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

1. Apr 25, 2013

### 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!

2. Apr 25, 2013

### Office_Shredder

Staff Emeritus
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)

3. Apr 25, 2013

### mihalisla

I ll try it . thank you very much!

4. Apr 26, 2013

### 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 .

5. Apr 26, 2013

### HallsofIvy

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

6. Apr 30, 2013

### 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 !

7. Apr 30, 2013

8. May 1, 2013

### mihalisla

Solved

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