Prove that $A^3-3A^2+4A-5=0$ for a given matrix $A$. Consider the following matrix $A$:
$$A=
\begin{bmatrix}
0 &1 &-1\\
1 &1 &1\\
-1 &0 &2\\
\end{bmatrix}
$$
$$\text{Prove that }\;A^3-3A^2+4A-5=0$$
I have no idea how to solve this.
Thanks.
 A: The problem is wrong. If one evaluates it all, one gets a very non-zero matrix.
A: Simply verify that $(A\cdot A \cdot A) - 3(A\cdot A) + 4A - 5I = 0$, where I here denotes the $3 \times 3$ identity matrix, and $0$ is the $3\times3$ zero matrix. 
Good luck. 
A: We can compute the characteristic polynomial of the matrix, which is 
$$q(x)=-x^3+3x^2-4$$
then divide the given polynomial $p(x)=x^3-3x^2+4x-5$ by the characteristic polynomial to get the remainder
$$p(x)=q(x)a(x)+r(x)$$
The remainder will have smaller degree. Since by Hamilton-Cayley the characteristic polynomial vanishes when evaluated at the matrix we get
$$p(A)=q(A)a(A)+r(A)=r(A)$$
then compute $r(A)$.
In our case $$p(x)=-q(x)+4x-9$$
Therefore $$p(A)=4A-9$$
So to evaluate the polynomial at $A$, we just need to multiply that matrix by $4$ and subtract later $9$ in the diagonal.
A: If you apply $A$ repeatedly starting at the second basis vector $e_2$ (which looked most promising to me) you get $(0,1,0)\overset{A}\mapsto(1,1,0)\overset{A}\mapsto(1,2,-1)\overset{A}\mapsto(3,2,-3)$. The first three vectors are clearly linearly independent and form a basis. The fourth one satisfies $(3,2,-3)=-4(0,1,0)+3(1,2,-1)$. Therefore $P=X^3-3X^2+4$ is the minimal degree monic polynomial such that $P[A](e_2)=0$. Since $P[A]$ also kills $A(e_2)$ and $A^2(e_2)$, it is actually the minimal (and characteristic) polynomial of $A$. Since it does not divide the given polynomial $X^3-3X^2+4X-5$, that polynomial evaluated in $A$ is not zero.
