Calculating the eigenvalues of an interesting matrix

[JohnLang]

Homework Statement

Let T be a square nxn matrix and let each entry t_ij=n(i-1)+j.

Calculate the eigenvalues of T of T when T is a 3x3 matrix and a 4x4 matrix.

Homework Equations

The Attempt at a Solution

For the homework I'm supposed to calculate the 3x3 case and the 4x4 case. Then it says find the eigenvalues for the general nxn case for extra credit.From MATLAB I'm able to see that the matrix T has rank 2 (If this helps). I'm sort of stuck on this part.

On MatLab when I've done up to order 6, each from order 3,4,5,6 have 2 non-zero eigenvalues and the other eigenvalues are 0 with algebraic multiplicity n-2

 

[JohnLang]

I feel like I need to exploit some properties of determinants.

 

[I like Serena]

Homework Statement

Let T be a square nxn matrix and let each entry t_ij=n(i-1)+j.

Calculate the eigenvalues of T of T when T is a 3x3 matrix and a 4x4 matrix.

Homework Equations

The Attempt at a Solution

For the homework I'm supposed to calculate the 3x3 case and the 4x4 case. Then it says find the eigenvalues for the general nxn case for extra credit.From MATLAB I'm able to see that the matrix T has rank 2 (If this helps). I'm sort of stuck on this part.

On MatLab when I've done up to order 6, each from order 3,4,5,6 have 2 non-zero eigenvalues and the other eigenvalues are 0 with algebraic multiplicity n-2

Welcome to PF, JohnLang!

To find the eigenvalues typically you need to solve the characteristic polynomial.
Have you tried to find the characteristic polynomial?

You can already observe some of its properties.
Since the rank of each matrix is 2, what does that mean for the form of the characteristic polynomial?

In particular you will need a way to calculate the second and third coefficient of the characteristic polynomial.
Do you know how to find those?
You can find some more information for instance here: http://en.wikipedia.org/wiki/Characteristic_polynomial

Btw, can you deduce that the rank is always 2?
To find the rank of a matrix, the typical way to do it, is to find how many independent column vectors there are.
This can be done by repeatedly subtracting one column from the other until no more columns become zero.

 

[JohnLang]

Hi, I Like Serena!

Thanks for welcoming me!

Anyway, so far I think I was able to show that the rank of the matrix which is 2 I used a really rough method

so I think what this means for the characteristic polynomial, it's going to look something like this

p(t)=t^n-2(t²-bt+c)

I'm kinda stuck on finding the coefficients I'm not quite sure. I calculated the tr(A)=(n³+n)/2

 

[Robert1986]

I'd try to see if you can somehow figure out a general way to row-reduce it into its Jordan Form or into a triangular form.

 

[JohnLang]

Robert:

So when I row reduced this matrix it looks something like this

[1 0 -1 -2 -3 ... 1-n;0 1 2 3 4 5 ... n-1;(remaining rows are 0)

which means that this matrix has rank 2

from here what should I do?

Also since the matrix T has rank 2, this means that there are non zero principal (2x2) minors from the referred literature from I Love Serena. I think that's how it's worded.

 

[I like Serena]

so I think what this means for the characteristic polynomial, it's going to look something like this

p(t)=t^n-2(t²-bt+c)

I'm kinda stuck on finding the coefficients I'm not quite sure. I calculated the tr(A)=(n³+n)/2

Yep. That looks correct.
In particular you have b=tr(A) and c is the sum of the principle minors.
And I see you already found a general formula for the trace of A.
Good!

That leaves a general formula for the sum of the principle minors.
But for the first part you don't need a general formula yet, you don't even have to use principle minors.
You just need the value of c for the 3x3 and the 4x4 matrix.
With that you can solve the characteristic polynomial and find the eigenvalues.

Robert:

So when I row reduced this matrix it looks something like this

[1 0 -1 -2 -3 ... 1-n;0 1 2 3 4 5 ... n-1;(remaining rows are 0)

which means that this matrix has rank 2

from here what should I do?

