Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

For a real, symmetric matrix $A$ and a real, rectangular matrix $X$, am looking for a matrix trace based representation of this simple linear algebraic expression $\sum_{i} A_{ii} ||X_{i.}||_\mathcal{H}^{2}$. Do note that $X_{i.}$ indicates the row $i$ of $X$ and the norm is a Hilbert norm (precisely a RKHS norm).

share|cite|improve this question
up vote 1 down vote accepted

Assuming, the dimensions conform, you will need the hadamard product (element-wise product of two matrices). $trace(A\circ B ) $ where $B=XX^T$ and $A\circ B$ denotes hadamard product, should do the job . Note that diagonal entries of $B$ are $||X_i||^2$. If you are insistent over avoiding the hadamard product also, it becomes a bit complicated. Let $\{J_i\}$ be set of all diagonal matrices with +1 or -1 in its diagonal. Note that there are $2^N$ such matrices (assuming dimension is $N\times N$). Now for any matrix $A$, we have \begin{align} D_A=\sum_{i}J_iAJ_i \end{align}
where $D_A$ denotes the diagonal matrix whose diagonal entries are same as that of $A$. Similarly for $B=XX^T$, we have \begin{align} D_B=\sum_{j}J_iBJ_i \end{align}
Note that $[D_A]_{ii}=A_{ii}$ and $[D_B]_{ii}=||X_i||^2$. Thus, the answer will be $trace(D_AD_B)$

share|cite|improve this answer
What are the off-diagonal entries of $B$ in the Hadamard product setting? In the non-Hadamard setting, is 'each' diagonal entry in $J$ 'either' -1 or 1 (or) are all entries in the diagonal either 1 or -1 and also what is the exact $J$ that we choose from the set ${J_i}$ when you say $trace(D_AD_B)$? Id like a more clearer definition and description. – qlinck Jan 28 '13 at 21:19
The fact that you say $2^N$ gives me an understanding that each entry in the diagonal can be -1 or 1. But the other questions-still remain and require clarity in the answer-i.e, what combination of $J$ do you choose?. – qlinck Jan 28 '13 at 21:48
$[B]_{ij}=<X_i,X_j>$. – dineshdileep Jan 29 '13 at 2:53
$D_A=J_1AJ_1+J_2AJ_2+J_3AJ_3+....J_KAJ_K$ where $K=2^N$. So you have to take all possible $J$'s and then take the sum as above. This will extract the diagonal elements of $A$ alone and will make all off-diagonal entries zero. – dineshdileep Jan 29 '13 at 2:55

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.