I am trying to understand how 'completing the square' method is used to determine mean and co-variance of multivariate Gaussian distribution but I am not able to make much sense of the following step $$−((x−μ)^TΣ^-(x − μ))/2 = −(x^TΣ^−x + x^TΣ^-μ + const)/2$$ where x and μ are vectors and Σ is co-variance matrix.

What does this step indicate and how is it derived?

P.S. I am not well versed with linear algebra which might be a reason for not being able to comprehend this fully, if you can point me to certain properties from linear algebra that are being used here, that will be very helpful


1 Answer 1


First, here's how you would complete the square with scalars (no matrices!)

$$ax^2 - 2bx = a\left(x-\frac{b}{a} \right)^2 - \frac{b^2}{a}$$

Please verify this for yourself.

Now, think intuitively about how you could rewrite this expression for matrices. Generally, a quadratic term $az^2$ for scalar values corresponds to a term $z^T A z$ when in matrix form. This is sometimes called a "quadratic form". Another thing is that we can't divide matrices, but we can multiply by the inverse, so a fraction like $\frac{b}{a}$ is usually converted to $A^{-1} b$.

Now we can rewrite the scalar complete-the-square equation in matrix form.

$$x^TA x - 2b^Tx = \left(x-A^{-1}b \right)^TA\left(x-A^{-1}b \right) - b^TA^{-1}b$$

Please expand this equation and verify that it is actually correct, and that the ad-hoc reasoning we did above actually works.

Now replace $b = \Sigma^{-1} \mu$ and $A = \Sigma$, and $\text{const} = b^TA^{-1} b$. You should arrive at something which resembles equation in the question.

Some linear algebra tips which might help: The matrix $A$ or $\Sigma$ is generally treated as symmetric without loss of generality, so $A = A^T$. Also I believe you wrote $\Sigma^-$ in your post but meant $\Sigma^{-1}$

  • $\begingroup$ (+1) Very nice explanation. BTW, $\Sigma^{-}$ often is used to indicate a pseudo-inverse of $\Sigma$ in case it might be singular. $\endgroup$
    – whuber
    Aug 14, 2017 at 17:31
  • $\begingroup$ It should be a $-\frac{b^2}{a}$ instead of $+\frac{b^2}{a}$ in the last term of the right hand side of the equation. Shouldn't it? $\endgroup$
    – Nip
    May 10, 2020 at 18:29
  • $\begingroup$ @Nip thanks for catching that, fixed $\endgroup$
    – shimao
    May 16, 2020 at 1:29

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