How can i solve the eigenvector equation for a $2 \times 2$ matrix with eigenvalues $\lambda = -1$ and $\lambda=3$

Question

where

$$A=\begin{bmatrix}1&1\\4&1\end{bmatrix}$$

(For clarification, the materials i am using are telling me to use the following formula and to input the appropriate values for a, b, c and d but i am getting terribly confused.)

(a-λ)x+by=0

cx+(d-λ)y=0

Answer 1

To find an eigenvector of $\lambda=3$ you need to solve : $$AX = \lambda X $$ $$\begin{bmatrix}1&1\\4&1\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=3\begin{bmatrix}x_1\\x_2\end{bmatrix}$$ $$x_1+x_2 = 3x_1 \text{ and }4x_1+x_2=3x_2 $$ Thus $X=\begin{bmatrix}1\\2\end{bmatrix}$ is an eigenvector for $\lambda=3$.

You can do the same reasonning for $\lambda=-1$.

Answer 2

It's pretty straightforward; I'll work out the case $\lambda = 3$; suppose

$\vec v = \begin{pmatrix} a \\ b \end{pmatrix} \tag 1$

is the eigenvector corresponding to $\lambda = 3$; then

$A \vec v = 3\vec v; \tag 2$

we write this equation out using the given entries of $A$, $\vec v$:

$\begin{bmatrix} 1 & 1 \\ 4 & 1 \end{bmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = 3 \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 3a \\ 3b \end{pmatrix}, \tag 3$

or

$a + b = 3a, \tag 4$

$4a + b = 3b; \tag 5$

from (4),

$b = 2a; \tag 6$

it is expeditious at this point to observe we cannot have an eigenvector with $a = 0$, lest by (6) $b = 0$ as well, and thus

$\vec v = 0; \tag 7$

but eigenvectors are non-vanishing by definition. Thus

$a \ne 0, \tag 8$

and since eigenvectors are scalable, that is, the quality of "eigenvector-ness" is invariant under scalar multiplication, we may assume

$a = 1; \tag 9$

then

$b = 2a = 2, \tag{10}$

$\vec v = \begin{pmatrix} 1 \\ 2 \end{pmatrix}; \tag{11}$

it is easy to see that

$4a + b = 4(1) + 2 = 6 = 3(2), \tag{12}$

so $\vec v$ also satisfies (5).

That's about it for the case $\lambda = 3$; I leave the case $\lambda = -1$ to the reader.

Answer 3

The characteristic polynomial of $A$ is $$ (\lambda-1)^2-4=(\lambda-1-2)(\lambda-1+2)=(\lambda-3)(\lambda+1) $$ Therefore $(A-3I)(A+I)=(A+I)(A-3I)=0$. So the columns of $(A-3I)$ are solutions of $(A+I)X=0$. And the columns of $(A+I)$ are solutions of $(A-3I)X=0$.

Concretely, $$ A-3I = \left[\begin{array}{cc}-2 & 1 \\ 4 & -2\end{array}\right] $$ So $$ (A+I)\left[\begin{array}{c} 1 \\ -2 \end{array}\right]=0 $$ And $$ A+I = \left[\begin{array}{cc} 2 & 1 \\ 4 & 2\end{array}\right] $$ So $$ (A-3I)\left[\begin{array}{c}1 \\ 2\end{array}\right] = 0. $$