# Thread: Finding eigen vectors of N dimensional linear transformation

1. ## Finding eigen vectors of N dimensional linear transformation

Definition of notation:
First let x for this problem denote the elementart tensor product.
Then let (AxB) acting on a vector v be defined by A<B,V> where this is the standard inner product.

Problem statement
With those defintions consider e and v members of Rn the vector space of N tuples allow A to be defined as A = exf +fxe
find the
eigen vectors
eigen spaces
specteral decompostition

Relevent equations
let the matrix that defines AxB be defined by (aibj) where ai and bj are the componets of the vectors

Attempt at a soloution
solve for Ag=g*lambda where we know g must be in the span{e,f}

we know that there are only 2 non-zero eigen values because there are N-2 mutually orthogonal components to e and f.

we can also know that all eigan values are real.

Thanks for any help

2. ## Re: Finding eigen vectors of N dimensional linear transformation

Originally Posted by carlosgrahm

Definition of notation:
First let x for this problem denote the elementart tensor product.
Then let (AxB) acting on a vector v be defined by A<B,V> where this is the standard inner product.

Problem statement
With those defintions consider e and v members of Rn the vector space of N tuples allow A to be defined as A = exf +fxe
find the
eigen vectors
eigen spaces
specteral decompostition

Relevent equations
let the matrix that defines AxB be defined by (aibj) where ai and bj are the componets of the vectors

Attempt at a soloution
solve for Ag=g*lambda where we know g must be in the span{e,f}

we know that there are only 2 non-zero eigen values because there are N-2 mutually orthogonal components to e and f.

we can also know that all eigan values are real.

Thanks for any help
If $A = e\otimes f + f\otimes e$ then $Ae = e\langle f,e\rangle + f\langle e,e\rangle$ and $Af = e\langle f,f\rangle + f\langle e,f\rangle.$ On the two-dimensional subspace spanned by e and f, the matrix of A (with respect to the basis {e,f}) is $M = \begin{bmatrix}\langle f,e\rangle&\langle f,f\rangle\\ \langle e,e\rangle&\langle e,f\rangle\end{bmatrix}.$

Find the eigenvalues and eigenvectors of that, just as you would for any 2x2 matrix, by solving the equation $\det(M-\lambda I) = 0.$

Note: if e and f are linearly dependent then the answer will be different (but easier).

3. ## Re: Finding eigen vectors of N dimensional linear transformation

Originally Posted by Opalg
If $A = e\otimes f + f\otimes e$ then $Ae = e\langle f,e\rangle + f\langle e,e\rangle$ and $Af = e\langle f,f\rangle + f\langle e,f\rangle.$ On the two-dimensional subspace spanned by e and f, the matrix of A (with respect to the basis {e,f}) is $M = \begin{bmatrix}\langle f,e\rangle&\langle f,f\rangle\\ \langle e,e\rangle&\langle e,f\rangle\end{bmatrix}.$

Find the eigenvalues and eigenvectors of that, just as you would for any 2x2 matrix, by solving the equation $\det(M-\lambda I) = 0.$

Note: if e and f are linearly dependent then the answer will be different (but easier).
$A = e\otimes f + f\otimes e$ if you look at this operation on any vector V
and $M = \begin{bmatrix}\langle f,e\rangle&\langle f,f\rangle\\ \langle e,e\rangle&\langle e,f\rangle\end{bmatrix}.$ operation on any vector v. they do not seem equivlent.

Should not A be taking vectors into the subspace spanned by e and f? But M's operation does not do that?

Thanks again for the only help I have gotten in days.

4. ## Re: Finding eigen vectors of N dimensional linear transformation

Originally Posted by carlosgrahm
$A = e\otimes f + f\otimes e$ if you look at this operation on any vector V
and $M = \begin{bmatrix}\langle f,e\rangle&\langle f,f\rangle\\ \langle e,e\rangle&\langle e,f\rangle\end{bmatrix}.$ operation on any vector v. they do not seem equivlent.

Should not A be taking vectors into the subspace spanned by e and f? But M's operation does not do that?
Let $W$ be the (two-dimensional) subspacce of $\mathbb{R}^n$ spanned by e and f, and let $W^\perp$ be the orthogonal complement of $W$. Then A maps $W$ into $W$, and it maps $W^\perp$ to {0}. That means that A splits up as the direct sum of the operator M on W, and the zero operator on $W^\perp.$ Thus $W^\perp$ is the (n–2)-dimensional eigenspace corresponding to the eigenvalue 0, and essentially all the action of A takes place in W. That is why you only have to look at the matrix M to find the nonzero eigenvalues and eigenvectors of A.

5. ## Re: Finding eigen vectors of N dimensional linear transformation

Originally Posted by Opalg
Let $W$ be the (two-dimensional) subspacce of $\mathbb{R}^n$ spanned by e and f, and let $W^\perp$ be the orthogonal complement of $W$. Then A maps $W$ into $W$, and it maps $W^\perp$ to {0}. That means that A splits up as the direct sum of the operator M on W, and the zero operator on $W^\perp.$ Thus $W^\perp$ is the (n–2)-dimensional eigenspace corresponding to the eigenvalue 0, and essentially all the action of A takes place in W. That is why you only have to look at the matrix M to find the nonzero eigenvalues and eigenvectors of A.
Oplag thank you so much for helping me out with this problem. I can not tell you how much it has helped me. Thank you