Exploring Algebraic Multiplicity: Solving 5a and b with Hints for 5c

[Dethrone]

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5a and b are quite trivial. I have no idea how to do 5c). Any hints?

 

[Nono713]

Seems like you should use the fact that the characteristic polynomial of $AB$ has degree $m$ while the characteristic polynomial of $BA$ has degree $n$. Since $m < n$, and using 5a and 5b, it means $BA$ must have "extra" distinct eigenvalues in addition to those of $AB$, and in fact that extra eigenvalue can only be zero (possibly repeated). It follows from the degrees of the polynomials (recall the eigenvalues of $AB$ have total algebraic multiplicity $m$) that this zero eigenvalue must have algebraic multiplicity $n - m$. So this is what needs to be proved apparently, not sure how I would approach the proof yet though.

(not sure why it says "at least $n - m$"; seems to me it must be exactly $n - m$, but I could be wrong here, it's been a while since I touched eigenstuff)

 

[Dethrone]

Hey Bacterius :D

Can you elaborate why 5a, b implies that $BA$ must have "extra" eigenvalues than $AB$?
I just scanned the section on multiplicities so I have not fully grasped it yet.

 

[Nono713]

Hey Bacterius :D

Can you elaborate why 5a, b implies that $BA$ must have "extra" eigenvalues than $AB$?
I just scanned the section on multiplicities so I have not fully grasped it yet.

Well, 5a and 5b say that $BA$ has all eigenvalues of $AB$. But $BA$ is $n \times n$ while $AB$ is $m \times m$, so $BA$ can have at most $m$ eigenvalues while $BA$ can (must?) have more. Actually, I am not really sure anymore - I know that the eigenvalues of a complex $n \times n$ matrix always have a total algebraic multiplicity of $n$ (by the fundamental theorem of algebra). But it could be less for real matrices because some eigenvalues could be complex - actually the fact that $AB$ may have repeated eigenvalues (i.e. of algebraic multiplicity > 1) might explain the "at least" part.

An eigenvalue $\lambda $ of algebraic multiplicity $m$ just means that in your characteristic polynomial you have a factor of $(x - \lambda)^m$. Geometric multiplicity is something different.

 

[Dethrone]

I think I understand your argument (it's a bit late here...), but I still have some questions. So $AB$ has $m$ eigenvalues (possible repeated), and $BA$ contains all those eigenvalues and more, since $m<n$ and $BA$ has $n$ eigenvalues. How do we ascertain that those extra eigenvalues in $BA$ must all be the zero eigenvalue? I understand that if that is true, then $BA$ must have at least $n-m$ eigenvalues that are $0$ and more depending on the algebraic multiplicity of the $0$ eigenvalue in $AB$.

I hope what I'm saying even makes sense...:p

Well, 5a and 5b say that $BA$ has all eigenvalues of $AB$. But $BA$ is $n \times n$ while $AB$ is $m \times m$, so $BA$ can have at most $m$ eigenvalues while $BA$ can (must?) have more.

You mean $AB$? and can have at most $m$ distinct eigenvalues?

 

[Nono713]

Yes, I meant AB sorry. And it can have at most $m$ distinct eigenvalues, for if it had more, then the characteristic polynomial of $AB$ would have degree greater than $m$, which is impossible (recall eigenvalues are precisely the roots of the characteristic polynomial of a matrix). So it certainly can't have more. It could have less if you only consider real eigenvalues, though.

As for your question, well that's what I would like to know too :p I haven't actually tried proving it (I have to go soon, so I might give the question an actual shot tomorrow). But that is the approach I would take I think.

 

[I like Serena]

Hey Rido! ;)

Here's how I would do it.

A has a null space with dimension of at least n-k.
Therefore A has at least n-k vectors that transform to 0.
It follows that BA has at least those same vectors as eigenvectors for eigenvalue 0.
Qed. (Wasntme)

 

[Dethrone]

Hey Rido! ;)

Here's how I would do it.

A has a null space with dimension of at least n-k.
Therefore A has at least n-k eigenvectors with eigenvalue 0.
It follows that BA has at least those same eigenvectors.
Qed. (Wasntme)

Hey ILS! (Wave)

So the dimension of the null space of a matrix is the algebraic multiplicity of the eigenvalue $0$? How do we know that BA has the all the eigenvectors of $A$. Is there a theorem I should be aware of?

