A four square matrix and $A^5$(a raised to the power of 5)$=0$. Then $A^4=?$

Question

A four square matrix and $A^5$(a raised to the power of 5)$=0$. Then $A^4=$

1. $I$(identity matrix)
2. $-I$
3. $0$
4. $A$

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My attempt:

You can use the characteristic equation $$ A^2-Tr(A)A+I_2\det{A}=O_2,$$ where $Tr(A)$ is the sum of the elements on the first diagonal while $\det{A}$ is the determinant of $A$ and it is zero because $A^5=0$. Hence $ A^2-Tr(A)A=O_2$. If $Tr(A)=0$ you are done. If not multiply the previous equation by$A^3$ and you will get $A^4=O_2$ and so on till $A^2=O_2$. ref @ Proving $A^2 = 0$ given $A^5 = 0$

Can you explain, please?

Answer 1

You need to use the Cayley Hamilton theorem. $A$ satisfies its own characteristic polynomial which has degree 4. If $A^5=0$, then the minimal polynomial $p(A)$ divides $A^5$. It is a polynomial of degree at most 4 since it has to divide the degree 4 characteristic polynomial. So the minimal polynomial $p(A)$ is either $A$, $A^2$, $A^3$, or $A^4$. In any case, $A^4=0$.

Answer 2

Answer is $0_{4\times 4}$. Because, the possible maximum degree of the minimal polynomial is $4$.

Answer 3

Expansion (if you want to learn more)

It can be any nilpotent matrix. Nilpotency of the matrix $\bf A$ means that ${\bf A}^k = {\bf 0}$ for some $k \in \mathbb{N_+}$. All of it's eigenvalues needs to be $0$, it can also be written as ${\bf A= VUV}^{-1}$, where $\bf U$ is a strictly upper-triangular matrix. Such a matrix gets one more superdiagonal of 0 for each higher exponent.

$${\bf U} = \left[\begin{array}{cccc} 0&6&4&4\\ 0&0&5&9\\ 0&0&0&5\\ 0&0&0&0 \end{array}\right], {\bf U}^2 = \left[\begin{array}{cccc} 0&0&30&74\\ 0&0&0&25\\ 0&0&0&0\\ 0&0&0&0 \end{array}\right], {\bf U}^3 = \left[\begin{array}{cccc} 0&0&0&150\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0 \end{array}\right]$$ If $k$ in ${\bf U}^k$is large enough, we see that the whole matrix will have become 0.

Answer 4

Let $V_k\subseteq \mathbb R^4$ be the image of $A^k$. Then $V_0\supseteq V_1\supseteq V_2\supseteq \ldots$. Clearly, $\dim V_0=4$. If $\dim V_4>0$ then $\dim V_k\in\{1,2,3,4\}$ for $k\in\{0,1,2,3,4\}$, hence at some $k\in\{0,1,2,3\}$ we must have $\dim V_k=\dim V_{k+1}$. But then $V_{k+1}=V_k$ and in fact $V_k=V_{k+1}=V_{k+2}=\ldots$ and specifically $V_k=V_4=V_5=0$. Consequently, $A^4=0$. (It may also be the case that $A^4=A$, but of course only if $A=0$).