# If $\Sigma^{-1}=(A^{-1})^TA^{-1}$, then why does $|A^{-1}|=|\Sigma|^{-1/2}$?

In the derivation of the joint pdf of $$f_\textbf{X}(\pmb{x})$$, where $$\textbf{X}=\pmb\mu+A\pmb Z$$ and $$\textbf{X}\sim~N_n(\pmb\mu,\Sigma)$$, there is a step I do not understand.

In particular, it is the fact that if $$\Sigma^{-1}=(A^{-1})^TA^{-1}$$, then $$|A^{-1}|=|\Sigma|^{-1/2}$$. This shows up during the derivation of the joint pdf as follows:

$$f_{\textbf{X}}(\textbf{x})=|A^{-1}|(2\pi)^{-n/2}exp\bigg(-\frac{1}{2} (\pmb x-\pmb \mu)^T(A^{-1})^TA^{-1}(\pmb x-\pmb \mu)\bigg)$$

Now since $$\Sigma=AA^T$$ is invertible we have $$\Sigma^{-1}=(AA^T)^{-1}=(A^T)^{-1}A^{-1}=(A^{-1})^TA^{-1}$$ and thus,

$$f_{\textbf{X}}(\textbf{x})=(2\pi)^{-n/2}|\Sigma|^{-1/2}exp\bigg(-\frac{1}{2} (\pmb x-\pmb \mu)^T\Sigma^{-1}(\pmb x-\pmb \mu)\bigg)$$

I can't understand how $$|\Sigma|^{-1/2}=|A^{-1}|$$?

• Please insert the definition of $|B|$ for a given matrix $B$... – dan_fulea May 15 at 16:27
• That's what you get taking determinant of both sides. – StubbornAtom May 15 at 16:28
• Hint: a matrix and its transpose have the same determinant – David Reed May 15 at 16:33
• I guess $A$ is square matrix, then $|\Sigma|^{-1/2}=|A^{-1}|$ follows from $|A|=|A^T|$ – kludg May 15 at 16:35
• It's also a theorem that $|A^{-1}| = |A|^{-1}$. This is actually easy enough to prove in a comment: $1 = |I| = |A^{-1}A| = |A^{-1}||A| \to |A^{-1}| = |A|^{-1}$ – David Reed May 15 at 16:49