I was going through this lecture related to convex optimization. It was proved that logdet function is concave. However, I didn't get the derivation at a part

enter image description here

I didn't get how the step marked in red in the given picture was derived. Which property was used


1 Answer 1


There are three properties being used here:

  • the determinant of a product of square matrices of equal size is equal to the product of the determinants of its square, equally sized factors
  • the logarithm of a product is equal to the sum of the logarithms of its factors
  • the determinant of a matrix is equal to the product of its eigenvalues.

Let's start from

\begin{align} g(t) = \log \det (Z^{1/2}(I + tZ^{-1/2}VZ^{-1/2})Z^{1/2}). \end{align}

The determinant product property yields

\begin{align} g(t) = \log\Big(\det(Z^{1/2})\det(I + tZ^{-1/2}VZ^{-1/2})\det(Z^{1/2})\Big), \end{align}

and by commutativity of multiplication, we can rearrange it so that

\begin{align} g(t) &= \log\Big(\det(Z^{1/2})\det(Z^{1/2})\det(I + tZ^{-1/2}VZ^{-1/2})\Big), \\ &= \log\Big(\det(I + tZ^{-1/2}VZ^{-1/2})\det(Z)\Big). \end{align}

Using the product-sum property of logarithms, this expression becomes

\begin{align} g(t) &= \log\det(I + tZ^{-1/2}VZ^{-1/2}) + \log\det(Z), \end{align}

and we're most of the way there with the $\det(Z)$ term already separated out.

If the eigenvalues of $Z^{-1/2}VZ^{-1/2}$ are $\lambda_{i}$, then:

  • the eigenvalues of $tZ^{-1/2}VZ^{-1/2}$ are $t\lambda_{i}$, because the eigenvectors of each matrix are the same
  • the eigenvalues of $I + tZ^{-1/2}VZ^{-1/2}$ are $(1 + t\lambda_{i})$ because the eigenvectors of $Z^{-1/2}VZ^{-1/2}$ will also be eigenvectors for $I$.

This aside about eigenvalues is important, because to complete the derivation, we need to use the property that the determinant of a matrix is the product of its eigenvalues:

\begin{align} g(t) &= \log\left(\prod_{i=1}^{n}(1+t\lambda_{i})\right) + \log\det(Z). \end{align}

To complete the derivation, we use the product-sum property of logarithms (again), and get

\begin{align} g(t) = \sum_{i=1}^{n} \log(1 + t\lambda_{i}) + \log\det(Z). \end{align}

  • $\begingroup$ I didn't get it. They have mentioned that they replaced the concave term by the convex one. However, I didn't get how the terms $\endgroup$
    – user34790
    Mar 10, 2013 at 2:29

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.