Solving Square Matrix Similarity to Diagonal Matrix

[Maybe_Memorie]

Homework Statement

A square matrix A (of some size n x n) satisfies the condition A^2 - 8A + 15I = 0.

(a) Show that this matrix is similar to a diagonal matrix.
(b) Show that for every positive integer k >= 8 there exists a matrix A
satisfying the above condition with tr(A) = k.

Homework Equations

A^2 - tr(A)A + det(A)I = 0

The Attempt at a Solution

I'm not entirely sure what to do but here's the attempt..

I subtracted the given formula from the equation, obtaining
(8 - tr(A))A + (15 - det(A))I = 0

So either tr(A) = 8 or A = cI, where c = [15 - det(A)]/tr(A) - 8

So since I've shown that A = cI, c being a constant, is this enough to show that A is similar to a diagonal matrix?

For (b), I'm completely lost... If it means use A^2 - kA + 15I = 0
then surely the result will follow straight from the last part? And presumably having k<8 will result in something impossible?

 

[talolard]

Do you think you could write the original equation as (A-3I)(A-8I).
If you could justify writing it out like that, and you looked at the determinant of that expression, what would you find out?

 

[I like Serena]

Interesting problem.
I do not have a solution yet.
Since I do not know your relevant equation, I decided to analyze it.

If I take a 1x1 matrix A = (a), I get:

(a)² - tr[(a)] (a) + det[(a)] I = (a²) - a . (a) + a . (1) = (a)

However, this is not equal to zero!?

For a 2x2 matrix it does check out, always.

For a 3x3 matrix it is not generally true, although there are solutions.

Where did you get your equation?

 

[micromass]

Do you know about "minimal polynomials" and the Cayley-Hamilton theorem??

Can you prove (using this) that the characteristic equation of A is exactly $x^2-8x+15$?

 

[I like Serena]

@micromass: I did not know the Cayley-Hamilton theorem yet, nor "minimal polynomials".
Thanks for the reference.
However, as far as I can tell now, it is only applicable to this problem for n=2, as is your characteristic equation.
Am I missing something?

@OP: I suspect these theorems are beyond the level you are at. Am I wrong?

 

[hunt_mat]

How about this, let P be an invertible matrix, the:
$$P^{-1}(A^{2}-8A+15I)P=0\Rightarrow P^{-1}APP^{-1}AP-8P^{-1}AP+15I=0\Rightarrow (P^{-1}AP)^{2}-8P^{-1}AP+15I=0$$
All you need to do now is show that A has a full set of eigenvectors and you can see that A is similar to a diagonal matrix.

To examine the eigenvalues, write $Ax=\lambda x$ and examine $A^{2}x$ and $Ix$, this should leave you with three equations which can manipulated to give you an equation for $\lambda$

 

[micromass]

@micromass: I did not know the Cayley-Hamilton theorem yet, nor "minimal polynomials".
Thanks for the reference.
However, as far as I can tell now, it is only applicable to this problem for n=2, as is your characteristic equation.
Am I missing something?

No, you're not missing anything I failed to see that it had to be an arbitrary size nxn

 

[I like Serena]

Errr... "All you need to do now is show that A has a full set of eigenvectors"?
You make it sound almost easy...
Any suggestions how to do that?
Because I've got no clue.

 

[hunt_mat]

Errr... "All you need to do now is show that A has a full set of eigenvectors"?
You make it sound almost easy...
Any suggestions for that?
Because I've got no clue.

I see what you're saying, I provided some suggestions to examine the eigenvalues. I will have another think about it.

 

[micromass]

Errr... "All you need to do now is show that A has a full set of eigenvectors"?
You make it sound almost easy...
Any suggestions for that?
Because I've got no clue.

Well, you can do it with minimal polynomials (even for arbitrary sizes).
Here's a sketch:
First show that the minimal polynomail of A divides $x^2-8x+15$ (this is true by definition).
Since the roots of the minimal polynomial are exactly the eigenvectors, one can show that 3 and 5 are the only possible eigenvectors.

So if n>2, then trace(A)>8, and a technique as in the OP shows that the matrix is diagonalizable. So only n=2 remains.

 

[Maybe_Memorie]

@OP: I suspect these theorems are beyond the level you are at. Am I wrong?

The equation I supplied at the start may only be for a 2x2 matrix, as I found it in a solution my lecturer had for a different problem on a 2x2 matrix, and incorrectly assumed it was valid for an nxn matrix.

