1. ## Eigenvector

Hello everyone, this is the following:

Given the matrix
$\displaystyle \begin{bmatrix}{a}&{1}\\{0}&{3}\end{bmatrix}$ to determines the values of $\displaystyle a$ such that the vector $\displaystyle (7, 8)$ is an eigenvector of the matrix

I know the value found should be a multiple of the vector given, I have done this:

Multiplying

$\displaystyle \begin{bmatrix}{a}&{1}\\{0}&{3}\end{bmatrix}$*$\displaystyle (7, 8)$
get

$\displaystyle (7a + 8, 21)$

$\displaystyle (7a + 8, 21) = k(7, 8)$ , where k is the value to establish multiple.

I do not know how to finish the exercise.

or if someone tells me if there is another way to do it.

2. You've started correctly. One math error, it should be
(7a+8,24)=k(7,8).

Then:
7a+8 = 7k
24 = 8k.

Solve for k (2nd eqn), then solve for a(1st eqn).

3. Originally Posted by Dogod11
Hello everyone, this is the following:

Given the matrix
$\displaystyle \begin{bmatrix}{a}&{1}\\{0}&{3}\end{bmatrix}$ to determines the values of $\displaystyle a$ such that the vector $\displaystyle (7, 8)$ is an eigenvector of the matrix

I know the value found should be a multiple of the vector given, I have done this:

Multiplying

$\displaystyle \begin{bmatrix}{a}&{1}\\{0}&{3}\end{bmatrix}$*$\displaystyle (7, 8)$
get

$\displaystyle (7a + 8, 21)$

$\displaystyle (7a + 8, 21) = k(7, 8)$ , where k is the value to establish multiple.

I do not know how to finish the exercise.

or if someone tells me if there is another way to do it.

You did great! Now just end the task: from $\displaystyle (7a+8,21)=k(7,8)$ it follows that $\displaystyle 21=8k\,\Longrightarrow\,k=\frac{21}{8}$ , and then $\displaystyle 7a+8=7k=\frac{147}{8}$ you get an easy equation from where you get the value of $\displaystyle a$

Tonio

Ps. BTW, it is $\displaystyle (7a+8,24)$ and not $\displaystyle (7a+8,21)$ ...but the development is the same.

4. O.K. thank you very much for the two

The value found is $\displaystyle 13 / 7.$

Find then the eigenvectors of the matrix

$\displaystyle \begin{bmatrix}{13/7}&{1}\\{0}&{3}\end{bmatrix}$

And I find the eigenvectors are

$\displaystyle (1, 0),$ and $\displaystyle (1, 8/7)$

Any idea why not give me exactly $\displaystyle 7, 8?$

Thank you very much

Greetings

5. Originally Posted by Dogod11
O.K. thank you very much for the two

The value found is $\displaystyle 13 / 7.$

Find then the eigenvectors of the matrix

$\displaystyle \begin{bmatrix}{13/7}&{1}\\{0}&{3}\end{bmatrix}$

And I find the eigenvectors are

$\displaystyle (1, 0),$ and $\displaystyle (1, 8/7)$

Any idea why not give me exactly $\displaystyle 7, 8?$

Thank you very much

Greetings

First, if $\displaystyle u$ is an eigenvector of a linear operator/matrix, then also $\displaystyle tu$ is an eigenvector, for any scalar $\displaystyle k\ne 0$...so, can you see something interesting now?

6. Yes, I understand,

Thank you very much again.

Greetings