find the eigenvalue of A ???

Question

Let $A$ be a $10 \times 10$ matrix defined by $A=(a_{ij})$ where $(a_{ij})=1-(-1)^{i-j}$

$$A= \begin{pmatrix} 0 & 2 & 0 & 2 & \cdots & 0 &2 \\ 2 & 0 & 2 & 0 & \cdots &2 &0 \\ \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 2 & 0 & 2 & \cdots &0 &2\\ 2 & 0 & 2 & 0 & \cdots &2 &0\\ \end{pmatrix}$$ Now find the eigenvalue of $A$ ???

My attempts : i can find the eigenvalue of $2\times 2 $ and $3\times 3 $matrixes ..But now i can not able to find the eigenvalue of $10\times 10$ matrixes how can i find ???

pliz help me

Answer 1

I will provide an alternate formulation, one that may be more useful in other problems but is also useful here.

Your matrix happens to be this matrix $M = \begin{pmatrix} 0 \quad 2 \\ 2 \quad 0 \end{pmatrix}$, just repeated five times across and five times down. There is actually an operation to describe this, the Kronecker product. Look it up on Wikipedia or Wolfram Mathworld.

In particular, if $\mathbf{[1]}_5$ is the $5 \times 5$ matrix consisting of all $1$s, then the matrix that you have, is nothing but: $\mathbf{[1]}_5 \otimes M$.

Here's the big result :

The set of eigenvalues of $A \otimes B$ is formed by taking all possible products of eigenvalues of $A$ with eigenvalues of $B$.

So, the recipe is simple : find all the eigenvalues of $\mathbf{[1]}_5$, all the eigenvalues of $M$, and take products.

The point about $\mathbf{[1]_5}$ is that even though it is $5 \times 5$, there is some pattern to its characteristic polynomial, which one may derive by induction. For example, just look at square matrices of small dimensions with all $[1]_s$.

The characteristic polynomial of the $1 \times 1$ matrix $[1]$ is $x - 1$. For $\mathbf{[1]}_2$ it is $x(x-2)$, for $\mathbf{[1]_{3}}$ it is $x^2(x-3)$. In fact, you can prove that it is $x^{n-1}(x-n)$ for $[1]_n$.

Hence, the eigenvalues of $[1]_5$ are the roots of $x^4(x-5) = 0$ i.e. the eigenvalues are $0$ and $5$.

The eigenvalues of $M$ can be found from its characteristic polynomial, which is $x^2 - 4 = 0$ so $x = \pm 2$.

You can verify from here : http://www.wolframalpha.com/input/?i=Kronecker+product+%5B%5B%5B1,1,1,1,1%5D,%5B1,1,1,1,1%5D,%5B1,1,1,1,1%5D,%5B1,1,1,1,1%5D,%5B1,1,1,1,1%5D%5D,%5B%5B0,2%5D,%5B2,0%5D%5D%5D

That the eigenvalues of matrix you want are : $0 \times 2$ , $0 \times -2$, $5 \times 2$ and $5 \times -2$, which gives the set $\{0,10,-10\}$.

Answer 2

HINT : What is the rank of your matrix ? What about $X=\left(1,1,1,1,1,1,1,1,1,1\right)$.

Answer 3

The rank of $A$ is clearly $2$. Besides, $(1,1,1,1,1,1,1,1,1,1)$ and $(1,-1,1,-1,1,-1,1,-1,1,-1)$ are eigenvectors of $A$ with eigenvalue $10$ and $-10$ respectively. So, the eigenvalues are $10$ (with multiplicity $1$), $-10$ (also with multiplicity $1$) and $0$ (with multiplicity $8$).