Eigenvalue Factorization and Matrix Substitution

[the_dialogue]

In my literature reviews I found a few things that I can't quite understand.

Homework Statement

I have the following equation:
http://img717.yfrog.com/img717/6416/31771570.jpg

I'm told that by using the eigenvalue factorization:
http://img89.yfrog.com/img89/760/83769756.jpg

, I can change the first equation to:
http://img28.imageshack.us/img28/5023/84802099.jpg

2. The attempt at a solution

I tried changing Equation 2 to just be (A^T)A and then subbing into the first equation, but I can't quite do anything with those inverses.

Also, what does the exponent of '-2' mean in the context of a 4x4 matrix? Lastly, what is matrix U?

Thank you!

 

[gabbagabbahey]

I tried changing Equation 2 to just be (A^T)A and then subbing into the first equation, but I can't quite do anything with those inverses.

I think it's probably easiest to start from \textbf{p}^{T}(\mathbf{\Lambda}+\lambda\textbf{I})^{-2}\textbf{q}=0 and work your way backwards instead.

Also, what does the exponent of '-2' mean in the context of a 4x4 matrix? Lastly, what is matrix U?

\textbf{C}^{-2}\equiv\textbf{C}^{-1}\textbf{C}^{-1}

You simply square the inverse of the matrix.

 

[the_dialogue]

I'll give it a try gabbagabbahey. Thanks.

Any idea what the matrix "U" is?

 

[gabbagabbahey]

Any idea what the matrix "U" is?

It's the invertible matrix which relates the matrix \textbf{A}^{T}\textbf{A}\mathbf{\Sigma} to the diagonal matrix \mathbf{\Lambda} via a similarity transform. Its columns will be the eigenvectors of \textbf{A}^{T}\textbf{A}\mathbf{\Sigma}.

See http://en.wikipedia.org/wiki/Diagonalizable_matrix for a refresher on matrix diagonalization.

 

[the_dialogue]

Yes I recall now. Thanks!