MathOverflow is a question and answer site for professional mathematicians. 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 equation:

$Y = [ C \bullet R ] \times X \times[C^T \bullet R^T ]$

In my textbook, this conveniently rearranges to:

$Y = [ C \times X \times C^T ] \bullet [R \bullet R^T ]$

where $\times$ denotes matrix multiplication, $\bullet$ is Hadamard (pointwise multiplication), and $R^T$ is the transposition of $R$.

How do they manage to rearrange this?

I've played around with it, and it seems that they're using the following identity:

$[ A \bullet B ] \times C = A \bullet [B \times C]$

I've tried proving it true, without success. For example, if

$D = [ A \bullet B ] \times C$ and $E = A \bullet [B \times C]$,


$D_{00} = \sum_{k=0}^m A_{0k} B_{0k}C_{k0}$

$E_{00} = A_{00} \sum_{k=0}^m B_{0k}C_{k0}$

As $D_{00} \neq E_{00}$, the identity is false.

Where have I gone wrong?


I'm working with $4 \times 4$ matrices here.


Here are the matrices I'm dealing with:

$ C = \begin{pmatrix} 1 & 1 & 1 & 1\\\\ 2 & 1 & -1 & -2\\\\ 1 & -1 & -1 & 1\\\\ 1 & -2 & 2 & -1\\\\ \end{pmatrix} $

$ R = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\\\\ \frac{1}{\sqrt{10}} & \frac{1}{\sqrt{10}} & \frac{1}{\sqrt{10}} & \frac{1}{\sqrt{10}}\\\\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\\\\ \frac{1}{\sqrt{10}} & \frac{1}{\sqrt{10}} & \frac{1}{\sqrt{10}} & \frac{1}{\sqrt{10}}\\\\ \end{pmatrix} $

$X$ is an arbitrary matrix with integer-only values.

share|cite|improve this question
Both identities are false; more or less any $2\times 2$ example will fail. – Federico Poloni May 9 '11 at 6:47
Do you mean to say that the example in the textbook is wrong? – misha May 9 '11 at 7:26
What textbook are you using? – J. M. May 9 '11 at 7:53
As it is stated, it seems wrong. Did you maybe forget to specify that some of the matrices you're using are diagonal? – Federico Poloni May 9 '11 at 8:28
@J. M. -- it's from "The H.264 advanced video compression standard" by Richardson – misha May 9 '11 at 8:37
up vote 2 down vote accepted

With these matrices the identity is true.

In order to prove it, notice that $R=DE$, where $D=\operatorname{diag}(\frac12,\frac1{\sqrt{10}},\frac12,\frac1{\sqrt{10}})$ and $E$ is the matrix of all ones. You can juggle around the diagonal factor $D$ (your proof shows that the "triple product identity" holds if $A$ is diagonal) and in particular you can reabsorb it into $C$, which is particularly convenient since now $Q=DC$ is orthogonal (that's where all those $\sqrt{10}$ come from, I guess). So after moving the diagonal factors the two terms become $$ Y_1=(Q \bullet E)\times X\times (Q^T\bullet E^T) $$ $$ Y_2=(Q \times X \times Q^T) \bullet (E\bullet E^T) $$ Here all the remaining Hadamard products have $E$ as one factor and thus are trivial.

Anyway, it definitely looks like someone is overcomplicating things in that book; they could have removed those $R$s and those Hadamard products completely: you can replace every $R\bullet M$ with $D\times M$ and every $M\bullet R^T$ with $M\times D$.

share|cite|improve this answer

What is true, though, is that the diagonal entries of $(A \circ B)C^{T}$ and $(A \circ C)B^{T}$ coincide. (See Lemma 5.1.4 on p. 305 of Topics in Matrix Analysis by Horn and Johnson).

share|cite|improve this answer

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.