 

[I like Serena]

Hey ILS! (Wave)

So the dimension of the null space of a matrix is the algebraic multiplicity of the eigenvalue $0$? How do we know that BA has the all the eigenvectors of $A$. Is there a theorem I should be aware of?

Almost:
Nullity = geometric multiplicity $\le$ algebraic multiplicity

Only the theorem which is the definition of an eigenvector:
Suppose $Av=0$, then $BA v = 0\cdot v$. (Wasntme)

 

[Dethrone]

Almost:
Nullity = geometric multiplicity $\le$ algebraic multiplicity

Only the theorem which is the definition of an eigenvector:
Suppose $Av=0\cdot v$, then $BA v = 0\cdot v$. (Wasntme)

So in general, any matrix that has a nullspace greater than 0 has an eigenvalue $0$, and the dimension of that nullspace gives us the geometric multiplicity of the eigenvalue $0$? (If true, I will need to prove this for myself. :p)

 

[I like Serena]

So in general, any matrix that has a nullspace greater than 0 has an eigenvalue $0$, and the dimension of that nullspace gives us the geometric multiplicity of the eigenvalue $0$? (If true, I will need to prove this for myself. :p)

Theorem Rido12.1
A matrix has nullity greater than $0$ iff it has eigenvalue $0$.

Theorem Rido12.2
The nullity of a matrix is equal to the geometric multiplicity of the eigenvalue $0$.

Theorem Rido12.3
The algebraic multiplicity of eigenvalue $0$ can be greater than the corresponding geometric multiplicity.I'll leave the proofs up to you. :p

 

[Dethrone]

Theorem Rido12.1
My proof goes both ways...but the backwards direction is easier. Suppose $A$ has eigenvalue $0$, then its characteristic equation is $c_A=\det (\lambda I-A)=\det (-A)=(-1)^n \det A=0$, where $A$ is n by n. Then $A$ is singular and must have nullspace > 0.

Can you give me hints on Theorem Rido12.2?

 

[I like Serena]

Theorem Rido12.1
My proof goes both ways...but the backwards direction is easier. Suppose $A$ has eigenvalue $0$, then its characteristic equation is $c_A=\det (\lambda I-A)=\det (-A)=(-1)^n \det A=0$, where $A$ is n by n. Then $A$ is singular and must have nullspace > 0.

Good! (Nod)

Can you give me hints on Theorem Rido12.2?

Let's list the appropriate definitions:

• The null space of a matrix A is the space of all vectors $v$ that satisfy $Av=0$.
• An eigenvector is a vector $v \ne 0$ such that $Av = \lambda v$, where $\lambda$ is the corresponding eigenvalue.
• The geometric multiplicity of an eigenvalue is the number of independent eigenvectors for that eigenvalue.

Can you combine them? (Wondering)

 

[Dethrone]

Theorem Rido 12.2

Suppose that $\lambda =0$ is an eigenvalue. Then to find the eigenspace corresponding to $\lambda=0$, we must solve $Av=\lambda v$, or $Av=0$. That means, we need to find $\operatorname{null}A$ (or $\operatorname{null}(-A)$ if we continue from $c_A=\det(0-A)$).
That means the eigenspace corresponding to $\lambda=0$ is the nullspace of A, and the geometric multiplicity of A is the nullity of $A$!

 

[I like Serena]

Theorem Rido 12.2

Suppose that $\lambda =0$ is an eigenvalue. Then to find the eigenspace corresponding to $\lambda=0$, we must solve $Av=\lambda v$, or $Av=0$. That means, we need to find $\operatorname{null}A$ (or $\operatorname{null}(-A)$ if we continue from $c_A=\det(0-A)$).
That means the eigenspace corresponding to $\lambda=0$ is the nullspace of A, and the geometric multiplicity of A is the nullity of $A$!

