# Left Eigenvectors

• November 27th 2011, 08:02 AM
page929
Left Eigenvectors
I have two problems dealing with left eigenvectors:

(1) Show that a left eigenvector y correcpinding to the eigenvalue http://www.physlink.com/Images/symbols/lambda.gif of A belonging to $M_n$ is a right eigenvector of A* corresponding to http://www.physlink.com/Images/symbols/lambda.gif (with a bar over it).

(2) Show that if y is a left eigenvector corresponding to the eigenvalue http://www.physlink.com/Images/symbols/lambda.gif of A belonging to $M_n$ then y(with a bar over it) is a right eigenvector of $A^T$ corresponding to http://www.physlink.com/Images/symbols/lambda.gif.

I know that the definition of left eigenvector states...the vector y ≠ 0 is said to be a left eigenvector of A if it satisfies y* A = http://www.physlink.com/Images/symbols/lambda.gify*. Then http://www.physlink.com/Images/symbols/lambda.gif is called a left eigenvector of A.
• November 27th 2011, 08:41 AM
Deveno
Re: Left Eigenvectors
if y is a left eigenvector of A, then y*A = λy*. therefore, (y*A)* = (λy*)*, that is:

A*y = λ*y.

(writing λ* for the complex conjugate of λ).

the trick is noticing that if:

$y^* = \begin{bmatrix}\overline{y_1}\\ \overline{y_2}\\ \vdots\\ \overline{y_n}\end{bmatrix}$ then:

$\lambda y^* = \begin{bmatrix}\lambda \overline{y_1}\\ \lambda \overline{y_2}\\ \vdots\\ \lambda \overline{y_n}\end{bmatrix}$ so:

$(\lambda y^*)^* = \begin{bmatrix} \overline{\lambda} y_1, \overline{\lambda} y_2, \dots, \overline{\lambda} y_n \end{bmatrix} = \overline{\lambda}y$

for the second question, note that y*A = λy*, means that y*(A - λI) = 0^t (the 0 row-vector), so

$(y^*(A - \lambda I))^T = 0$ that is: $(A - \lambda I)^T\overline{y} = 0$ so that:

$A^T\overline{y} = \lambda\overline{y}$
• November 27th 2011, 12:44 PM
page929
Re: Left Eigenvectors
Must the left eigenvectors and right eigenvectors of a matrix be the same? Prove or give counter example.

I want to say no, but am not sure if I am even correct let alone proving. Can someone help me out? Thanks!