Finding the eigenvector given eigenvalue and $A$?

Question

I am studying for an exam for work which is covering some linear algebra which I am having some difficulty with.

I have the following $(2 \times 2)$-matrix: $$ A = \begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix} $$ And I have the eigenvalue $5$.

How would I go about solving for the eigenvector? I know the following formula: $$ A(\text{eigenvalue}) = (\text{eigenvalue})(\text{eigenvector}) $$

I was thinking that since I have the $A$ and the eigenvalue, I can divide the product of that by the eigenvalue (divide each value in the matrix by $1/5$)?

Any help would be great!

Answer 1

Let $A=\left(\begin{smallmatrix}3&1\\2&4\end{smallmatrix}\right)$. Then, since you were told (correctly) the $5$ is an eigenvalue, your approach is fine: solve the system $A\left(\begin{smallmatrix}x\\y\end{smallmatrix}\right)=5\left(\begin{smallmatrix}x\\y\end{smallmatrix}\right)$, that is, the system$$\left\{\begin{array}{l}3x+y=5x\\2x+4y=5y,\end{array}\right.$$which is equivalent to$$\left\{\begin{array}{l}-2x+y=0\\2x-y=0.\end{array}\right.$$Take $\left(\begin{smallmatrix}1\\2\end{smallmatrix}\right)$ as a solution, for instance.

Answer 2

Generally speaking, given a matrix $M$ and an eigenvalue $\lambda$ of $M$, there is not only one associated eigenvector. The set of all eigenvectors associated with $\lambda$ is a linear subspace (of $\mathbb{R}^2$ here).

$X=\begin{pmatrix}x \\ y\end{pmatrix}\in\mathbb{R}^2$ is an eigenvector of $M$ associated with the eigenvalue $\lambda$ $\Leftrightarrow M\cdot X = \lambda X$ $$\Leftrightarrow(M-I_2\lambda)\cdot \begin{pmatrix}x \\ y\end{pmatrix}=0_{\mathbb{R}^2}$$ Which is a system of equations whose set of solutions is indeed a linear subspace of $\mathbb{R}^2$.

So in your cases, you have : $$\Leftrightarrow\begin{pmatrix}-2 & 1 \\ 2 & -1\end{pmatrix}\cdot\begin{pmatrix}x \\ y\end{pmatrix}=0_{\mathbb{R}^2}$$ $$\Leftrightarrow -2x+y=0$$ $$\Leftrightarrow 2x=y$$ $$\Leftrightarrow \begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}x \\ 2x\end{pmatrix}=x\begin{pmatrix}1 \\ 2\end{pmatrix}$$

So if we call $E_5(M)$ the linear space of all eigenvectors associated with the eigenvalue 5, you know that $E_5(M)\subset \text{span}\left(\begin{pmatrix}1 \\ 2\end{pmatrix}\right)$.

$E_5(M)$ being a linear subspace of a linear space whose dimension is 1, it is either the null space, or equal to span$\left(\begin{pmatrix}1 \\ 2\end{pmatrix}\right)$

And since $\text{dim}(E_5(M))\geq1$ (because 5 is indeed an eigenvalue of M), then $E_5(M)= \text{span}\left(\begin{pmatrix}1 \\ 2\end{pmatrix}\right)$.