Eigen vector of a 2x2 matrix

[mathdaemon]

Hello everyone,

How do I find the eigen vector of 1 cos x
cosx 1
This is a bit confusing so I request your help.
The answering options are
a) a^2 1 b) a^n n c) a^n na^(n-1) d) a^n na^(n-1)
0 a^n 0 a^n 0 a^n -n a^n

From the characteristic equation, the roots are 1+cos x and 1-cos x.
Hope its correct.
How do I get an answer that matches the form above.

Appreciate your help
Thank You

 

[DrPhil]

Hello everyone,

How do I find the eigen vector of 1 cos x
cosx 1
This is a bit confusing so I request your help.
The answering options are
a) a^2 1 b) a^n n c) a^n na^(n-1) d) a^n na^(n-1)
0 a^n 0 a^n 0 a^n -n a^n

From the characteristic equation, the roots are 1+cos x and 1-cos x.
Hope its correct.
How do I get an answer that matches the form above.

Appreciate your help
Thank You

I agree with the eigenvalues you found.

$\displaystyle \displaystyle A = \begin{pmatrix} 1 & \cos x \\ \cos x & 1 \end{pmatrix}$

If $\displaystyle \lambda$ is an eigenvalue and the corresponding eigenvector is $\displaystyle \displaystyle \begin{pmatrix}x \\ y\end{pmatrix}$, then

$\displaystyle \displaystyle A \begin{pmatrix}x \\ y\end{pmatrix} = \lambda \begin{pmatrix}x \\ y\end{pmatrix}$

$\displaystyle \displaystyle \begin{pmatrix} 1 - \lambda & \cos x \\ \cos x & 1 - \lambda \end{pmatrix} \begin{pmatrix}x \\ y\end{pmatrix} = 0$

which requires that the determinant of the left side be 0, which eventually leads to

$\displaystyle \lambda = 1 \pm \cos x$

Look first at eigenvalue $\displaystyle \lambda = 1 + \cos x$. Can you show that for that case, the eigenvector is of the form $\displaystyle \displaystyle \begin{pmatrix}x \\ 0\end{pmatrix}$? Not that the normalization of the eigenvector is arbitrary.

Then let $\displaystyle \lambda = 1 - \cos x$. Which of the proposed answers works? Again, the norma;ization is arbitrary.

Can you get further with these hints?

 

[mathdaemon]

I don't get you.I am sorry.Please give some example

 

[HallsofIvy]

Since we are using "x" in the matrix, I would NOT use "(x, y)" for the vector!

mathdaemon, the whole point of an "eigenvalue, eigenvector pair" is that $\displaystyle \lambda$ is an eigenvalue for matrixA, with eigenvector v, if and only if $\displaystyle Av= \lambda v$

Here, taking (a, b) as the eigenvector corresponding to eigenvalue 1+ cos(x) we must have
$\displaystyle \begin{pmatrix}1 & cos(x) \\ cos(x) & 1\end{pmatrix}\begin{pmatrix}a \\ b\end{pmatrix}= (1+ cos(x))\begin{pmatrix}a \\ b\end{pmatrix}$

$\displaystyle \begin{pmatrix}a+ cos(x)b \\ cos(x)a+ b\end{pmatrix}= \begin{pmatrix}a+ cos(x)a \\ b+ cos(x)b\end{pmatrix}$

which means we must have a+ cos(x)b= a+ cos(x)a and cos(x)a+ b= b+ cos(x)b. Those two equations both give a= b. That is, any multiple of (1, 1) is an eigenvector corresponding to eigenvalue 1+ cos(x).

Similarly, if (a, b) is an eigenvector corresponding to eigenvalue 1- cos(x) must satisfy

$\displaystyle \begin{pmatrix}1 & cos(x) \\ cos(x) & 1\end{pmatrix}\begin{pmatrix}a \\ b\end{pmatrix}= (1- cos(x))\begin{pmatrix}a \\ b\end{pmatrix}$

$\displaystyle \begin{pmatrix}a+ cos(x)b \\ cos(x)a+ b\end{pmatrix}= \begin{pmatrix}a- cos(x)a \\ b- cos(x)b\end{pmatrix}$

which means we must have a+ cos(x)b= a- cos(x)a and cos(x)a+ b= b- cos(x)b which give a= -b. Any multiple of (1, -1) is an eigenvector corresponding to eigenvalue 1- cos(x).

 

[mathdaemon]

Thank you