Finding matrix for $A\left(I+A\right)=0$

Question

A is square matrix and $A\left(I+A\right)=0$. Then, obviously, A may be $0$ or $-I$.

I know that having $XY=0$ doesn't imply that $X=0$ or $Y=0$, though it may be the case. I suppose there are others matrices. But how to find them? Or how to prove that there is no matrix left?

Do you know the dimensions of $A$? Because there are finitely many possibilities (up to similarity) among $n\times n$ matrices for fixed $n$, but infinitely many total. — Alex Becker, Jun 28 '13 at 14:24
If you expand the product, you get $A^2 = -A$. What does that say about the eigenvalues of $A$? — Daniel Fischer, Jun 28 '13 at 14:24
Right. So a simple example of another such $A$ is a diagonal matrix with diagonal entries $-1$ or $0$, and at least one of both values. — Daniel Fischer, Jun 28 '13 at 14:34
You might also find this question useful: http://math.stackexchange.com/questions/379076/matrices-a-of-order-2-times-2-that-satisfy-the-equation-a2-5a6i-o/379173#379173 — Graham Hesketh, Jun 28 '13 at 17:33

MichalisN · Answer 1 · 2013-06-28T14:47:51.840

If the coefficients of the matrix are in $\mathbb{C}$ you can choose a basis so that $A$ is in Jordan normal form. Then automatically $1+A$ is in this form and the problem reduces to showing how the Jordan blocks must look. Let $J$ be such a block, i.e. a square matrix of dimension $n$ with $\lambda$ on the diagonal and $1$'s on the superdiagonal (the coefficients right above the diagonal). If $n>2$ multiplying $J$ with $J+1$ the entry in the first row and third column will always be $1$, so the result can't be zero. If $n=2$ we get $$ \begin{pmatrix}\lambda(\lambda+1)&&2\lambda+1\\0&&\lambda(\lambda+1)\end{pmatrix} $$ So again the result is never zero. We conclude that all the Jordan blocks have dimension $1$, so $A$ is diagonal and the entries on the diagonal can be either $0$ or $-1$.

Remark: A more elegant way of proving the above statement is realising that generalised eigenvectors of $A$ correspond to normal eigenvectors. A generalised eigenvector of $A$ is in the kernel of $A^n$ or $(A+I)^n$, since $A^2=-A$ implies that $A$ can have only $0$ or $-1$ as eigenvalues. But $\text{ker}(A^n)=\text{ker}(A)$ and, since $(A+I)^2=A^2+2A+I=A+I$ we have $(A+I)^n=A+I$, which implies $\text{ker}((A+I)^n)=\text{ker}(A+I)$.

score 0 · Answer 2 · answered Jun 28 '13 at 16:25

This question can be answered easily if you know the concept of minimal polynomial. Call the underlying field $\mathbb{F}$. As $A(A+I)=0$, the minimal polynomial $m(x)$ of $A$ divides $x(x+1)$. Since $\mathbb{F}$ must contain $0$ and $-1$, it follows that $m(x)$ can be factored into distinct linear factors over $\mathbb{F}$. Hence $A$ is diagonalisable over $\mathbb{F}$ and its eigenvalues can only be $0$ or $-1$. In other words, $A=PDP^{-1}$ where $P$ can be any invertible matrix over $\mathbb{F}$ and $D$ can be any diagonal matrix whose diagonal entries assume values in $\{-1,0\}$.

Finding matrix for $A\left(I+A\right)=0$

2 Answers2