Diagonalizability of Singular Matrices: Investigating Rank and Eigenvectors

[Dank2]

Homework Statement

Let A be a 3x3 singular Matrix that satisfy:
p(A+5I) < p(A)
p - is the rank of the matrix
I - is the identity matrix,
Is A Diagonalizable?

Homework Equations

The Attempt at a Solution

I know that A diagonalizable matrix can be Singular from every rank, even at 0 rank, so i can't see how i can conclude anything.

 

[fresh_42]

You can exclude many cases by the given assumptions. What does it mean for A to be singular?
Can A or A+5I equal the zero matrix?

 

[Dank2]

singular means it isn't invertible or the columns are linearly independent or the rank is lower than 3.
i think A + 5I cannot be the zero matrix, but how's that even related to the Diagnolization of A, since any rank of matrix can be diagonliazable.

 

[fresh_42]

There is only one possibility for the ranks of A and A + 5I. Assume that A is diagonalizable. What does it mean for the eigenvalues? This gives you a criterium.

 

[Dank2]

You mean the eigenvectors? they are independent. if not i don't know, i know that if A is diagonalizable it means that the amount of the same eigenvalues should be equal to the dimension of the eigenvectors space that is related to the current eigenvalue.

also the rank of A<3
and i can't see how to prove that A+5I > 0 it does look reasonable to me.
So A+5I > 0 ==> A+5I = > 1
and then rank of A = 2.
How is that helps me determine if A Diagonalizable

 

[Dank2]

Still need help

 

[fresh_42]

I meant the eigenvalues.
$p(A)=2$ and $p(A+5I)=1$ are the only possibilities for the ranks, since $p(A) < 3$ and neither matrix can be zero which is equivalent to rank zero. (If $A+5I = 0$ then $A = -5I$ of rank $3$. If $A=0$ then there is no way for another rank below that.)

Now assume $A$ is diagonalizable, i.e. there is a matrix $S$ with $SAS^{-1} = diag(α,β,γ)$.
Then $S(A+5I)S^{-1} = SAS^{-1} + 5I = diag(α+5,β+5,γ+5)$ and without loss of generality $γ=0$ because $p(A)=2.$
Now we still have $p(A+5I)=1$ which is only possible for $α = β = -5$.

 

[Dank2]

So if the rank of the matix nxn is k , then the amount of 0 eigenvalues is n-k ? .
and also if i get two eigenvalues that are 0, how can i know if the matrix is diagonal or not ?

 

[Ray Vickson]

You mean the eigenvectors? they are independent. if not i don't know, i know that if A is diagonalizable it means that the amount of the same eigenvalues should be equal to the dimension of the eigenvectors space that is related to the current eigenvalue.

also the rank of A<3
and i can't see how to prove that A+5I > 0 it does look reasonable to me.
So A+5I > 0 ==> A+5I = > 1
and then rank of A = 2.
How is that helps me determine if A Diagonalizable

Why do you say p(A) < 3? All you were told is that p(A+5I) < p(A), and so you can have p(A) = 3, p(A+5I) = 1 or 2.

 

[Dank2]

Why do you say p(A) < 3? All you were told is that p(A+5I) < p(A), and so you can have p(A) = 3, p(A+5I) = 1 or 2.

Because A is singular, that means it has linearly independent columns, and that means the rank must be less than 3.

 

[Dank2]

I think I've got it, the rank of P'AP = Rank of A .
Then Rank of A = Rank of it's Diagonal matrix . therefore the rank - n is equal to the number of 0 on the diagonal.
But, still haven't figured out if it's Diagonalizable or not.

 

[fresh_42]

So if the rank of the matix nxn is k , then the amount of 0 eigenvalues is n-k ? .

Amount isn't a good word here. In case of an eigenvalue $\lambda$ we speak of geometric multiplicity (dimension of the eigenspace of $\lambda$) and algebraic multiplicity (power of $t-\lambda$ in the characteristic polynomial $det (A-tI) = 0$) . If $A$ is diagonalizable then they are equal.
In general (in the situation asked) $n = p(A) + \dim \ker(A) = k + (n-k)$ and $\ker(A) = \{x \in V | Ax=0=0 \cdot x\}$ the eigenspace of eigenvalue $0$.

and also if i get two eigenvalues that are 0, how can i know if the matrix is diagonal or not ?

You can't. It heavily depends on the scalar domain that make up your coordinates.

 

[Dank2]

You can't. It heavily depends on the scalar domain that make up your coordinates.

Or maybe A is diagonalizable since the rank of A is 2, and ker of A=1, and it does have only one eigenvalue = 0 , and so the dim of the eigenspace of 0 is 1 .

How can i see the the dim of the Eigenspace of -5 is 2 ?

 

[fresh_42]

The question in my book says if A is Diagonalizable? Maybe they have mistake ?

If $A = diag(-5,-5,0)$ then $A$ is diagonal and all conditions are met. We only derived that there can't be other eigenvalues.

 

[Dank2]

If $A = diag(-5,-5,0)$ then $A$ is diagonal and all conditions are met. We only derived that there can't be other eigenvalues.

Thanks, think i got it.

 

[Dank2]

If $A = diag(-5,-5,0)$ then $A$ is diagonal and all conditions are met. We only derived that there can't be other eigenvalues.

I wondered one more thing, if i could show that the dim of the eigenspace of -5 is 2 ? or do i must have values of the matrix for that ?

 

[Ray Vickson]

Thanks, think i got it.

