Prob with calculating Eigen Vectors

[zak100]

[FONT=&quot]Hi,[/FONT]
[FONT=&quot]I have a following Matrix:[/FONT]

[FONT=&quot]r1= 2 7[/FONT]
[FONT=&quot]r2=-1 -6[/FONT]


[FONT=&quot]I have got following eigen values:[/FONT]
[FONT=&quot]lambda1 = -5 and lambda2=1[/FONT]

[FONT=&quot]A-lambda * I matrix is:[/FONT]
[FONT=&quot]r1 = 7 7[/FONT]
[FONT=&quot]r2 = -1 -1[/FONT]

[FONT=&quot]and x matrix is:[/FONT]
[FONT=&quot]r1 =x[/FONT]
[FONT=&quot]r2 =y[/FONT]

[FONT=&quot]They used the concept of augmented matrix to find the eigen vectors. [/FONT][FONT=&quot] [/FONT]



[FONT=&quot]I cant understand why we have to use augmented matrix in this case.File attached[/FONT]
[FONT=&quot]Some body please guide me.[/FONT]

[FONT=&quot]Zulfi.[/FONT]

 

[Steven G]

Hi,
I have a following Matrix:

r1= 2 7
r2=-1 -6


I have got following eigen values:
lambda1 = -5 and lambda2=1

A-lambda * I matrix is:
r1 = 7 7
r2 = -1 -1

and x matrix is:
r1 =x
r2 =y

They used the concept of augmented matrix to find the eigen vectors.



I cant understand why we have to use augmented matrix in this case.File attached
Some body please guide me.

Zulfi.

What is the x matrix you referred to?

 

[HallsofIvy]

You don't "have to" and, personally, I prefer to use the basic definition of "eigenvector". Since 1 is an eigenvalue, there must exist a vector \(\displaystyle \begin{bmatrix}x \\ y \end{bmatrix}\) such that x and y are not both 0 and \(\displaystyle \begin{bmatrix}2 & 7 \\ -1 & -6\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= 1\begin{bmatrix}x \\ y\end{bmatrix}\). That, of course, can be written \(\displaystyle \begin{bmatrix}2-1 & 7 \\ -1 & -1- 1 ]\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}=\)\(\displaystyle \begin{bmatrix} 1 & 7 \\ -1 & -2\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}\)\(\displaystyle = \begin{bmatrix}0 \\ 0 \end{bmatrix}\)


But, with a simple 2 by 2 matrix, I prefer to separate the two rows into two equations: \(\displaystyle \begin{bmatrix}2 & 7 \\ -1 & -6\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}2x+ 7y \\ -x- 6y\end{bmatrix}= 1\begin{bmatrix}x \\ y\end{bmatrix}\) gives 2x+ 7y= x and -x- 6y= y. Those quickly reduce x+ 7y= 0 and -x- 7y= 0. Since one is just -1 times the other, those two equations are dependent and there are an infinite number of solutions. (Of course. If <x, y> is an eigenvector so is any multiple of it.) Both equations reduce to x= -7y so that any eigenvector is a multiple of \(\displaystyle \begin{bmatrix}-7 \\ 1 \end{bmatrix}\).

If \(\displaystyle \begin{bmatrix} x \\ y \end{bmatrix}\) is an eigenvector corresponding to eigenvalue -5 then \(\displaystyle \begin{bmatrix} 2 & 7\\ -1 & -6 \end{bmatrix}= \begin{bmatrix}2x+ 7y \\ -x- 6y \end{bmatrix}= \begin{bmatrix}-5x \\ -5y\end{bmatrix}\). So we have 2x+ 7y= -5x and -x- 6y= -5y. Those equations reduce to 7x+ 7y= 0 and -x- y= 0 both of which further reduce to y= -x so that any eigenvector is a multiple of \(\displaystyle \begin{bmatrix}1 \\ -1 \end{bmatrix}\).

 

[zak100]

Thanks, Good work. Zulfi.