Determinant #2

G

George Saliaris

Junior Member
Joined
Dec 15, 2019
Messages
53
I am practicing for my exams in Linear Algebra and I came across this determinant...



I noticed that A (the matrix which has the determinant above) = 2B + I
where B has '1's in every column and row.I also got that B^2 = nB ,so if λ is an eigenvalue of B then λ^2 is an eigenvalue of Β^2.So λ = 0 οr λ = n..
Also,if C = 2B then if d is an eigenvalue of C,then d+1 is an eigenvalue of C+I..We also have that the determinant of a matrix is equal to the product of its eigenvalues..So maybe I am supposed to find the eigenvalues of I in terms of the eigenvalues of C,I ? If so,I do not know the multiplicity of each eigenvalue...How can I continue?
 

Attachments

  • Screenshot_9.png
    Screenshot_9.png
    115.7 KB · Views: 11
You screenshot shows the matrix but not the question. I am assuming you need to find eigenvalues of A, right?
 
You've proven that B can only have [imath]n[/imath] and 0 for eigenvalues, from which you can get eigenvalues of [imath]A[/imath]. You've probably worked out the eigenvalues for [imath]A[/imath], but haven't stated explicitly in your post.
As for multiplicity, I would look at eigenvectors of [imath]B[/imath]: there are [imath]n-1[/imath] eigenvectors with eigenvalue of 0 and only one with eigenvalue of [imath]n[/imath] --can you see it?
 
blamocur said:
You've proven that B can only have [imath]n[/imath] and 0 for eigenvalues, from which you can get eigenvalues of [imath]A[/imath]. You've probably worked out the eigenvalues for [imath]A[/imath], but haven't stated explicitly in your post.
As for multiplicity, I would look at eigenvectors of [imath]B[/imath]: there are [imath]n-1[/imath] eigenvectors with eigenvalue of 0 and only one with eigenvalue of [imath]n[/imath] --can you see it?
Click to expand...
Is this because the order of B is 1? (so there are n-1 eigenvectors with the eigenvalue 0,If so we have not learned that)...What I want to do is to calculate the product abcd... (where a,b,c,d,...... eigenvalues of A and I have found that if λ is an eigenvalue of B then 2λ+1 is also an eigenvalue of A)..But how can I get the multiplicity of each eigenvalue in B and A? Are they of same multiplicity or something else?
 
If [imath]B[/imath] has 1 eigenvalue of [imath]n[/imath] and [imath]n-1[/imath] eigenvalues of 0 then [imath]2B+I[/imath] has 1 eigenvalue of [imath]2n+1[/imath] and [imath]n-1[/imath] eigenvalues of 1, right ?

But what is your final goal? Finding the determinant or the eigenvalues?
 
The determinant can be computed without resorting to eigenvalues. I've managed to transform [imath]A[/imath] to an upper triangular form, at which point the determinant is simply the product of the diagonal elements.
 
blamocur said:
The determinant can be computed without resorting to eigenvalues. I've managed to transform [imath]A[/imath] to an upper triangular form, at which point the determinant is simply the product of the diagonal elements.
Click to expand...
To be more precise, consider the case [imath]n=4[/imath] and define two matrices, both of which have determinants of 1:

[math]U = \left (\begin{array}{rrrrr} 1 & 0 & 0 & 0 &\\ 0 & 1 & 0 & 0 &\\ 0 & 0 & 1 & 0 &\\ -2 & -2 & -2 & 1 &\\ \end{array}\right) V = \left (\begin{array}{rrrrr} 1 & 0 & 0 & -1 &\\ 0 & 1 & 0 & -1 &\\ 0 & 0 & 1 & -1 &\\ 0 & 0 & 0 & 1 &\\ \end{array}\right)[/math]Since [imath]\det(U) = \det(V) = 1[/imath] we know that [imath]\det(A) = \det (UVA)[/imath], but [imath]UVA[/imath] is an upper-triangular matrix whose determinant is the product of its diagonal elements.
Of course this can be defined for any [imath]n[/imath], not just 4 -- can you work out the rest?

P.S. Note that multiplication by those matrices is equivalent to subtracting the rows (scaled by 2 in case of [imath]U[/imath]) of the matrix, somewhat similar to the Gaussian elimination used for solving systems of linear equations. This is how I worked out the solution, but writing out the matrices seems simpler than describing the algorithm.
 
Have not done the calculations but what you mentioned might work..The thing is how could somebody come up with these matrices?
(Also Gaussia elimination was the first thing I tried but It did not get me far..
 
George Saliaris said:
Have not done the calculations but what you mentioned might work..The thing is how could somebody come up with these matrices?
(Also Gaussia elimination was the first thing I tried but It did not get me far..
Click to expand...
I started with Gaussian elimination, but the matrices provide more concise notation. But might find it easier to formally prove the result by using the elimination instead of the matrices.
 
George Saliaris said:
How did the Gaussian elimination worked for you?
Click to expand...
I used a process similar to the Gaussian elimination. For example, the [imath]VA[/imath] operation is equivalent to subtracting the last row of [imath]A[/imath] from every other row. This does not convert [imath]A[/imath] to the diagonal form yet, but the next step, i.e., multiplying by [imath]U[/imath], finishes the process.
 
Right.I understand the 1st part (the VA one)...But as I asked before,how could somebody come up with U?
(Also the exam is tomorrow and I have not figured a way .-.)
 
George Saliaris said:
Right.I understand the 1st part (the VA one)...But as I asked before,how could somebody come up with U?
(Also the exam is tomorrow and I have not figured a way .-.)
Click to expand...
The way I've figure ti out was by looking at [imath]VA[/imath], I.e., after the first round of elimination. Do you want to show what you get for [imath]VA[/imath] for small [imath]n[/imath],
e.g. [imath]n=4[/imath]?
 
George Saliaris said:
I get that VA = V (if my calculations are correct)...So detA = 1 ?
Click to expand...
This does not sound right. Since 'V' is invertible this would mean that [imath]A = I[/imath], i.e. [imath]A[/imath] would have to be an identity, which it is not. Want to try again?
 
I suggest you brush up on matrix multiplication. Meanwhile, here is what I get for VA:
[math]V\cdot A = \left( \begin{array}{rrrr} 1 & 0 & 0 & -1\\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 \\ \\ \end{array} \right) \cdot \left( \begin{array}{rrrr} 3 & 2 & 2 & 2\\ 2 & 3 & 2 & 2 \\ 2 & 2 & 3 & 2 \\ 2 & 2 & 2 & 3 \\ \\ \end{array} \right) = \left( \begin{array}{rrrr} 1 & 0 & 0 & -1\\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \\ 2 & 2 & 2 & 3 \\ \\ \end{array} \right)[/math]
You can now figure out [imath]U(VA)[/imath].
 
...I got the first 2 rows of VA same as V and then I got bored of the multiplication and assumed it would be the same .-.
 
You must log in or register to reply here.
Top