Calculate the eigenvalues and eigenvectors of 2 x 2 matrix

Question

Q5. Calculate the eigenvalues and eigenvectors of the following matrix

$$\left(\begin{matrix} 3 & \sqrt{2} \\ \sqrt{2} & 2 \end{matrix}\right)$$

It is $2 \times 2$ matrix and having square-root value.

Answer 1

The characteristic polynomial is $$(3-\lambda)(2-\lambda)-2.$$ Can you see how to get that?

Now set this polynomial equal to zero and solve for $\lambda$ to get the eigenvalues. If we let $A$ be your matrix. Then you can find the eigenvectors by solving $(A-\lambda I)v=0$, which you can do by augmenting a column of zeroes onto the matrix $A-\lambda I$ and doing some row reduction.

Answer 2

We have for the eigenvalues:

$$ \text{det} (\begin{bmatrix} 3 & \sqrt{2} \\ \sqrt{2} & 2 \\ \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix})$$

So then we get $\lambda^2-5\lambda+4=0$ so then $\lambda=1,4$ which are our eigenvalues.

Then you can try the last bit for the eigenvectors we have $A-\lambda I=0$, where $A$ is the original matrix and $\lambda$ are our two eigenvalues values we have found. Then you solve this to get your eigenvectors!

Answer 3

Find eigenvalues from the characteristic polynomial:

$\left|\begin{matrix} 3-\lambda & \sqrt{2} \\ \sqrt{2} & 2-\lambda \end{matrix}\right| =\lambda^2-5*\lambda+4=(\lambda-1)*(\lambda-4)$

$\lambda_1=1\;\; \lambda_2=4$

For every λ we find its own vector(s):

$\lambda_1=1$

$A-\lambda_1I=\left(\begin{matrix} 2 & \sqrt{2} \\ \sqrt{2} & 1 \end{matrix}\right)$

$(A-\lambda I)v=0$

So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination:

$\left(\begin{matrix} 2 & \sqrt{2} & 0 \\ \sqrt{2} & 1 & 0 \end{matrix}\right)$

$\begin{matrix} x_1 & +\frac{\sqrt{2}}{2}*x_2 & = & 0 \end{matrix}$

General Solution: $X=\left(\begin{matrix} \frac{-\sqrt{2}}{2}*x_2 \\ x_2 \end{matrix}\right)$

Let $x_2=1,\; v_1=\left(\begin{matrix} \frac{-\sqrt{2}}{2} \\ 1 \end{matrix}\right)$

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$\lambda_2=4$

$A-\lambda_2I=\left(\begin{matrix} -1 & \sqrt{2} \\ \sqrt{2} & -2 \end{matrix}\right)$

$(A-\lambda I)v=0$

So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination:

$\left(\begin{matrix} -1 & \sqrt{2} & 0 \\ \sqrt{2} & -2 & 0 \end{matrix}\right)$

$\begin{matrix} x_1 & -\sqrt{2}*x_2 & = & 0 \end{matrix}$

General Solution: $X=\left(\begin{matrix} \sqrt{2}*x_2 \\ x_2 \end{matrix}\right)$

Let $x_2=1, v_2=\left(\begin{matrix} \sqrt{2} \\ 1 \end{matrix}\right)$