That's enough.
This proves that the rank is 2, meaning the characteristic polynomial has the form you already mentioned.

What's left is to find b and c in all cases and solve it for the eigenvalues.

After that you can try to find a general formula for c.
But let's take one step at a time.

 

[I like Serena]

what do you mean by finding b and c in all cases?

Well, your problem asks for the eigenvalues of the 3x3 matrix and the 4x4 matrix that are filled with the consecutive numbers 1, 2, 3, etcetera.

So I mean these 2 cases.

Btw, to find the pattern, I'd also do the 2x2 matrix.

 

[JohnLang]

I'm having problems when I'm calculating 4x4 matricies, my principal 3x3 minors are all 0
which makes sense since the rank of A is 2 so in this case rref(A) will have 2 rows of zeros which implies the determinant of A is 0.

Any further suggestions?

 

[I like Serena]

Ah, but the principal minors are not zero.

Can you calculate the following determinant and reorder it into a polynomial?
Then you should see what "c" is in this case.
$$\text{det} \begin{bmatrix}
1 - λ & 2 & 3 \\
4 & 5 - λ & 6 \\
7 & 8 & 9 - λ
\end{bmatrix}$$
If you can, perhaps you can try to keep track how you effectively calculate c?

 

[Robert1986]

Robert:

So when I row reduced this matrix it looks something like this

[1 0 -1 -2 -3 ... 1-n;0 1 2 3 4 5 ... n-1;(remaining rows are 0)

which means that this matrix has rank 2

from here what should I do?

Also since the matrix T has rank 2, this means that there are non zero principal (2x2) minors from the referred literature from I Love Serena. I think that's how it's worded.

If this is the correct row reduction (I suspect there is an error somewhere because the first row has n+1 entries) then it *might* be easy to find two eigenvectors that have different eigenvalues. These will be the two eigenvalues you need (note the eigenvectors are not the eigenvectors of the original matrix.) And, as you said, there are only two non-zero eigenvalues.

 

[JohnLang]

so when I do it this way I get the characteristic polynomial, p(λ)=-λ(λ²-15λ-18)

and that gives me the correct eigenvalues

the principal minors are

[5 6;8 9]

[1 3;7 9]

[1 2;4 5]

the sum of the principal minors give me -18, furthermore I see that the trace is b. but when I do the 4x4 case, the principal minors of the 4x4 matrix are

[6 7 8;10 11 12; 14 15 16]

[1 3 4;9 11 12;13 15 16]

[1 2 4;5 6 8;13 14 16]

[1 2 3;5 6 7;9 10 11]

and all their determinants are 0

I'm saying that the 4x4 char poly is

p(λ)=λ²(λ²-65λ-C)

I guess I'm lost on the C part.

 

[I like Serena]

The 3x3 case is correct.

But the trace of the 4x4 is not 65.
And the principal minors are supposed to be 2x2 matrices (much easier to calculate ;)).

 

[JohnLang]

sorry 34, also if I'm going to calculate the principal minors for 2x2 that means there will be 6 of them for the 4x4 case right? 4 choose 2?

 

[I like Serena]

Yup.

 

[JohnLang]

WOOT! ok the 4x4 case worked, let's see if I can get it going for the general case, at least I know there are n choose 2 principal 2x2 minors

 

[I like Serena]

 

[JohnLang]

is it fair to say that only the first 2 rows of T will be considered in the general case of finding the 2x2 principal submatricies since the rank of T is 2 meaning there are a lot of zero rows?

 

[I like Serena]

Hmm, I don't know.
I don't think so.

Anyway, I'm off to bed. :zzz:

 

[JohnLang]

haha thanks for all the help!

 

[Dick]

is it fair to say that only the first 2 rows of T will be considered in the general case of finding the 2x2 principal submatricies since the rank of T is 2 meaning there are a lot of zero rows?

