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I'm trying to find the eigenvector/eigenvalues of the $2\times2$ matrix: \begin{pmatrix}4 & 2 \\ 2 & 3 \end{pmatrix}

This is my work:

$$\det(A-\lambda I) = \lambda^2-7 \lambda+8=0 \iff \lambda=\frac{7+\sqrt{17}}{2} \ \lor \ \lambda= \frac{7-\sqrt{17}}{2}$$

$x_1$ (eigenvector)=\begin{pmatrix} (1+\sqrt17)/4 \\ k \end{pmatrix} , where k is any number. How do I "NORMALISE" this eigenvector? Can someone check my working because I'm getting weird answers.

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  • $\begingroup$ I can't read this, please visit the latex manual to make this readable, and edit your post. For now I'll edit what i think to understand. $\endgroup$
    – Bob
    May 9, 2013 at 13:46
  • $\begingroup$ Your eigenvector should have 2 elements, not just one. $\endgroup$
    – Kaster
    May 9, 2013 at 13:58
  • $\begingroup$ @Anon Perhaps if you explain from where you get $\begin{pmatrix} (1+\sqrt17)/4 \\ k \end{pmatrix}$, you might get a better explanation why it is incorrect. $\endgroup$ May 9, 2013 at 16:51

3 Answers 3

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If $\mathbf{x}$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $A\mathbf{x}=\lambda\mathbf{x}$ and $(A-\lambda I)\mathbf{x}=\mathbf{0}$.

First, find the eigenvector corresponding to the eigenvalue $λ=\frac{7+\sqrt{17}}{2}$:

$$\begin{align*} &\quad\quad\quad\quad\left(\begin{array}{c|c} A-\lambda I & 0 \end{array}\right)\quad\quad\text{insert your }A\text{ and }\lambda\\ &=\left(\begin{array}{cc|c} 4-\tfrac{7+\sqrt{17}}{2} & 2 & 0 \\ 2 & 3-\tfrac{7+\sqrt{17}}{2} & 0 \end{array}\right)\quad\quad\text{compute the differences}\\ &\implies \left(\begin{array}{cc|c} \tfrac{1-\sqrt{17}}{2} & 2 & 0 \\ 2 & \tfrac{-1-\sqrt{17}}{2} & 0 \end{array}\right)\quad\quad\text{multiply the first row by }\tfrac{4}{1-\sqrt{17}}\\ &\implies \left(\begin{array}{cc|c} 2 & \tfrac{8}{1-\sqrt{17}} & 0 \\ 2 & \tfrac{-1-\sqrt{17}}{2} & 0 \end{array}\right)\quad\quad\text{multiply the first fraction by }1+\sqrt{17}\\ &\implies \left(\begin{array}{cc|c} 2 & \tfrac{8(1+\sqrt{17})}{-16} & 0 \\ 2 & \tfrac{-1-\sqrt{17}}{2} & 0 \end{array}\right)\quad\quad\text{simplify the first fraction}\\ &\implies \left(\begin{array}{cc|c} 2 & \tfrac{-1-\sqrt{17}}{2} & 0 \\ 2 & \tfrac{-1-\sqrt{17}}{2} & 0 \end{array}\right)\quad\quad\text{subtract the first row from the second}\\ &\implies \left(\begin{array}{cc|c} 4 & -1-\sqrt{17} & 0 \\ 0 & 0 & 0 \end{array}\right)\quad\quad\text{deduce the solution}\\ &\implies \mathbf{x}=k\pmatrix{1+\sqrt{17}\\4}\end{align*}$$

Now, normalize it by

$$\hat{\mathbf{x}}=\frac{\mathbf{x}}{||\mathbf{x}||}$$

and do the same thing for the second eigenvalue.

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  • $\begingroup$ Could you explain your steps please. $\endgroup$
    – Anon
    May 9, 2013 at 14:38
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    $\begingroup$ @Anon I have added some explanation. Please, ask if something remains unclear. $\endgroup$
    – Librecoin
    May 9, 2013 at 15:00
  • $\begingroup$ I don't understand this we have not been taught this notation... $\endgroup$
    – Anon
    May 9, 2013 at 15:06
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    $\begingroup$ @Anon I have used augmented matrices. There is an example at the end of the article. $\endgroup$
    – Librecoin
    May 9, 2013 at 15:11
  • $\begingroup$ I need to do it the way I was taught in lectures though and we didn't use augmented matrices to do it please help I'm so confused. Souldn't your last line have a -1 instead of +1 too? $\endgroup$
    – Anon
    May 9, 2013 at 15:40
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HINT: normalized vector $\vec n$ from given vector $\vec v$ can be found with this formula: $$\vec n=\frac{\vec v}{||\vec v||}$$ where $||\vec v||=\sqrt{\vec v\cdot\vec v}$ is the norm of $\vec v$.

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  • $\begingroup$ When I try this k comes out as a complex number!? Have you actually tryed this? $\endgroup$
    – Anon
    May 9, 2013 at 14:10
  • $\begingroup$ How come you are trying to get $k$ when you should get a vector corresponding to your source $\vec x_1$? $k$ should be a parameter here. $\endgroup$
    – Ruslan
    May 9, 2013 at 14:13
  • $\begingroup$ I thought to normalise the vector its modulus should equal 1 so I need a k that makes that happen... $\endgroup$
    – Anon
    May 9, 2013 at 14:15
  • $\begingroup$ Well, then you are not trying to normalize a given vector, but instead find $k$ so that its norm appears 1. Looks like you've misformulated your problem. $\endgroup$
    – Ruslan
    May 9, 2013 at 14:17
  • $\begingroup$ So how do I "Normalise" the eigenvector?! Question says find normalised eigenvectors of matrix shown above. $\endgroup$
    – Anon
    May 9, 2013 at 14:18
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You can use WolframAlpha to check your result.

I wanted to mention, that it is often useful to apply what you know about eigenvalues; you can use $\lambda_1+\lambda_2=7$, $\lambda_1\lambda_2=-8$, $\lambda_1^2-7\lambda_1+8=0$ when trying to solve the system.

For example:
$A-\lambda_1 I= \begin{pmatrix} 4-\lambda_1 & 2 \\ 2 & 3-\lambda_1 \end{pmatrix}\sim \begin{pmatrix} 4-\lambda_1 & 2 \\ 2 & 3-\lambda_1 \end{pmatrix}\sim \begin{pmatrix} 12-7\lambda_1+\lambda_1^2 & 2(3-\lambda_1)\\ 2 & 3-\lambda_1 \end{pmatrix}= \begin{pmatrix} 4 & 2(3-\lambda_1)\\ 2 & 3-\lambda_1 \end{pmatrix}\sim \begin{pmatrix} 2 & 3-\lambda_1\\ 0 & 0 \end{pmatrix}$ (Note that if you are using row vectors, you would transpose the matrix $A-\lambda I$ to get the linear system you have to solve. But from your post it seems that you are using column vectors.)

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  • $\begingroup$ What does this symbol $\sim$ mean here? Does that mean $\begin{pmatrix} 4 & 2(3-\lambda_1)\\ 2 & 3-\lambda_1 \end{pmatrix}\sim$ is not exactly equal to $\begin{pmatrix} 2 & 3-\lambda_1\\ 0 & 0 \end{pmatrix}$? $\endgroup$
    – JJJohn
    Jul 6, 2019 at 10:02
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    $\begingroup$ @baojieqh The symbol $\sim$ stands for row equivalence of matrices. $\endgroup$ Jul 6, 2019 at 10:28

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