If
A = \pmatrix{-5 &amp; 1 &amp; 0 \\ 0 &amp; -5 &amp; 0 \\ 0 &amp; 0 &amp; 0}
then all the conditions are met and $A$ cannot be diagonalized.

 

[Dank2]

You mean can be ? if not :
I = Identity matrix
I' = Inverse
I'AI = A
which means A is diagonlaziable. We just found matrix I that Diagonalize A.

 

[Ray Vickson]

You mean can be ? if not :
I = Identity matrix
I' = Inverse
I'AI = A
which means A is diagonlaziable. We just found matrix I that Diagonalize A.

No, it was deliberately chosen to be NON-diagonalizable. Look again: the element $a_{12} \neq 0$. The matrix $A$ is a 3x3 Jordan canonical form for eigenvalues -5,-5,0, with a non-diagonal Jordan block in the top left 2x2 location.

The eigenvalue -5 has algebraic multiplicity 2 but geometric multiplicity 1; that means that you cannot find three linearly-independent eigenvectors, only two. You need to include a so-called "generalized" eigenvector in order to have a full 3-dimensional basis. That is why the matrix cannot be diagonalized.

 

[Dank2]

No, it was deliberately chosen to be NON-diagonalizable. Look again: the element $a_{12} \neq 0$. The matrix $A$ is a 3x3 Jordan canonical form for eigenvalues -5,-5,0, with a non-diagonal Jordan block in the top left 2x2 location.

Where did that a₁₂ =1 came from ?
and how do you know that the Geometric multiplicity of -5 is 1 ?

 

[fresh_42]

Where did that a₁₂ =1 came from ?
and how do you know that the Geometric multiplicity of -5 is 1 ?

This $a_{12}=1$ has been willingly chosen, as I have chosen $diag(-5,-5,0).$
However, as @Ray Vickson has said: it is the Jordan normal form for matrices (basically achieved by choosing a basis in which a matrix can be written as the sum of a diagonal and a nilpotent (strict upper triangular) matrix) which allow you to see the dimensions of eigenspaces. Diagonalizability depends on whether the geometric and algebraic multiplicities of eigenvalues coincide or not (see above).

These are two examples of possible matrices $A$, one of which is diagonal and one is not diagonalizable.
(I haven't done it so far but it might be possible to make a finite conclusion if we take additionally the fact that $p(A+5I)=1$ into account?)

 

[Dank2]

This $a_{12}=1$ has been willingly chosen, as I have chosen $diag(-5,-5,0).$
However, as @Ray Vickson has said: it is the Jordan normal form for matrices (basically achieved by choosing a basis in which a matrix can be written as the sum of a diagonal and a nilpotent (strict upper triangular) matrix) which allow you to see the dimensions of eigenspaces. Diagonalizability depends on whether the geometric and algebraic multiplicities of eigenvalues coincide or not (see above).

These are two examples of possible matrices $A$, one of which is diagonal and one is not diagonalizable.
(I haven't done it so far but it might be possible to make a finite conclusion if we take additionally the fact that $p(A+5I)=1$ into account?)

We don't have nilpotent in our course, is possible to see that the geometric multiplicity of -5 is 2 or 1 ?

 

[Dank2]

If
A = \pmatrix{-5 &amp; 1 &amp; 0 \\ 0 &amp; -5 &amp; 0 \\ 0 &amp; 0 &amp; 0}
then all the conditions are met and $A$ cannot be diagonalized.

but Rank of A+5I is = Rank of A

 

[fresh_42]

We don't have nilpotent in our course, is possible to see that the geometric multiplicity of -5 is 2 or 1 ?

Therefore I added "strict upper triangular".
Assume you have a matrix $S$ for which $SAS^{-1} = B$ where $B$ is either a diagonal matrix or of the form Ray suggested, i.e. has only two eigenvectors, one to the eigenvalue $0$ and one for $-5$. Then examine $A+5I$ and remember it has to be of rank one.

 

[Dank2]

Therefore I added "strict upper triangular".
Assume you have a matrix $S$ for which $SAS^{-1} = B$ where $B$ is either a diagonal matrix or of the form Ray suggested, i.e. has only two eigenvectors, one to the eigenvalue $0$ and one for $-5$. Then examine $A+5I$ and remember it has to be of rank one.

see post #23,

 

[fresh_42]

see post #23,

You already know the defect of $A+5I$ that is the dimension of its kernel. What happens to a basis of this kernel, if you apply $A$?

 

[Dank2]

You already know the defect of $A+5I$ that is the dimension of its kernel. What happens to a basis of this kernel, if you apply $A$?

Sorry , dimension of it's kernel? you mean the dim of the nullspace of A+5I ?

 

[fresh_42]

Sorry , dimension of it's kernel? you mean the dim of the nullspace of A+5I ?

Yes. Take basis vectors $v_i$ of it and calculate $Av_i.$

 

[Dank2]

basis nullspace of A is v1 = (1,0,0),v2 = (0,1,0)
Av1 = -5*v1
Av2 = -5*v2
That is for A =
-5 0 0
0 -5 0
0 0 0
Because A as in post 23 doesn't satisfy p(A+5I) = 1

 

[fresh_42]

basis nullspace of A is v1 = (1,0,0),v2 = (0,1,0)
Av1 = -5*v1
Av2 = -5*v2
That is for A =
-5 0 0
0 -5 0
0 0 0
Because A as in post 23 doesn't satisfy p(A+5I) = 1

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.)

 

[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.