Eigenvalue of a special matrix

Question

A = \begin{vmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{vmatrix}

I want to find the eigenvalue of this matrix A. I know how to find its eigenvalues by using rule of Sarrus, and they are 2,1 and 0. But I believe there is definitely a easier method to get its eigenvalue.

Thank you guys.

Answer 1

This is one specific case where eigenvectors are easier to spot. Note that $A$ sends the vector $(x,y,z)^T$ to $(x+z, y, x+z)^T$. In the special case $x=z=0$, namely the vector $e_2$ goes to itself, so 1 is an eigenvalue. Another special case $x=z, y=0$, for example $(1,0,1)^T$ goes to $(2,0,2)^T$ which is twice the input, hence 2 is an eigenvalue. Also the matrix is singular (two of its rows are same) and so 0 is an eigenvalue.

Answer 2

Hint: use the characteristic polynomial approach

$$|A - \lambda I | = 0 \implies -\lambda ^3+3 \lambda ^2-2 \lambda = 0 \implies -(\lambda -2) (\lambda -1) \lambda = 0 \implies \lambda_{1,2,3} = 0, 1, 2 $$

If you are looking for iterative methods, see the list of methods.

Answer 3

Since it seems you want some method for finding the eigenvalues without using what Amzoti did, here is one approach.

It is rather quick to see that $$\left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}\right]\left[\begin{matrix} a \\ b \\ c \end{matrix}\right]=\left[ \begin{matrix} a+c \\ b \\ a+c \end{matrix}\right]$$ Then we want $\lambda a= a+c$, $\lambda b=b$, $\lambda c=a+c$. The only solutions to the middle equation are $\lambda=0,1$ or $b=0$; combining the first and last tells us that $a=c$ when $\lambda\neq 0$, and thus $2a=\lambda a$. When $a,c\neq 0$, this gives us the final eigenvalue 2.