# eigen values of transpose

• Jun 22nd 2011, 06:53 AM
transgalactic
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

??
• Jun 22nd 2011, 08:45 AM
FernandoRevilla
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

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

Quote:

and has the same eigen vectors??
Not necessarily.
• Jun 22nd 2011, 08:56 AM
transgalactic
Re: eigen values of transpose
thanks :)
• Jun 22nd 2011, 11:23 AM
transgalactic
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)
• Jun 22nd 2011, 01:52 PM
topspin1617
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.
• Jun 22nd 2011, 01:57 PM
transgalactic
Re: eigen values of transpose
i am asked to prove that their eigen vector are the same too .,for evey eigen value
• Jun 22nd 2011, 02:16 PM
FernandoRevilla
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 $A=\begin{bmatrix}{0}&{1}\\{0}&{0}\end{bmatrix}$

The eigenvectors of $A$ are different from the eigenvectors of $A^t$ .
• Jun 22nd 2011, 02:18 PM
HallsofIvy
Re: eigen values of transpose
Well, you can't. As you have been told that is not true.

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

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