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 $
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 $
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
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.
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 $0, 1,-1$.