Yay! (Party)Btw, note that $\det(\lambda I - A)=0$ merely follows from $Av=\lambda v$.
It's the latter that is the actual definition, while the first only holds if the determinant is defined (which is not the case for your unusual $\mathbb P_2$ vector space). (Nerd)

 

[Dethrone]

Yay! (Party)Btw, note that $\det(\lambda I - A)=0$ merely follows from $Av=\lambda v$.
It's the latter that is the actual definition, while the first only holds if the determinant is defined (which is not the case for your unusual $\mathbb P_2$ vector space). (Nerd)

haha, thanks so much for the help, ILS! (Cool)

Btw, is there an easy proof that the geometric multiplicity must not exceed the algebraic multiplicity? The proof in my notes is quite convoluted.

 

[I like Serena]

haha, thanks so much for the help, ILS! (Cool)

Btw, is there an easy proof that the geometric multiplicity must not exceed the algebraic multiplicity? The proof in my notes is quite convoluted.

Easiest is to transform a matrix to Jordan Normal Form.
Then it becomes instantly obvious (when you know how to read the Jordan Normal Form).
Of course then we would need to prove we can get to the Jordan Normal Form, but I prefer to take that as a given.

Btw, it is possible for the algebraic multiplicity to be higher.
To prove that, it suffices to find an example.
Can you find one for, say, eigenvalue 0? (Wondering)

 

[Dethrone]

What class is the Jordan Normal form generally taught? In an advanced linear algebra class or in something like abstract algebra?

So, to find a situation when the algebraic multiplicity is greater than the geometric for the 0 eigenvalue?
I am trying to find one for the 2 by 2 case but it's not working; should I try a higher dimension?

i.e $c_A=\det(\lambda I-A)=\lambda ^2$
Working backwards, the only $A$ that will work is the zero matrix...
I guess I can easily show that the geometric is less than the algebraic by increasing the rank of the matrix (i.e introducing linearly dependent rows).

 

[I like Serena]

What class is the Jordan Normal form generally taught? In an advanced linear algebra class or in something like abstract algebra?

I think it's typically in the first year of university - in a linear algebra class. (Thinking)
Definitely not in abstract algebra.

So, to find a situation when the algebraic multiplicity is greater than the geometric for the 0 eigenvalue?
I am trying to find one for the 2 by 2 case but it's not working; should I try a higher dimension?

I suggest to stick with the 2 by 2 case. (Mmm)

i.e $c_A=\det(\lambda I-A)=\lambda ^2$
Working backwards, the only $A$ that will work is the zero matrix...

Think again... (Sweating)

 

[Dethrone]

I think it's typically in the first year of university - in a linear algebra class. (Thinking)
Definitely not in abstract algebra.

Hmm, that isn't taught in my class, nor in nearby universities (engineering program) for first years. I suspect it might be taught to pure math students. Anyway, I've heard that it'll be taught in my DE course next year, because they have many applications to solving DE? Particularly diagonalizing matrices?

Anyway, I've tried again, this time in a more systematic approach. I have found that $A$ must have $a+d=0$ and $ad-bc=0$. A candidate is $\left[\begin{array}{1,-1}1 & -1 \\ 1 & -1 \end{array}\right]$. This will have algebraic multiplicity of $2$ and geometric multiplicity of $1$.

 

[I like Serena]

Hmm, that isn't taught in my class, nor in nearby universities (engineering program) for first years. I suspect it might be taught to pure math students. Anyway, I've heard that it'll be taught in my DE course next year, because they have many applications to solving DE? Particularly diagonalizing matrices?

It would be taught to all pure math and all physics students.

It makes sense to see it in DE, but that should merely be an application of linear algebra that should have already been taught by that time. Then again, the order of classes in university doesn't always make sense. :p

Anyway, I've tried again, this time in a more systematic approach. I have found that $A$ must have $a+d=0$ and $ad-bc=0$. A candidate is $\left[\begin{array}{1,-1}1 & -1 \\ 1 & -1 \end{array}\right]$. This will have algebraic multiplicity of $2$ and geometric multiplicity of $1$.

You found one! You found one! (Whew)

Note that $a+d$ is called the $\operatorname{Trace}$ of the matrix, which is also the sum of the eigenvalues.
And $ad-bc$ is the $\operatorname{Determinant}$, which is also the product of the eigenvalues. (Nerd)

Btw, a simpler matrix is $\begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}$.
You can read from it that the corresponding eigenvector is $(^1_0)$, which is the only eigenvector, even though it has an algebraic multiplicity of 2. (Wasntme)