Let $A$ be a $5 \times 5$ matrix with dim of solution space of $Ax=0$ is at least $2$ then what can we say about the rank of $A^2$ and $det(A^2)$?

Question

Let $A$ be a $5 \times 5$ matrix with dimension of solution space of $Ax=0$ is at least $2$ then what can we say about the rank of $A^2$ and $\det(A^2)$?

What I am getting is that nullity of matrix $A$ is $2,3,4,5$; then by Rank Nullity theorem Rank of A will be $3,2,1,0$. Then $\det A$ is definitly $0$. Then $\det A^2=\det A$, $\det A=0$.

Am I correct? I can not conclude anything about rank of $A^2$. Please help me.

You probably mean that the dimension of the solution space is at least 2. — Magdiragdag, Apr 16 '20 at 10:02
Does this answer your question? How the determinant of $A^2$ is $0$? — nmasanta, Feb 06 '21 at 09:16

score 2 · Answer 1 · answered Apr 16 '20 at 08:54

2

Since the kernel of $A$ is contained in the kernel of $A^2$, we obtain that the rank of $A^2$ is at most the rank of $A$, hence $\mathrm{rank}(A^2) \in \{0,1,2,3\}$. Depending on $A$, the rank of $A^2$ can take any of these values. If we let $A$ to be diagonal, e.g.

$$A = \mathrm{diag}(1,1,1,0,0),$$

then $A^2 = A$, so they have the same rank. The same works if the number of $1$'s on the diagonal is one of $0,1,2$.

answered Apr 16 '20 at 08:54

Marktmeister

1,648

why kernel of A is in kernel o A^2? – tret Apr 17 '20 at 04:33
@tret If $v$ is such that $Av = 0$, then $A\cdot Av = A^2v = 0$. – Marktmeister Apr 17 '20 at 06:30

Let $A$ be a $5 \times 5$ matrix with dim of solution space of $Ax=0$ is at least $2$ then what can we say about the rank of $A^2$ and $det(A^2)$?

1 Answers1