About rank of matrix

Question

Let $A \in M_{6}(\mathbb{R})$ and $A^{3}-2 A^{2}-15 A=0$. If $\operatorname{tr}(A)=4,$ find $\operatorname{rank}(A)$.

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How we can solve this? I think we have $x(x^2-2x-15)=0$ so $x=-3 ,0, 5$ so eigenvalues of $A$ are $3,5,0$.

Answer 1

Hint: Note the following:

• Because $A^3 - 2A^2 - 15A = 0$, the eigenvalues of $A$ must be solutions to $$ x^3 - 2x^2 - 15x = 0 \implies x(x+3)(x-5) = 0. $$ However, we cannot conclude from this that each such solution is an eigenvalue. Because this polynomial has no repeated factors, $A$ must be diagonalizable.

• Note that $6 - \operatorname{rank}(A) = \dim \ker(A)$, where $\ker(A)$ denotes the kernel (AKA nullspace) of $A$. Note that $\ker(A)$ is the eigenspace of $A$ associated with $0$. For a diagonalizable matrix, the dimension of this eigenspace is equal to the (algebraic) multiplicity of the eigenvalue $0$.

• The trace of a matrix is the sum of its eigenvalues (up to algebraic multiplicity).

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If $m$ is the multiplicity of eigenvalue $-3$ and $n$ is the multiplicity of eigenvalue $5$, then the trace of $A$ is $-3m + 5n$. We need to find non-negative integers $m,n$ for which $-3m + 5n = 4$ and $m + n \leq 6$.

Once we have done this, we find that the multiplicity of the eigenvalue $0$ is $6 - m - n$, and we can find the rank accordingly.

Answer 2

Given, $A$ satisfy the polynomial $x(x+3)(x-5)$.

Must remember, $A$ satisfy it's own minimal polynomial. Suppose, $p(x)$ is minimal polynomial of $A$.

So, $A$ can satisfy the polynomial $p(x)q(x)$, where $q(x)$ is some other polynomial.

So, we can take minimal polynomial of $A$ as in $7$ ways. So, $p(x)$ can be $(x)$ or $(x-5)$ or $(x+3)$ or $x(x+3)$ or $x(x-5)$ or $(x+3)(x-5)$ or $x(x+3)(x-5)$.

Clearly, for each case, $A$ must be diagonalizable,as each factor in minimal polynomial of $A$ appear only with degree $1$.

For the case , when $p(x)=x$, then $A$ is null matrix, so, rank($A$)=0.

For the case , when $p(x)=x-5 $, then $A=diag(5,5,5,5,5,5)$ So, rank is $6$

For the case , when $p(x)=x+3 $ ,then $A=diag(-3,-3,-3,-3,-3,-3)$ ,so ,rank is $6$.

For the case , when $p(x)=x(x+3) $, as, $A$ is diagonalizable, if $0$ has A.M. $k$ ($1\le k \le 5 $), then $-3$ has A.M. is $(6-k)$. Rank of $A$ is $(6-k)$.

For the case, when $p(x)=x(x+3)(x-5) $ , if A.M. of $0$ is $m$,($1\le m \le 4 $), then rank($A$)=$(6-m)$.

I think you can proceed for other cases in similar way.