No, but you can take an alternate route skipping the whole submatrix thing that's kind of like that. Since you know that the rank is 2, if you presume there are two nonzero eigenvalues then their eigenvectors span the range R of T. Further T(R)=R. So you can figure out everything you want to know by analyzing how T maps the two dimensional space R to R. How easy that is depends on how clever you can be about picking a basis for R.

 

[JohnLang]

Hmm, this is interesting, since T has 2 distinct eigenvalues in which I'll presume, furthermore since the other eigenvalues are all zero, is this enough to say that T is diagonalizable? Would that help in me in analyzing how T maps R?

 

[Dick]

Hmm, this is interesting, since T has 2 distinct eigenvalues in which I'll presume, furthermore since the other eigenvalues are all zero, is this enough to say that T is diagonalizable? Would that help in me in analyzing how T maps R?

That's too abstract to really help. You do have to get your hands dirty. Take the 3x3 case. You want to pick two basis vectors for R that are easy to work with. Try messing around.

 

[JohnLang]

btw each of the non zero eigenvalues have a corresponding eigenvector because geometric multiplicity is always less than or equal to algebraic multiplicity right?

 

[Dick]

btw each of the non zero eigenvalues have a corresponding eigenvector because geometric multiplicity is always less than or equal to algebraic multiplicity right?

Well, every eigenvalue has at least one eigenvector. But I wouldn't worry too much about proving these things exist. The idea here is to actually find the eigenvalues and potentially eigenvectors. At which point there really won't be any question they exist.

 

[Dick]

Don't give up on this, ok? I spent some time on this and hate to see it go to waste. R is spanned by the column vectors of T. You can tough it out by choosing any two columns of T and crunching the numbers. But it's better if you find a clever basis. Think about linear combinations of the columns. Here's a big hint. For the 3x3 case pick v=[-1,0,1] and use the basis {v,T(v)}. Try to find the 2x2 matrix of T over R in that basis. You can generalize this without it turning into a complete nightmare for the nxn case.

 

[JohnLang]

OK I'll give it a shot. I did give up on the problem because it seemed really annoying

 

[Dick]

OK I'll give it a shot. I did give up on the problem because it seemed really annoying

I'm going to peddle some more hints, since I've got nothing else to do with them. T acts on the subspace R by mapping v->T(v) and T(v)->T(T(v)). T(v) is a pretty simple vector, work it out. This is the 'clever choice of basis' bit. Now you can find the 2x2 matrix M=[[a,b],[c,d]] of T over R by solving T(v)=a*v+b*T(v) and T(T(v))=c*v+d*T(v). It's pretty easy, in fact the first equation is trivial. The charcteristic polynomial of M is your missing quadratic factor in the characteristic polynomial of T. Interested again?

 

[lurflurf]

An interesting thing about this matrix is that
(let tr(M) be the trace of the matrix M)

$$tr(A^T A)=\sum_{k=1}^{n^2} k^2=\frac{1}{6} n^2(n^2+1)(2n^2+1)$$

though that does not appear to be of direct use.

If x and y are the non zero eigenvalues we have
x+y=tr(A)
x^2+y^2=tr(A^2)
xy=(1/2)(tr(A)^2-tr(A^2))

where the we can predict the above are polynomials in n of degrees 3, 6, and 5 respectively.

From which the characteristic polynomial and x and y are easily found.

As a side note I am also pretty sure there is a slick way to do this using a change of basis or row reduction, and a way using minor expansion.

 

[I like Serena]

For the extra credit of the problem, the only thing left is a general expression for the "b" and the "c" in the characteristic polynomial.
The coefficient "b" is the trace, while "c" is the sum of the 2x2 minor principals.

In other words:
$$b = \textrm{tr}(t_{ij}) = \sum_i t_{ii} = t_{11} + t_{22} + ... + t_{nn}$$
$$c = \sum_i \sum_{j > i} \det(\textrm{principal submatrix of i and j}) = \sum_i \sum_{j > i} t_{ii} t_{jj} - t_{ij} t_{ji}$$
You already found a general expression for the trace.
Now if you substitute the expression that is given for $t_{ij}$, and write out the summation, you should be able to find a general expression.