eigen values of transpose

A is a square matrix over F field

if k is the eigen value of A

prove that k is eigen value of A^t too

and has the same eigen vectors

??

eigen vectors are the solution space P(A)

is found by solving (A-kI)x=0

dim P(A)=dim n -dim (ro(a))

rho(a)=rho(a^t)

|A|=|A^t|

these are the laws i maneged to come up with to solve it

??

Re: eigen values of transpose

Quote:

Originally Posted by

**transgalactic** A is a square matrix over F field if k is the eigen value of A prove that k is eigen value of A^t too

$\displaystyle \chi(\lambda)=\det (A-\lambda I)=\det ((A-\lambda I)^t)=\det(A^t-\lambda I)$ so, $\displaystyle A$ and $\displaystyle A^t$ have the same characteristic polynomial.

Quote:

and has the same eigen vectors??

Not necessarily.

Re: eigen values of transpose

Re: eigen values of transpose

why its not nessesery?

why if we have the same eigen values then we have the same eigen vector(our matrices differ)

Re: eigen values of transpose

Quote:

Originally Posted by

**transgalactic** why its not nessesery?

why if we have the same eigen values then we have the same eigen vector(our matrices differ)

What are you trying to ask??

Just because two different matrices have an eigenvalue in common doesn't mean that the eigenvector is the same in both cases. Just try it on some small examples to see that the vector can change.

By the way..... spell check really is a wonderful thing. It really helps us answer your questions if they are written correctly and neatly.

Re: eigen values of transpose

i am asked to prove that their eigen vector are the same too .,for evey eigen value

Re: eigen values of transpose

Quote:

Originally Posted by

**transgalactic** i am asked to prove that their eigen vector are the same too .,for evey eigen value

Choose $\displaystyle A=\begin{bmatrix}{0}&{1}\\{0}&{0}\end{bmatrix}$

The eigenvectors of $\displaystyle A$ are different from the eigenvectors of $\displaystyle A^t$ .

Re: eigen values of transpose

Well, you can't. As you have been told that is not true.

For example, if $\displaystyle A= \begin{bmatrix}5 & -2 \\ 6 & -2\end{bmatrix}$ has eigenvalues 1 and 2 with corresponding eigenvectors $\displaystyle \begin{bmatrix}1 \\ 2\end{bmatrix}$ and $\displaystyle \begin{bmatrix}2 \\ 5\end{bmatrix}$, respectively.

Its transpose is $\displaystyle A^*= \begin{bmatrix}5 & 6 \\ -2 & -2\end{bmatrix}$ which also has eigenvalues 1 and 2 but with corresponding eigenvectors $\displaystyle \begin{bmatrix}3\\ -2\end{bmatrix}$ and $\displaystyle \begin{bmatrix}-2 \\ 1\end{bmatrix}$, respectively. Those eigenvectors are not multiples of the eigenvectors of A.