Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have the following problem:

I need to implement a program that doesn't accept the matrix quadratic form $B^T\times B$ but it accepts the scalar quadratic form instead. Actually I need to find a simple way to write $trace((B^T\times B)^{-1})$ in scalar quadratic form where $B\in M_{(n,m)}$.

I know that the scalar quadratic form of $trace(B^T\times B)=\sum_{j=1}^mb_j^T\times b_j$ where $b_j$ is the $j^{th}$ column of $B$.

for instance if we have $trace([\begin{array}{cc} 1 & 2 \\ 3 & 4 \\ \end{array}] \times [\begin{array}{cc} 1 & 3 \\ 2 & 4 \\ \end{array}])= [\begin{array}{cc} 1 & 2 \\ \end{array}]\times [\begin{array}{c} 1 \\ 2 \\ \end{array}]+ [\begin{array}{cc} 3 & 4 \\ \end{array}]\times [\begin{array}{c} 3 \\ 4 \\ \end{array}]=30$

And so I wonder if there is a similar representation for $trace((B^T\times B)^{-1})$.

If not I'm interested in finding a matrix factorization for a symmetric positive definite matrix that gives me $S=B^T\times B$. In other words, if we have $S$ we can find $B$ and vice versa. But here I'm excluding the Cholesky matrix decomposition since it deals only with triangular matrices.

Many thanks!

share|cite|improve this question
Better give some 2 by 2 and 3 by 3 examples, your use of the terminology is incorrect. – Will Jagy Jan 27 '13 at 0:39
I just added an example.. – user2987 Jan 27 '13 at 1:02
Assuming $B$ is nonsingular, your $B^T B$ is positive definite, the trace is the sum of the eigenvalues. The trouble is that the eigenvalues of the inverse are the reciprocals of those eigenvalues, so you want to take the sum of the reciprocals and do something with it. I'm still not sure what you want, of course. – Will Jagy Jan 27 '13 at 1:07
yes I know that for symmetric matrices the trace is the sum of the eigenvalues, therefore, for the trace inverse we would have the the sum of the inverse of the eigenvalues.. But here I'm interested in finding how to write $B^TB$ differently in such a way my program does accept it. I did the scalar quadratic form for $trace(B^TB)$ but I couldn't find a way to write $trace[(B^TB)^{-1}]$ in different form. – user2987 Jan 27 '13 at 1:15
No idea what your program is supposed to do. Try to give an example, from start to finish but in purely mathematics notation, of a successful "way to write..." – Will Jagy Jan 27 '13 at 1:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.