Could anybody help me with solving such matrix problem:

Find all matrices $A$ (non zero and identity matrices), which corresponds to this equation:

$$ A^2=A^3 $$

  • 1
    $\begingroup$ Use Jordan normal form (presumably, you are talking about complex matrices). $\endgroup$ – Alex B. Jun 3 '11 at 8:58
  • $\begingroup$ @Alex B. Yes, we have arbitrary matrices (in common they are complex). Could you show me, how to use it? $\endgroup$ – user11697 Jun 3 '11 at 9:04

For every Jordan block of $A$, denoted as $J=\lambda I+N$, where $N$ is a nilpotent matrix with the property $N^n=0, N^k\ne0(k<n)$, $J$ also satisfies the equation. Now from $J^2=J^3$ one can obtain: $$\lambda^2(\lambda-1)I+(3\lambda^2-2\lambda)N+(3\lambda-1)N^2+N^3=0$$

So the order of $J$ must be smaller than $4$ for $N^3=0$. If $order=1$, $J=0\text{ or }1$. If $order=2$, $J=\begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}$. And no $\lambda$ can be found when $order=3$. Finally, such $J$'s (or their transpositions) direct sum provides an instance of $A$. For example,

$$A=\begin{pmatrix} 1&0&0&0\\ 0&0&1&0\\ 0&0&0&0\\ 0&0&0&0 \end{pmatrix}$$


The given matrix $A$ describes a linear transformation ${\mathbb R}^n\to {\mathbb R}^n$ which we denote by $A$ as well.

So we now are talking about a linear transformation $A:\ X\to X$ of some vector space $X$ that satisfies $A^2=A^3$. It follows that $(A^2)^2 = A (A^3) = A(A^2) = A^3 = A^2$, so $A^2$ is a projection. Let ${\rm im}(A^2) =: U$ and ${\ker}(A^2)=V$; then "by general principles" $X=U\oplus V$. For any $y\in U$ there is an $x\in X$ with $y=A^2 x$. From this we get $Ay=A^3 x =A^2 x= y$, i.e., $A$ is the identity on $U$. Let $W:={\rm ker} A \subset V$ and chose a subspace $W'\subset V$ such that $V=W \oplus W'$. $A$ restricted to $W'$ is injective, but for any $y\in W'$ we have $A (Ay)=0$, whence $Ay\in W$.

Now choose a basis $(e_i)_{1\leq i\leq n}$ for $X$ as follows: The first $r$ vectors $e_i \ (1\leq i\leq r)$ form a basis of $U$, the $d$ vectors $e_{r+ 2k-1} \ (1 \leq k\leq d)$ form a basis of $W'$, and the $d$ vectors $e_{r+ 2k}$ $\ (1 \leq k\leq d)$ are defined as $e_{r+2k}:=A e_{r+2k-1}$ $\ (1 \leq k\leq d)$. The latter form a linearly independent set in $W$. Finally choose $e_i \in W$ $\ (r+2d+1\leq i\leq n)$ so that altogether we have a basis of $W$, whence of $V$.

If we now look at the matrix of $A$ with respect to this basis then we see first $r$ ones in the main diagonal, then $d$ $2\times2$-boxes $\left[\matrix{0 & 0\cr 1 & 0\cr}\right]$ along the main diagonal, and everything else is $0$.

So we have proven the following: If a given matrix $A$ satisfies $A^2=A^3$ then by choosing a suitable new basis of ${\mathbb R}^n$ we can arrive at a matrix of the simple form just described.


Such a matrix has minimal polynomial that divides $t^3-t^2 = t^2(t-1)$. Therefore, the characteristic polynomial is of the form $t^n(t-1)^m$ for some nonnegative integers $n$ and $m$ that add up to the dimension of the matrix.

  • If the minimal polynomial is $(t-1)$, then the matrix is the identity.
  • If the minimal polynomial is $t$, then the matrix is the zero matrix.
  • If the minimal polynomial is $t(t-1)$, then the matrix is diagonalizable and the only eigenvalues are $0$ and $1$, so the matrix is a projection.
  • If the minimal polynomial is $t^2$, then the matrix is nilpotent: its Jordan canonical form has at least one $2\times 2$ block of the form $$\left(\begin{array}{cc}0&1\\0&0\end{array}\right),$$ no block of size greater than $2\times 2$, and all blocks associated to $\lambda=0$.
  • If the minimal polynomial is $t^2(t-1)$, then the Jordan canonical form of the matrix is as above, except that it also has at least one $1\times 1$ block associated to $\lambda = 1$, and all blocks associated to $\lambda=1$ are $1\times 1$.
  • $\begingroup$ Nice way to bring together the varied answers, Arturo! (+1) $\endgroup$ – Namaste Jun 3 '11 at 20:06

This is wrong! Sorry!

$A^3=A^2$ is the same as $A(A-I)(A+I)=0$. A solves it's own characteristic equation. So find all $A's$ whose eigenvalues are $0, 1,-1$.


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