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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

  • $-3,1,5$ are eigenvalues, using the characteristic polynomial we can conclude that matrix $A$ is $4 \times 4$

  • 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.

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  • $\begingroup$ What do you mean by $p(A+2I)+p(A+3I)+p(A-5I)=9$? How can a matrix be equal to $9$? Do you mean $9I$? $\endgroup$ – Ben Grossmann Oct 18 '14 at 0:25
  • $\begingroup$ @Omnomnomnom $p(A)$=rank of matrix A$ $\endgroup$ – JaVaPG Oct 18 '14 at 0:28
  • $\begingroup$ You should use \rho or some other symbol for the rank of a matrix. $\endgroup$ – JimmyK4542 Oct 18 '14 at 0:29
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    $\begingroup$ $-3$, not $ 3 $ $\endgroup$ – Belgi Oct 18 '14 at 0:31
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    $\begingroup$ Since $-2$ is not an eigenvalue, you have $\operatorname{rk} (A+2I) = 4$, which means $\operatorname{rk} (A+3I)+ \operatorname{rk} (A-5I) = 5$. $\endgroup$ – copper.hat Oct 18 '14 at 0:36
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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.

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  • $\begingroup$ Could I prove that $A+2I$ since 0 is not an eigenvalue $\rightarrow$ A is invertible, since 2 is not an eigenvalue $\rightarrow$ A+2I is invertible? $\endgroup$ – JaVaPG Oct 18 '14 at 0:45
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    $\begingroup$ @JaVaPG I find your wording confusing. It's not $2$ that matters here, it's $-2$. Since $-2$ is not an eigenvalue of $A$, it follows that $A-(-2I)$ is invertible. $\endgroup$ – Git Gud Oct 18 '14 at 0:58
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Hint: with such a characteristic polynomial, the matrix is congruent to either $$ \left( \begin{array}{ccc} -3 & 1 & 0 &0\\ 0 & -3 & 0 &0\\ 0 & 0 & 1 &0\\ 0 & 0 & 0 &5\end{array} \right) $$ or $$ \left( \begin{array}{ccc} -3 & 0 & 0 &0\\ 0 & -3 & 0 &0\\ 0 & 0 & 1 &0\\ 0 & 0 & 0 &5\end{array} \right)$$

Then in each case, compute $P(A)$ in this new base.

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Hint: The rank of $A - \lambda I$ is $n$ minus (the number of Jordan blocks of $A$ that are associated with $\lambda$).

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