Log Concavity of Determinants

[Euge]

Let $A, B\in M_n(\mathbb{R})$ be positive definite matrices. Prove that for $0\le t\le 1$, $$\det(tA + (1 - t)B) \ge \det(A)^t\det(B)^{1-t}$$

 

[julian]

Spoiler

The case $t=0$ is considered separately. Obviously

$$
\det B \geq (\det A)^0 \det B .
$$

We now consider the case $0 < t \leq 1$. I start with

$$
\det (t A + (1-t) B) \geq (\det A)^t (\det B)^{1-t}
$$

and work backwards to an inequality that can be seen to hold. Take the log of both sides

$$
\ln \det ( t A + (1-t) B) \geq t \ln \det A + (1-t) \ln \det B
$$

where we have used that $\det A > 0$ and $\det B > 0$. As $\det A \not= 0$ the matrix $A$ has an inverse and the above inequality becomes

$$
\ln \det [tA ( \mathbb{1} + t^{-1} (1-t) A^{-1} B)] \geq t \ln \det A + (1-t) \ln \det B
$$

or

$$
\ln \det (t A) + \ln \det ( \mathbb{1} + t^{-1} (1-t) A^{-1} B) \geq t \ln \det A + (1-t) \ln \det B
$$

or

$$
nt + \ln \det ( \mathbb{1} + t^{-1} (1-t) A^{-1} B) \geq (1-t) (\ln \det B - \ln \det A)
$$

or

$$
nt + \ln \det ( \mathbb{1} + t^{-1} (1-t) A^{-1} B) \geq (1-t) \ln \det A^{-1} B
$$

The matrix $A^{-1} B$ is not necessarily symmetric and so can't in general be diagonalised, but there is a similarity transformation that puts it into Jordan normal form. So that

$$
nt + \ln \det [S^{-1} ( \mathbb{1} + t^{-1} (1-t) A^{-1} B) S] \geq (1-t) \ln \det (S^{-1} A^{-1} B S)
$$

becomes

$$
nt + \ln \det ( \mathbb{1} + t^{-1} (1-t) J) \geq (1-t) \ln \det J
$$

Which, in terms of eigenvalues is

$$
nt + \ln \prod_{i=1}^n ( 1 + t^{-1} (1-t) \lambda_i^{-1} \mu_i) \geq (1-t) \ln (\prod_{i=1}^n \lambda_i^{-1} \mu_i)
$$

where $\lambda_i$ are the eigenvalues of $A$ and $\mu_i$ are the eigenvalues of $B$. These eigenvalues are positive and so the logarithm makes sense. The above inequality can be rewritten as

$$
nt + \sum_{i=1}^n ( 1 + t^{-1} (1-t) \lambda_i^{-1} \mu_i) \geq (1-t) \sum_{i=1}^n \lambda_i^{-1} \mu_i
$$

or

\begin{align*}
nt + \sum_{i=1}^n 1 & \geq [(1-t) - t^{-1} (1-t)] \sum_{i=1}^n \lambda_i^{-1} \mu_i
\nonumber \\
& = - (1-t)^2 t^{-1} \sum_{i=1}^n \lambda_i^{-1} \mu_i
\end{align*}

which obviously holds as the RHS is non-positive for $t \in (0,1]$ as $\lambda_i^{-1} \mu_i$ ($i = 1 , \dots , n$) is positive.

 

[julian]

I think I spotted an error in the argument of my previous post, but I think I know how to fix it.

EDIT: on 9th Sept I have added further explanation.

Spoiler

In my previous post I demonstrated that

$$
\det (t A + (1-t) B) \geq (\det A)^t (\det B)^{1-t}
$$

reduces to

$$
nt + \ln \det ( \mathbb{1} + t^{-1} (1-t) A^{-1} B) \geq (1-t) \ln \det A^{-1} B \qquad (*)
$$

A positive definite matrix has a positive definite square-root matrix. Proof:

The matrix $B$ is positive definite and so there is a similarity transformation such that $S^{-1} B S = D$ where $D$ is the diagonal matrix. Where the matrix $S$ is chosen to be the matrix with the $n$ eigenvectors as columns. The elements of $D$ are eigenvalues of $B$ and so are positive. When a matrix is symmetric, the diagonalizing matrix $S$ can be made an orthogonal matrix by suitably choosing the eigenvectors (then $S^{-1} = S^T$). As such we can define a positive definite square-root matrix $B^{1/2} = S D^{1/2} S^T$ ($B^{1/2}$ has the same eigenvectors as $B$ but with eigenvalues that are the square-root of the eigenvalues of $B$).
Q.E.D.

Now, the matrix $A^{-1} B$ can be written as $B^{-1/2} (B^{1/2} A^{-1} B^{1/2}) B^{1/2}$. As such the matrix $A^{-1} B$ has the same eigenvalues as the positive definite matrix $B^{1/2} A^{-1} B^{1/2}$ because:

$$
\det (\mathbb{1} \lambda - A^{-1} B) = \det (\mathbb{1} \lambda - B^{-1/2} (B^{1/2} A^{-1} B^{1/2}) B^{1/2}) = \det (\mathbb{1} \lambda - B^{1/2} A^{-1} B^{1/2})
$$

As such the matrix $A^{-1} B$ has positive eigenvalues, let us denote them as $\nu_i$.