And yes, I know the Cayley-Hamilton theorem, it is on my course, but I haven;t been exposed to it enough to fully understand it

 

[I like Serena]

And yes, I know the Cayley-Hamilton theorem, it is on my course, but I haven;t been exposed to it enough to fully understand it

Yeah, well, I may have been exposed to it as well, but if I have been, I've long forgotten it!
Lack of enough exposure I guess.

 

[Maybe_Memorie]

Yeah, well, I may have been exposed to it as well, but if I have been, I've long forgotten it!
Lack of enough exposure I guess.

Or lack of going to lectures.

So, from what I've gathered, since the minimal polynomial of A is x^2 - 8x + 15, and the roots of this are 3 and 5, then the eigenvalues are 3 and 5.

Since it has eigenvalues it presumambly has a jordan normal form and is thus similar to a diagonal matrix?

 

[micromass]

So, from what I've gathered, since the minimal polynomial of A is x^2 - 8x + 15, and the roots of this are 3 and 5, then the eigenvalues are 3 and 5.

Not necessarily. It can happen that A only has 3 as eigenvalue, for example.

 

[I like Serena]

Well, you can do it with minimal polynomials (even for arbitrary sizes).
Here's a sketch:
First show that the minimal polynomail of A divides $x^2-8x+15$ (this is true by definition).
Since the roots of the minimal polynomial are exactly the eigenvectors, one can show that 3 and 5 are the only possible eigenvectors.

So if n>2, then trace(A)>8, and a technique as in the OP shows that the matrix is diagonalizable. So only n=2 remains.

Hold on, I get (after some research) that 3 and 5 are the only possible eigenvalues, but how would you know that A has n eigenvalues?

Btw, for n=2 we can solve the system, finding either the 2 eigenvalues {3, 5}, meaning A is diagonizable, or finding 2 specific solutions for A, which are diagonal matrices.

 

[micromass]

Hold on, I get (after some research) that 3 and 5 are the only possible eigenvalues, but how would you know that A has n eigenvalues?

It has n eigenvalues up to multiplicity
Hmm, the case n=1 and n=2 is easy. But the higher order cases still elude me...

 

[I like Serena]

It has n eigenvalues up to multiplicity
Hmm, the case n=1 and n=2 is easy. But the higher order cases still elude me...

Now that I think about it, any nxn matrix has n eigenvalues, although they may be complex.
Since 3 and 5 are the only possible eigenvalues, the possibly complex eigenvalues would have to be either 3 or 5 as well!

 

[micromass]

Since it has eigenvalues it presumambly has a jordan normal form and is thus similar to a diagonal matrix?

Ah, you have seen Jordan normal forms?? Good!

No, a Jordan normal form is in general not a diagonal matrix. But in this case it is!

Remember that the sizes of the largest Jordan block is the multiplicity of the eigenvalue in the minimal polynomial. And here the multiplicity is always one, thus the Jordan blocks all have size 1.

 

[I like Serena]

Since it has eigenvalues it presumambly has a jordan normal form and is thus similar to a diagonal matrix?

A Jordan normal form is not necessarily similar to a diagonal matrix.
If the Jordan normal form contains a block, it isn't.

Edit: aarrrrgggh

 

[I like Serena]

Remember that the sizes of the largest Jordan block is the multiplicity of the eigenvalue in the minimal polynomial. And here the multiplicity is always one, thus the Jordan blocks all have size 1.

Huh? :uhh:

If an eigenvalue has a multiplicity of m, that only means it occurs m times in the Jordan normal form in blocks of size 1.
However, if the minimal polynomial would have 2 complex (conjugate) roots, that would correspond to a block of size 2.

 

[Maybe_Memorie]

Ah, you have seen Jordan normal forms?? Good!

No, a Jordan normal form is in general not a diagonal matrix. But in this case it is!

Remember that the sizes of the largest Jordan block is the multiplicity of the eigenvalue in the minimal polynomial. And here the multiplicity is always one, thus the Jordan blocks all have size 1.

Sorry, I was getting diagonal confused with triangular!
Right, so since we need to show that A is similar to a diagonal matrix, we need to show that it has n distinct eigenvalues. Yes?
But I;m confused as I thought the only possible eigenvalues were 3 and 5?

 

[micromass]

Huh? :uhh:

If an eigenvalue has a multiplicity of m, that only means it occurs m times in the Jordan normal form in blocks of size 1.
However, if the minimal polynomial would have 2 complex (conjugate) roots, that would correspond to a block of size 2.

Uuh, I'm not sure if we're talking about the same Jordan normal form here. If we haver 2 conjugate complex roots, then we have two blocks: one for each root.

If x has multiplity m, then we know that there is a Jordan block of size m. Or am I totally missing the point here?

