Eigen values and Eigenvectors for a special case of a symmetric matrix

[mihalisla]

Hey guys if i have a vector x=[x1,x2, ... xn]
what are the eigenvectors and eigenvalues of X^T*X ?
I know that i get a n by n symmetric matrix with it's diagonal entries in
the form of Ʃ xii^2 for i=1,2,3,. . . ,n

Thank you in advance once again!

 

[Office_Shredder]

You can calculate them fairly directly. Let A = x^t x Then for a vector v = (v_1,...,v_n)
Av^t = (x^t x) v^t = x^t (x v^t) = x^t \left( x\cdot v \right)
where x\cdot v is the dot product of x and v. Based on this you should be able to spot that there can only be one eigenvector with a nonzero eigenvalue (and what the eigenvectors and eigenvalues are)

 

[mihalisla]

I ll try it . thank you very much!

 

[mihalisla]

I tried it but i get a nx1 in size matrix. Aren't eigenvalues and eigenvectors for nxn matrices . . . ?
How can I get the values ?
Thank you .

 

[HallsofIvy]

Yes, Office Shredder said that would give the eigenvectors, not a matrix.

 

[mihalisla]

Sorry but i still don't get it. What is the eigenvector in the second part of the equation? Could you provide the solution beyond that first step! Thank you !

 

[mihalisla]

How can I define the rank of that matrix that is always one ? Then I can get on from this forum from
https://www.physicsforums.com/showthread.php?t=682216

 

[mihalisla]

Solved

Solved ! lamda=x*x' and the corresponding eigenvector is x