2n x 2n matrices without real eigenvalues

In summary: For an arbitrary positive integer ##n##, give a ##2n## x ##2n## matrix ##A## with non-real eigenvalues.
  • #1
pyroknife
613
3

Homework Statement


For an arbitrary positive integer ##n##, give a ##2n## x ##2n## matrix ##A## without real eigenvalues.

Homework Equations

The Attempt at a Solution


First of all, I am having some trouble interpreting this problem. I do not know if it is generalized where I am supposed to find a ##2n## x ##2n## matrix A without real eigenvalues for ANY ##n##, or just for one specific ##n## value, which I can pick. I assume it is the former.

Since it is 2*n, we know that the number of rows/columns is even for any ##n##. If it was only ##n##, without it being multiplied it by 2, then the number of rows/columns could be odd, and thus there will always be at least one real eigenvalue.

Since there's an even # of rows/columns, it is possibly that matrix ##A## not have any real eigenvalues.

This is about as far as I got and I am having a very hard time figuring out how to generalize this problem for any arbitrary value of ##n##.
 
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  • #2
pyroknife said:

Homework Statement


For an arbitrary positive integer ##n##, give a ##2n## x ##2n## matrix ##A## without real eigenvalues.

Homework Equations

The Attempt at a Solution


First of all, I am having some trouble interpreting this problem. I do not know if it is generalized where I am supposed to find a ##2n## x ##2n## matrix A without real eigenvalues for ANY ##n##, or just for one specific ##n## value, which I can pick. I assume it is the former.

The former is correct.

Since it is 2*n, we know that the number of rows/columns is even for any ##n##. If it was only ##n##, without it being multiplied it by 2, then the number of rows/columns could be odd, and thus there will always be at least one real eigenvalue.

Since there's an even # of rows/columns, it is possibly that matrix ##A## not have any real eigenvalues.

This is about as far as I got and I am having a very hard time figuring out how to generalize this problem for any arbitrary value of ##n##.

I would start small. Can you do it for 2x2 matrix? A 4x4? Maybe that will give you some ideas.
 
  • #3
LCKurtz said:
The former is correct.
I would start small. Can you do it for 2x2 matrix? A 4x4? Maybe that will give you some ideas.
2x2 matrix would be simple as the following would be one example
##\begin{bmatrix}
0 & 1\\
-1 & 0
\end{bmatrix}##

I am trying to think of a 4x4.
 
  • #4
Good so far. Think of that as a building block for your 4x4.
 
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  • #5
LCKurtz said:
Good so far. Think of that as a building block for your 4x4.
I think part of the problem I am having is writing down the determinant of ##(A-\lambda I)## for larger systems.
 
  • #6
Well, lots of zero's helps. If your matrix were diagonal its determinant would be easy, but that won't work. Think in terms of diagonal blocks.
 
  • #7
LCKurtz said:
Well, lots of zero's helps. If your matrix were diagonal its determinant would be easy, but that won't work. Think in terms of diagonal blocks.
Okay let's see...
I'm not too familiar with the term "diagonal blocks."
I assume I can stack the 2 x 2 matrix I previously had along the tri-diagonal of a 4 x 4 matrix such that it looks like:
##\begin{bmatrix}
0 & 1 & 0 & 0\\
-1 & 0 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & -1 & 0
\end{bmatrix}##

I just checked and this has all imaginary eigenvalues.
 
  • #8
Ahh yes, it looks like I can place the original 2x2 matrix I had along the tri-diagonal band of any ##2n## x ##2n## matrix and obtain a matrix with only non-real eigenvalues.
 
  • #9
Heh heh. Looks like you are on to something, eh? You can probably find some theorems about determinants of matrices such as this. They are called block matrices.
 
  • #10
I have to hit the sack. I'm guessing you can take it from here. If not, see you tomorrow.
 
  • #12
LCKurtz said:
I have to hit the sack. I'm guessing you can take it from here. If not, see you tomorrow.
Thanks! I'm just trying to figure out how to write this succintly.
 
  • #13
Could you think of physical simultaneous linear d.e.s that have to give 4 nonreal eigenvalues (vibrational modes) and get their eigenvalue equation, can choose to try find simplest or find most general?
 
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  • #14
LCKurtz said:
I have to hit the sack. I'm guessing you can take it from here. If not, see you tomorrow.
Hey LCKurtz, is there some terminology of method to succinctly write these matrices for an arbitrary n?
What I came up with is a bit wordy...
 
  • #15
epenguin said:
Could you think of physical simultaneous linear d.e.s that have to give 4 nonreal eigenvalues (vibrational modes) and get their eigenvalue equation, can choose to try find simplest or find most general?
Hmmm, I have not thought about placing this problem in the context of differential equations. I can see how that would help, but I have only just touched the surface in both subjects (linear algebra and differential equations) that connecting the two is bit outside of my knowledge.
 
  • #16
pyroknife said:
Hey LCKurtz, is there some terminology of method to succinctly write these matrices for an arbitrary n?
What I came up with is a bit wordy...
Look here:
http://en.wikipedia.org/wiki/Block_matrix#Block_diagonal_matrices

and look at the first picture under Block Diagonal Matrices (3 in the contents). You could probably work with something like that.
 
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  • #18
pyroknife said:
Hmmm, I have not thought about placing this problem in the context of differential equations. I can see how that would help, but I have only just touched the surface in both subjects (linear algebra and differential equations) that connecting the two is bit outside of my knowledge.

A connection to remember to make later. I wonder if linear algebra as "subject" would be on most scientists' curriculum without it?
 
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1. What are 2n x 2n matrices without real eigenvalues?

2n x 2n matrices without real eigenvalues are square matrices with a dimension of 2n, where n is a positive integer, and they do not have any real eigenvalues. This means that there are no real numbers that can be multiplied by the matrix to give a result that is equal to the original matrix multiplied by that number.

2. Can 2n x 2n matrices without real eigenvalues have complex eigenvalues?

Yes, 2n x 2n matrices without real eigenvalues can have complex eigenvalues. This means that the eigenvalues of the matrix are complex numbers, which have a real part and an imaginary part. However, these matrices do not have any real eigenvalues, which means that the real part of the complex eigenvalues is equal to 0.

3. How do we determine if a 2n x 2n matrix has real eigenvalues or not?

To determine if a 2n x 2n matrix has real eigenvalues or not, we can use the characteristic polynomial of the matrix. If the coefficients of the polynomial are all real numbers and the polynomial has no real roots, then the matrix does not have any real eigenvalues. Alternatively, we can also use the fact that the determinant of a matrix with real entries and complex eigenvalues is always a complex number.

4. What is the significance of 2n x 2n matrices without real eigenvalues?

2n x 2n matrices without real eigenvalues are important in the study of complex numbers and complex analysis. They are also used in various applications in physics, engineering, and computer science. These matrices can represent systems with non-real solutions and help us understand and analyze complex systems and phenomena.

5. Can 2n x 2n matrices without real eigenvalues be diagonalized?

No, 2n x 2n matrices without real eigenvalues cannot be diagonalized. Diagonalization is a process of transforming a matrix into a diagonal matrix, where all the non-diagonal elements are equal to 0. However, since these matrices do not have any real eigenvalues, they cannot be transformed into a diagonal matrix, which requires the existence of real eigenvalues.

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