 

[micromass]

Sorry, I was getting diagonal confused with triangular!
Right, so since we need to show that A is similar to a diagonal matrix, we need to show that it has n distinct eigenvalues. Yes?
But I;m confused as I thought the only possible eigenvalues were 3 and 5?

No, you're confusing things. If we have n distinct eigenvalues, then it is diagonalizable. But the converse is not true. So we can even have only one eigenvalue, but still be diagonalizble...

 

[Maybe_Memorie]

I'm not sure how you're getting for n>2 tr(A)>8

 

[micromass]

I'm not sure how you're getting for n>2 tr(A)>8

It doesn't really matter anymore...

But recall that the trace is the sum of the eigenvalues. The only eigenvalues that you can sum are 3 and 5. And if you add up three numbers in {3,5}, then you get something >8.

But anyway, that doesn't help us further.

What is the relationship between the minimal polynomial and the Jordan blocks?

 

[I like Serena]

Uuh, I'm not sure if we're talking about the same Jordan normal form here. If we haver 2 conjugate complex roots, then we have two blocks: one for each root.

If x has multiplity m, then we know that there is a Jordan block of size m. Or am I totally missing the point here?

My bad, I need to brush my knowledge of Jordan normal forms off.
Not enough lectures (lately) I guess.

I've just been reading up on this knowledge and found that an eigenvalue with multiplicity m does not mean there is a block of size m necessarily.
There could be 1 up to m blocks.
What we need, is that there are m blocks of size 1, which is the precondition for the matrix to be diagonizable (that is, all blocks in the matrix have to be size 1).

This also means we're not done yet.

 

[micromass]

I've just been reading up on this knowledge and found that an eigenvalue with multiplicity m does not mean there is a block of size m necessarily.

To my knowledge, it is true. The multiplicity in the minimal polynomial is the size of the largest Jordan block. Let me find a reference.

 

[micromass]

Wiki knows all:

Minimal polynomial

The minimal polynomial of a square matrix A is the unique monic polynomial of least degree, m, such that m(A) = 0. Alternatively, the set of polynomials that annihilate a given A form an ideal I in C[x], the principal ideal domain of polynomials with complex coefficients. The monic element that generates I is precisely m.

Let λ1, ..., λq be the distinct eigenvalues of A, and si be the size of the largest Jordan block corresponding to λi. It is clear from the Jordan normal form that the minimal polynomial of A has degree ∑si.

While the Jordan normal form determines the minimal polynomial, the converse is not true. This leads to the notion of elementary divisors. The elementary divisors of a square matrix A are the characteristic polynomials of its Jordan blocks. The factors of the minimal polynomial m are the elementary divisors of the largest degree corresponding to distinct eigenvalues.

The degree of an elementary divisor is the size of the corresponding Jordan block, therefore the dimension of the corresponding invariant subspace. If all elementary divisors are linear, A is diagonalizable.

 

[Maybe_Memorie]

What is the relationship between the minimal polynomial and the Jordan blocks?

The degree of the minimal polynomial tells us the amount of times each eigenvalue appears in a jordan block.. It's difficult for me to explain. For example,
if the characteristic polynomial was (t-3)^3 and the minimal polynomial was also (t-3)^3 then the jordan normal form would be

3 1 0
0 3 1
0 0 3

but if the minimal polynomial was (t-3)^2 the jordan normal form is

3 1 0
0 3 0
0 0 3

If the minimal polynomial is only of degree 1 then the jordan normal form must be diagonal.

 

[I like Serena]

The degree of the minimal polynomial tells us the amount of times each eigenvalue appears in a jordan block.. It's difficult for me to explain. [...]

Soon I will not be helping you any more... you will have to help me!

 

[I like Serena]

Well, I believe we have the answer for (a), don't you?

As for (b), I believe we can use the fact that any diagonal matrix with only 3's and 5's on the diagonal should be a solution for the equation...

 

[micromass]

Well, I believe we have the answer for (a), don't you?

As for (b), I believe we can use the fact that any diagonal matrix with only 3's and 5's on the diagonal should be a solution for the equation...

Aah, thus (b) turns out to the following problem: for each k>8, prove that k can be written as the sum of 3's and 5's.

This should be fun

 

[I like Serena]

Aah, thus (b) turns out to the following problem: for each k>8, prove that k can be written as the sum of 3's and 5's.

This should be fun

Hey! What about k=8?

 

[micromass]

Hey! What about k=8?

Aaaaah, of course, k=8 is the most important one!

Hint for the OP: first try to write k=8,9,...,15 as the sum of 3's and 5's. The rest of the numbers are easy (so I claim)

 

[Maybe_Memorie]

I'm confused as to what k actually is