How to find eigen values of such kind of matrices ( Matrices which are bigger than $ 3\times 3$)

Question

The following question is part of a masters exam for which I am preparing and I don't have any methodology on how this type of matrices are solved.

This was question asked:

I don't know how to find eigenvalues of matrices ( larger than 3 $\times$ 3).

So, it's my humble request if you can tell some details or source of results and observations which are useful for dealing with such matrices.

I have studied linear algebra from David C lay.

Thanks a lot!!

Answer 1

Let $A$ be that matrix. Then, we have $$A(x_1,x_2,x_3,x_4,x_5,x_6)^T = (x_4,x_5,x_6,x_1,x_2,x_3)^T,$$ i.e. $A$ acts on a vector by swapping the 1st/4th entries, the 2nd/5th entries, and the 3rd/6th entries.

So for a vector $v$ to be an eigenvector of $A$, swapping the 1st/4th entries, the 2nd/5th entries, and the 3rd/6th entries of $v$ should yield a scalar multiple of $v$. As an example, $$A(1,0,0,1,0,0)^T = (1,0,0,1,0,0)^T,$$ so $(1,0,0,1,0,0)^T$ is an eigenvector with eigenvalue $1$. Can you find six linearly independent eigenvectors and their corresponding eigenvalues?

Just a note: In general, it is hard/tedious to find the eigenvalues of an arbitrary large matrix. However, if a large matrix has a "pattern" to the entries, it may be possible to find the eigenvectors/eigenvalues "by inspection". That's probably what this question is expecting you to do.

Answer 2

Denote the given $6 \times 6$ matrix by $A$.

Hint One option is to observe that the square of the given matrix is the identity matrix, or, rearranging, that $A^2 - I = 0$. So, the minimal polynomial of $A$ divides $t^2 - 1 = (t + 1) (t - 1)$. (Alternatively, any eigenvector $v$ of $A$, say of eigenvalue $\lambda$, satisfies $v = A^2 v = \lambda^2 v$.)

Hint for another method Consider as usual the polynomial $\lambda I - A$; the first and fourth columns of this matrix are, respectively $$\pmatrix{\lambda\\\cdot\\\cdot\\-1\\\cdot\\\cdot}, \pmatrix{-1\\\cdot\\\cdot\\\lambda\\\cdot\\\cdot}.$$ What values of $\lambda$ make these two columns linearly dependent?

Remark Both of these methods are specific to this matrix and ones that have a similar structure. Like JimmyK45542 wrote in his good answer, the question is really testing your ability to find and execute a method for computing those eigenvalues that takes advantage of the features that make the given matrix special.

Answer 3

In addition to the existing answers I'd like to say that it's not hard to calculate the characteristic polynomial using Laplace expansion:

\begin{align}p(x)&=\det\begin{bmatrix} x & 0 & 0 & -1 & 0 & 0\\ 0 & x & 0 & 0 & -1 & 0\\ 0 & 0& x& 0 & 0& -1\\ -1 & 0 &0&x& 0 & 0\\ 0 & -1 & 0 & 0&x &0\\ 0 & 0& -1 & 0 & 0 &x \end{bmatrix}\\ &=x\det\begin{bmatrix} x & 0 & 0 & -1 & 0\\ 0& x& 0 & 0& -1\\ 0 &0&x& 0 & 0\\ -1 & 0 & 0&x &0\\ 0& -1 & 0 & 0 &x \end{bmatrix}+1\det\begin{bmatrix} 0 & 0 & -1 & 0 & 0\\ x & 0 & 0 & -1 & 0\\ 0& x& 0 & 0& -1\\ -1 & 0 & 0&x &0\\ 0& -1 & 0 & 0 &x \end{bmatrix}\\ &=x\cdot x\det\begin{bmatrix} x & 0 & -1 & 0\\ 0& x& 0 & -1\\ -1 & 0 &x &0\\ 0& -1 & 0 &x \end{bmatrix}-1\det\begin{bmatrix} x & 0 & -1 & 0\\ 0& x & 0& -1\\ -1 & 0 &x &0\\ 0& -1 & 0 &x \end{bmatrix}\\ &=(x^2-1)\det\underbrace{\begin{bmatrix} x & 0 & -1 & 0\\ 0& x & 0& -1\\ -1 & 0 &x &0\\ 0& -1 & 0 &x \end{bmatrix}}_{=M} \end{align} Now it's your turn to prove $\det(M)=(x^2-1)^2$ using the same method (Laplace expansion) as above.

The final result is

$p(x)=(x^2-1)^3$