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