 at 12:41