I saw the following statement in the book "Pattern Recognition and Machine Learning": "

"We note that the matrix $\Sigma$ can be taken to be symmetric, without loss of generality, because any antisymmetric component would disappear from the exponent.

The statement is about Gaussian distribution: [

Could someone explain me why exactly we can take $\Sigma$ to be symmetric, and how antisymmetric components would disappear from the exponent?


  • $\begingroup$ The statement would make a lot of sense if $x$ and $\mu$ are column-vectors. Are you sure about the dimensions of those matrices? Can we say that $x$ and $\mu$ are $n \times 1$? $\endgroup$ – Omnomnomnom Nov 6 '17 at 10:24
  • $\begingroup$ Yes, you are write. I took the image from the google search, and I had to cut the part showing dimensions. I will fix it. The question remains, however. $\endgroup$ – Valeria Nov 6 '17 at 12:15

Suppose that $M$ is an $n \times n$ matrix and $y$ is an $n \times 1$ column vector. Let $S = \frac 12 (M + M^T)$ and $A = \frac 12 (M - M^T)$. Notably, $S$ and $A$ are sometimes called the symmetric and antisymmetric parts of $M$; in particular, verify that $M = S + A$, $S^T = S$, and $A^T = -A$.

Because $A$ is antisymmetric, we note that since $y^TAy$ is $1 \times 1$, we have $$ y^TAy = (y^TAy)^T = y^TA^Ty = y^T(-A)y = -y^TAy $$ from which we may conclude that $y^TAy = 0$. Thus, we note that $$ y^TMy = y^T(S + A)y = y^TSy + y^TAy = y^TSy $$ So, replacing $M$ with the symmetric matrix $S$ will not change the value of $y^TMy$.

For the purposes of your expression, we can take $M = \Sigma^{-1}$ and $y = x - \mu$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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