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In reading a paper I came across this expression which I don't quite understand: $$ \frac{\lambda_1}{2N}\operatorname{tr}\left((\mathbf{H}^M\mathbf{H}^M)^T\right) $$ For context, $\lambda_1$ and $N$ are scalars and $\mathbf{H}^M$ ($H$ henceforth) is a $MxN$ matrix.

The paper claims that this value is related to the variance of the column vectors which make up H, but either there is a typo or (quite likely) a linear algebra concept I don't know. The reason I am confused is that trace requires square input, and if H were to be made square then the transpose would have to apply to one of the two $H$s, not both after multiplying them (which itself doesn't make sense to me because only a square matrix can be multiplied by itself in the first place).

The paper is here, the mentioned expression is at the bottom of the third page in equation (5).

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  • $\begingroup$ Could it be $\mathbf{H}^M (\mathbf{H}^M)^T$ instead? That would be square for any matrix $\mathbf{H}^M$. $\endgroup$ – Matthew Leingang Jul 25 '18 at 18:36
  • $\begingroup$ That is what I had been thinking, but assuming that I haven't been able to figure out what the relationship to the variance of the column vectors would be. At that point it becomes a stats question more than a linalg one. If that were a typo then the trace results in the sum of the element-wise square of H, which to me seems more correlated to magnitude of values than variance. $\endgroup$ – Sean Hastings Jul 25 '18 at 18:40
  • $\begingroup$ Good point, added. $\endgroup$ – Sean Hastings Jul 25 '18 at 18:44
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It looks like a typo. In the paper that you have linked, the equation(12) on page(4) rewrites

$$ arg.min .J = ... -\frac{\lambda_1}{2}\Bigl(tr\bigl(\frac{1}{N}H^M(H^M)^T\bigr)+\alpha tr(\Sigma_B-\Sigma_W)\Bigr) + ... $$

with the transpose correctly placed.

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