Finding normalised eigenvectors...

Question

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.

Answer 1

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.

Answer 2

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 $\vec v$.

Answer 3

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.)