Matrix $A$ is defined over real number.
Characteristic polynomial : $p(x)=(x+3)^2(x-1)(x-5)$
It also known that :
$$\text{rank}(A+2I)+\text{rank}(A+3I)+\text{rank}(A-5I)=9$$
- prove $A$ diagonalize.
My solution
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$-3,1,5$ are eigenvalues, using the characteristic polynomial we can conclude that matrix $A$ is $4 \times 4$
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Since eigenvalue $1,5$ has algebraic multiplicity of $1$, we can conclude that geometric multiplicity is also $1$ hence:
$$\text{rank}(I+A)=3$$
$$\text{rank}(5I-A)=3$$
I don't find a way to continue from here.
Best Answer
Picking up from where you left off.
From $\text{rank}(A-5I)=3$ and from $\text{rank}(A+2I)+\text{rank}(A+3I)+\text{rank}(A-5I)=9$ you get $$\text{rank}(A+2I)+\text{rank}(A+3I)=6.$$
Now prove that $A+2I$ is invertible. What does that tell you about the rank of $A+2I$?
Infer that $\text{rank}(A+3I)=2$. What does that tell you about the geometric multiplicity of $-3$?
Conclude.