# Thread: Eigenvalues, Eigenvectors and Diagonalization

1. ## Eigenvalues, Eigenvectors and Diagonalization

I have the matrix A=|7 2|
|12 -3|
I found the eigenvalues λ=9 and λ=-5
Now i have to find the eigenvector, using Gauss.
I set
A-9I = |-2 2 0| R1
|12 -12 0| R2
and used Gauss this way:
R2+5*R1
R1/-2 and R2/2
R2/R1
R2-R1
R2/2
R1+R2
until i got the vector
|1 0 0|
|0 1 0|
I cant really find out what to do now. And then i have to find a matrix, P, that diagonalizes the matrix A.

In the other example i had the vector
A= |-1 -2 -2|
|1 2 1|
|-1 -1 0|
and i solved for the eigenvalues and got
-λ^3+λ^2-3λ-1
But if i solve it with my texas instrument, i get λ=-0,295598 wich doesn't sound so right. So how do i solve it?

Thank you

2. Originally Posted by laugalitla
I have the matrix A=|7 2|
|12 -3|
I found the eigenvalues λ=9 and λ=-5
Now i have to find the eigenvector, using Gauss.
I set
A-9I = |-2 2 0| R1
|12 -12 0| R2
and used Gauss this way:
R2+5*R1
Okay, that would give $\displaystyle \left[\begin{array}{ccc}-2 & 2 & 0 \\ 2 & -2& 0 \end{array}\right]$

R1/-2 and R2/2
and this will give you $\displaystyle \left[\begin{array}{ccc}1 & -1 & 0\\ 1 & -1 & 0\end{array}\right]$

R2/R1
I don't know what you mean by this. Dividing one row by another is not a "row-operation".

R2-R1
If I ignore the previous line and subtract R1- R2, I get $\displaystyle \left[\begin{array}{ccc}1 & 1 & 0 \\ 0 & 0 & 0\end{array}\right]$ and I can stop here. An eigenvector is $\displaystyle \left[\begin{array}{c} 1 \\ 1 \end{array}\right]$.

R2/2
R1+R2
until i got the vector
|1 0 0|
|0 1 0|
First, that is not a vector, it is a matrix. Also you should have realized that there was something wrong. 9 was an eigenvalue because det(A- 9I)= 0 which means that A is not invertible and cannot be row reduced to the identity matrix as you have.

[quot]I cant really find out what to do now. And then i have to find a matrix, P, that diagonalizes the matrix A.[/quote]

A more fundamental way of doing this: from the definition of "eigevalue" there exist a non-zero vector v such that $\displaystyle \left[\begin{array}{cc}7 & 2 \\ 12 & -3\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right]= 9\left[\begin{array}{c}x \\ y\end{array}$.
Multiplying that out gives the three equations 7x+ 2y= 9x, 12x- 3y= 9y or 2x+ 2y= 0 and 3x- 3y= 0 both of which are satified by any x,y such that y= x. Again, that says that $\displaystyle \left[\begin{array}{c} 1 \\ 1 \end{array}\right]$ is an eigenvector.

In the other example i had the vector
A= |-1 -2 -2|
|1 2 1|
|-1 -1 0|
and i solved for the eigenvalues and got
-λ^3+λ^2-3λ-1
But if i solve it with my texas instrument, i get λ=-0,295598 wich doesn't sound so right. So how do i solve it?

Thank you
You don't show how you got that equation but that's not what I get.