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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?

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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

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

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  • $\begingroup$ So for the second eigenvalue, would you just make 3- λ zero and then add 1+1 = 2 making 2 the second eigenvalue? $\endgroup$ – david mah Nov 26 '16 at 21:04
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    $\begingroup$ @davidmah Use trace and determinant. $\endgroup$ – hamam_Abdallah Nov 26 '16 at 21:13
  • $\begingroup$ Well I am using the determinate but that gives me equations like ($λ^2 - 6λ+ 20) + (λ^3 - λ^2 - 30λ + 40$) and I have 100% no idea how I can factor those two polynomials so I can find out what lambda is. That's kind of what my question was asking. Is there an easier way I can find the determinate so I can find the eigenvalue. $\endgroup$ – david mah Nov 26 '16 at 21:18
  • $\begingroup$ And that will work for every 3x3 matrix? $\endgroup$ – david mah Nov 26 '16 at 21:23
  • $\begingroup$ AND THAT IS WHAT I AM ASKING! At times I get very complicated polynomials that I'm trouble factoring and my question is, is there an easier way to do this. I'm not asking for this specific problem but just in general for any 3x3 matrix. $\endgroup$ – david mah Nov 26 '16 at 21:27
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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

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

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  • $\begingroup$ Where did you get -2 + λ from? $\endgroup$ – david mah Nov 26 '16 at 21:01
  • $\begingroup$ So how did you factor that polynomial? $\endgroup$ – david mah Nov 26 '16 at 21:04
  • $\begingroup$ Again, I did the row operations and I really have no idea where you got -2 + λ from. $\endgroup$ – david mah Nov 26 '16 at 21:06
  • $\begingroup$ @davidmah Note that $(1,1,3-\lambda)-(1,3-\lambda,1)=(0,-2+\lambda,2-\lambda).$ $\endgroup$ – mfl Nov 26 '16 at 21:08
  • $\begingroup$ That doesn't answer my question. How did you go from $-\lambda^3+9\lambda^2-24\lambda+20$ to $-(\lambda-5)(\lambda-2)^2$? $\endgroup$ – david mah Nov 26 '16 at 21:10
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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 $$ \color{blue}{ v_{2} = \left[ \begin{array}{r} -1 \\ 1 \\ 0 \\ \end{array} \right], \quad v_{3} = \left[ \begin{array}{r} -1 \\ 0 \\ 1 \\ \end{array} \right]} $$

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