# Eigenvector

• Dec 7th 2009, 09:15 AM
Dogod11
Eigenvector
Hello everyone, this is the following:

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

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

Multiplying

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

$(7a + 8, 21)$

$(7a + 8, 21) = k(7, 8)$ , 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.
• Dec 7th 2009, 09:38 AM
qmech
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).
• Dec 7th 2009, 09:47 AM
tonio
Quote:

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

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

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

Multiplying

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

$(7a + 8, 21)$

$(7a + 8, 21) = k(7, 8)$ , 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 $(7a+8,21)=k(7,8)$ it follows that $21=8k\,\Longrightarrow\,k=\frac{21}{8}$ , and then $7a+8=7k=\frac{147}{8}$ you get an easy equation from where you get the value of $a$

Tonio

Ps. BTW, it is $(7a+8,24)$ and not $(7a+8,21)$ ...but the development is the same.
• Dec 7th 2009, 10:24 AM
Dogod11
O.K. thank you very much for the two

The value found is $13 / 7.$

Find then the eigenvectors of the matrix

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

And I find the eigenvectors are

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

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

Thank you very much

Greetings
• Dec 7th 2009, 10:35 AM
tonio
Quote:

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

The value found is $13 / 7.$

Find then the eigenvectors of the matrix

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

And I find the eigenvectors are

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

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

Thank you very much

Greetings

First, if $u$ is an eigenvector of a linear operator/matrix, then also $tu$ is an eigenvector, for any scalar $k\ne 0$...so, can you see something interesting now?
• Dec 7th 2009, 10:39 AM
Dogod11
Yes, I understand,

Thank you very much again.

Greetings(Smirk)