Diagonalizability of Singular Matrices: Investigating Rank and Eigenvectors

[Dank2]

p(A+5I) = 1

 

[Dank2]

Almost. It only guarantees you that $SAS^{-1}$ is of this form, i.e. $A$ is diagonalizable. And recalculate $p(A+5I)$ in this case. (It is also sufficient to state the linear independency of $v_1$ and $v_2$. No need for coordinates.)

I got confused a bit now.
We took a diagonalized matrix and we proved that it can be diagonalized ? still I am not sure if there is a matrix that is not diagonalized that stratify this.

 

[fresh_42]

What is the criterion you use by saying $A$ is diagonalizable?

 

[Dank2]

What is the criterion you use by saying $A$ is diagonalizable?

If it has 3 linearly independent eigenvectors,
or if it has distinct eigenvalues
or if geometric multiplicity is equal to the algebraic multiplicity.

 

[fresh_42]

So, the second isn't applicable. How many eigenvectors of $A$ arise from the dimension of the nullspace of $A+5I$?
And how many from the nullspace of $A$ itself? Are all of these linearly independent? And why?

The former considerations showed us which eigenvalues there have to be and which possibilities exist. They are not really necessary but were easy to do.

 

[Dank2]

So, the second isn't applicable. How many eigenvectors of $A$ arise from the dimension of the nullspace of $A+5I$?
And how many from the nullspace of $A$ itself? Are all of these linearly independent? And why?

The former considerations showed us which eigenvalues there have to be and which possibilities exist. They are not really necessary but were easy to do.

I understand it if we assume A is the matrix that has it's eigenvalues on the diagonal. But what if A other matrix? in your question you assume A = diag(-5,-5,0) ?

 

[fresh_42]

I understand it if we assume A is the matrix that has it's eigenvalues on the diagonal. But what if A other matrix? in your question you assume A = diag(-5,-5,0) ?

No, I only assume the equation $3 = dim V = p(A+5I) + \dim \ker(A+5I) = 1 + 2$.
That gives me two linear independent vectors $x,y$ in the nullspace of $A+5I$. Both have to be eigenvectors of $A$ as well, namely to the eigenvalue $-5$.
Another vector $z$ spans the nullspace of $A$ by the same equation for $A$ instead.
It only remains to show that all three are linear independent.
So assume $0 = ax+by+cz$. Why are all coefficients $a,b,c$ equal to zero?

 

[Dank2]

Both have to be eigenvectors of AAA as well

why ?

 

[fresh_42]

why ?

Come on! $(A+5I)x = 0 = Ax +5x ⇔ Ax = -5x$
Edit: You got to reset yourself.

 

[Dank2]

So assume 0=ax+by+cz0=ax+by+cz0 = ax+by+cz. Why are all coefficients a,b,ca,b,ca,b,c equal to zero?

Well you've just shown that the algebraic multiplicity is equal to the geometric multiplicity, so it's done. then A is diagonalizable, and the eigenvectors are linearly independent. therefore a=b=c=0

 

[Dank2]

Come on! $(A+5I)x = 0 = Ax +5x ⇔ Ax = -5x$
Edit: You got to reset yourself.

Just so i would know where i stand, how hard would you rate this question from 1-10

 

[Dank2]

Edit: You got to reset yourself.

I'm just bad at math, with very poor backround

 

[fresh_42]

Knowing the solution: easy. It actually contained two parts: ruling out other combinations of ranks and then investigate the various eigenvectors and -values.

However, as you could have seen by my comments I started to think way too complicated by myself.
This is a usual phenomena. As long as you don't know something, it appears hard, whereas it turns out to be easy if you know how to do it - often, but not always. Unfortunately there are theorems that will stay hard, even if one knows how to prove them. But this doesn't matter: you learned something about the Jordan normal form, the algebraic and geometric multiplicity and so on. So even longer ways to a solution can be no lost effort.