How to find the eigenvalues with repeated eigenvectors of this $3\times3$ matrix

Question

So I need to find the eigenvectors and eigenvalues of the following matrix: $\begin{bmatrix}3&1&1\\1&3&1\\1&1&3\end{bmatrix}$. I know how to find the eigenvalues however for a 3x3 matrix, it's so complicated and confusing to do. So I start by writing it like this: $\begin{bmatrix}3-λ&1&1\\1&3-λ&1\\1&1&3-λ\end{bmatrix}$ and then I figure out what lambda is by finding it's determinate. I started by performing a row operation to put a zero in the first column: $\begin{bmatrix}3-λ&1&1\\1&3-λ&1\\0&-2+λ&2-λ\end{bmatrix}$ so then I expand the bottom row. Therefore the first column would be zero and for the second column I got (-λ² + 4).

For the third column I have to multiply (3-λ)(3-λ) and then after subtracting 1 from it, I then have to multiply it by (2-λ) which would get me a polynomial that has λ³ in it and that can be difficult to factor.

So is there any easier way that I can do this?

Answer 1

Hint

the sum is the same for each line

$1+1+3=5$

so $(1,1,1)$ is an eigenvector with $5$ as eigenvalue.

the other eigenvalues $\lambda_1$ and $\lambda_2$ are such that

$$\lambda_1+\lambda_2+5=Tr(A)=9$$

and

$$5\lambda_1.\lambda_2=det(A)=20$$

Answer 2

Note that you have a typo in the step $$\begin{vmatrix}3-λ&1&1\\1&3-λ&1\\1&1&3-λ\end{vmatrix}\underbrace{=}_{\rm{incorrect}}\begin{vmatrix}3-λ&1&1\\1&3-λ&1\\0&-2-\lambda&2-λ\end{vmatrix}.$$

It should be $$\begin{vmatrix}3-λ&1&1\\1&3-λ&1\\1&1&3-λ\end{vmatrix}=\begin{vmatrix}3-λ&1&1\\1&3-λ&1\\0&-2+\lambda&2-λ\end{vmatrix}.$$

Also

$$\begin{vmatrix}3-λ&1&1\\1&3-λ&1\\1&1&3-λ\end{vmatrix}=-\lambda^3+9\lambda^2-24\lambda+20=-(\lambda-5)(\lambda-2)^2.$$

Answer 3

Problem statement

Resolve the eigensystem for

$$ \mathbf{A} = % \left[ \begin{array}{ccc} 3 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 3 \\ \end{array} \right] $$

Eigenvalues

The matrix to manipulate is $$ \mathbf{A} - \lambda \mathbf{I}_{3} = \left[ \begin{array}{ccc} \boxed{3-\lambda} & \boxed{1} & \boxed{1} \\ 1 & 3-\lambda & 1 \\ 1 & 1 & 3-\lambda \\ \end{array} \right] $$ Define characteristic equation $$ p(\lambda) = \det \left( \mathbf{A} - \lambda \mathbf{I}_{3} \right) $$ Construct determinant from minors $$ % \begin{align} % \det \left( \mathbf{A} - \lambda \mathbf{I}_{3} \right) % &= \boxed{\left( 1 - \lambda \right)} \left| \begin{array}{cc} 3-\lambda & 1 \\ 1 & 3-\lambda \\ \end{array} \right| % - \boxed{1} \left| \begin{array}{cc} 1 & 1 \\ 1 & 3-\lambda \\ \end{array} \right| % + \boxed{1} \left| \begin{array}{cc} 1 & 3-\lambda \\ 1 & 1 \\ \end{array} \right| \\[3pt] % &= -\lambda ^3 + 9 \lambda ^2 - 24 \lambda + 20 \\[2pt] % &= -(\lambda -5) (\lambda -2)^2\\ % \end{align} % $$ The roots $p\left( \lambda \right) = 0$ are the eigenvalues.

The eigenvalue spectrum is $$ \color{blue}{\lambda \left( \mathbf{A} \right) = \left\{ 5, 2, 2 \right\}} $$

Eigenvectors

Solve $$ \left( \mathbf{A} -\lambda_{k} \mathbf{I}_{3} \right) u = 0$$

$ \lambda_{1} = 5$: $$ % \begin{align} % \left[ \begin{array}{c} -2u_{1} + u_{2} + u_{3} \\ u_{1} - 2 u_{2} + u_{3} \\ u_{1} + u_{2} - 2 u_{3} \\ \end{array} \right] % = % \left[ \begin{array}{c} 0 \\ 0 \\ 0 \\ \end{array} \right] % \qquad \Rightarrow \qquad % \color{blue}{v_{1} = \left[ \begin{array}{c} 1 \\ 1 \\ 1 \\ \end{array} \right]} % \end{align} % $$

$ \lambda_{2} = 2$: Repeated root $$ \mathbf{A} - 2 \mathbf{I}_{3} = \left[ \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{array} \right] $$ Find two null space vectors for this matrix. Manipulate the real variables and look for solutions of the form $$ \left[ \begin{array}{c} \alpha \\ 1 \\ 0 \\ \end{array} \right], \quad \left[ \begin{array}{c} \beta \\ 0 \\ 1 \\ \end{array} \right] $$ The eigenvectors are $$ \color{blue}{ v_{2} = \left[ \begin{array}{r} -1 \\ 1 \\ 0 \\ \end{array} \right], \quad v_{3} = \left[ \begin{array}{r} -1 \\ 0 \\ 1 \\ \end{array} \right]} $$