Finding the eigenvalue for a matrix

Question

I have the matrix of $A = \begin {pmatrix} 2 && 1 && 1 \\ 1 && 2 && 1 \\ 1 && 1 && 2 \end {pmatrix}$ and am trying to find its eigenvalue.

My work so far:

$A-\lambda = \begin {pmatrix} 2-\lambda && 1 && 1 \\ 1 && 2-\lambda && 1 \\ 1 && 1 && 2-\lambda \end {pmatrix}$

Using laplace expansion:

$(2-\lambda)((2-\lambda)^2-1) - ((2-\lambda)-1) + (1-(2-\lambda))$

Using difference of squares:

$(2-\lambda)(((2-\lambda)-1)(((2-\lambda)+1)) - ((2-\lambda)-1) + (1-(2-\lambda))$

and this is where I got stuck. I can't get it to the point where I can set each point to $0$ to get the eigenvalues. Any help would be appreciated.

Every row sums to $4$ so you know the all ones vector is a eigenvector with eigenvalue $4$. — Countingstuff, Apr 24 '18 at 16:05
Just from looking at the matrix you can say $\lambda = 1$ is an eigenvalue, divide your (characteristic) equation by $\lambda - 1$ and you'll get what you want, that's one way — Naweed G. Seldon, Apr 24 '18 at 16:07
Of course you can do it by the Laplace expansion. Obviously you'll get 3 answers of which 2 are the same. — poyea, Apr 24 '18 at 16:09

score 1 · Accepted Answer · answered Apr 24 '18 at 16:18

Why don't you try opening the brackets as well? You have $$(2-\lambda)((2-\lambda)^2-1) - ((2-\lambda)-1) + (1-(2-\lambda))\\ = (2-\lambda)^3 - (2-\lambda) - (2-\lambda) + 1 + 1 - (2-\lambda)\\ = (2-\lambda)^3 - 3(2-\lambda) + 2$$

Take $\mu = 2 - \lambda$. You get $\mu^3 - 3\mu + 2$, and $\mu - 1$ is a divisor of the equation. $$\mu^3 - 3\mu + 2 = (\mu^2 + \mu - 2) (\mu - 1)=(\mu + 2)(\mu - 1)^2$$ Therefore $\mu = -2,1,1 \implies \lambda = 4,1,1$

score 0 · Answer 2 · answered Apr 24 '18 at 16:24

Taking the determinant, we get $p(\lambda) = (2-\lambda)^3 + 2 - 3(2-\lambda)$ = $-\lambda^3 + 6\lambda^2 -12\lambda + 8 + 2 - 6 +3\lambda = \lambda^3+6\lambda^2-9\lambda + 4$. Factoring this, we get down to $(4-\lambda)(\lambda-1)^2$. So we have eigenvalues $\lambda_1 = 1$ with multiplicity 1 and $\lambda_2 = 1$ with multiplicity 2.

A nice trick for computing the determinant of a $3x3$ matrix is to: 1) rewrite the first two columns to the right of the matrix 2) multiply down the three diagonals, add these values 3) multiply up the three diagonals and subtract this value. This is equivalent to the determinant, i.e. for $\begin{bmatrix}a & b & c\\\ d&e&f\\\ g&h&i \end{bmatrix}$, we obtain $aei+bfg+cdh-ceg-afh-bdi$. It makes factoring to find the eigenvalues a bit easier to identify. Hope this helps!

Finding the eigenvalue for a matrix

2 Answers2