The matrix $A^{-1} B$ is not necessarily symmetric and so can't in general be diagonalised, but there is a similarity transformation that puts it into Jordan normal form. So that $(*)$ becomes

$$
nt + \ln \det ( \mathbb{1} + t^{-1} (1-t) J) \geq (1-t) \ln \det J
$$

where the diagonal elements of $J$ are the $\nu_i > 0$ ($i = 1 , \dots , n$). This reduces to

$$
nt + \ln \prod_{i=1}^n ( 1 + t^{-1} (1-t) \nu_i) \geq (1-t) \ln (\prod_{i=1}^n \nu_i)
$$

These eigenvalues are positive and so the logarithm makes sense. The above inequality can be rewritten as

$$
nt + \sum_{i=1}^n ( 1 + t^{-1} (1-t) \nu_i) \geq (1-t) \sum_{i=1}^n \nu_i
$$

or

\begin{align*}
nt + \sum_{i=1}^n 1 & \geq [(1-t) - t^{-1} (1-t)] \sum_{i=1}^n \nu_i
\nonumber \\
& = - (1-t)^2 t^{-1} \sum_{i=1}^n \nu_i
\end{align*}

which obviously holds as the RHS is non-positive for $t \in (0,1]$ as $\nu_i$ ($i = 1 , \dots , n$) is positive.

 

[julian]

I have made a mistake in the previous attempts (inattention!). I think this time I have done the problem.

Spoiler

I start with

$$
\det (t A + (1-t) B) \geq (\det A)^t (\det B)^{1-t}
$$

and work backwards to an inequality that can be proven to hold. Take the log of both sides

$$
\ln \det ( t A + (1-t) B) \geq t \ln \det A + (1-t) \ln \det B
$$

where we have used that $\det A > 0$ and $\det B > 0$. As $\det A \not= 0$ the matrix $A$ has an inverse and the above inequality becomes

$$
\ln \det [A ( t\mathbb{1} + (1-t) A^{-1} B)] \geq t \ln \det A + (1-t) \ln \det B
$$

or

$$
\ln \det A + \ln \det ( t \mathbb{1} + (1-t) A^{-1} B) \geq t \ln \det A + (1-t) \ln \det B
$$

or

$$
\ln \det ( t \mathbb{1} + (1-t) A^{-1} B) \geq (1-t) (\ln \det B - \ln \det A)
$$

or

$$
\ln \det ( t \mathbb{1} + (1-t) A^{-1} B) \geq (1-t) \ln \det A^{-1} B \qquad (*)
$$

A positive definite matrix has a positive definite square-root matrix. Proof:

The matrix $B$ is positive definite and so there is a similarity transformation such that $S^{-1} B S = D$ where $D$ is the diagonal matrix. Where the matrix $S$ is chosen to be the matrix with the $n$ eigenvectors as columns. The elements of $D$ are eigenvalues of $B$ and so are positive. When a matrix is symmetric, the diagonalizing matrix $S$ can be made an orthogonal matrix by suitably choosing the eigenvectors (then $S^{-1} = S^T$). As such we can define a positive definite square-root matrix $B^{1/2} = S D^{1/2} S^T$ ($B^{1/2}$ has the same eigenvectors as $B$ but with eigenvalues that are the square-root of the eigenvalues of $B$).
Q.E.D.

Now, the matrix $A^{-1} B$ can be written as $B^{-1/2} (B^{1/2} A^{-1} B^{1/2}) B^{1/2}$. As such the matrix $A^{-1} B$ has the same eigenvalues as the positive definite matrix $B^{1/2} A^{-1} B^{1/2}$ because:

$$
\det (\mathbb{1} \lambda - A^{-1} B) = \det (\mathbb{1} \lambda - B^{-1/2} (B^{1/2} A^{-1} B^{1/2}) B^{1/2}) = \det (\mathbb{1} \lambda - B^{1/2} A^{-1} B^{1/2})
$$

As such the matrix $A^{-1} B$ has positive eigenvalues, let us denote them as $\nu_i$.

The matrix $A^{-1} B$ is not necessarily symmetric and so can't in general be diagonalised, but there is a similarity transformation that puts it into Jordan normal form. So that $(*)$ becomes

$$
\ln \det ( t \mathbb{1} + (1-t) J) \geq (1-t) \ln \det J
$$

where the diagonal elements of $J$ are the $\nu_i > 0$ ($i = 1 , \dots , n$). This reduces to

$$
\ln \prod_{i=1}^n ( t + (1-t) \nu_i) \geq (1-t) \ln (\prod_{i=1}^n \nu_i)
$$

These eigenvalues are positive and so the logarithm makes sense. The above inequality can be rewritten as

$$
\sum_{i=1}^n \ln ( t + (1-t) \nu_i) \geq (1-t) \sum_{i=1}^n \ln \nu_i .
$$

In order to prove this inequality we introduce the function:

$$
f(t) = \sum_{i=1}^n \ln ( t + (1-t) \nu_i) - (1-t) \sum_{i=1}^n \ln \nu_i .
$$

First note that $f(0) = f(1) = 0$. The first derivative is

$$
f'(t) = \sum_{i=1}^n \frac{1- \nu_i}{\nu_i + t (1- \nu_i)} + \sum_{i=1}^n \ln \nu_i
$$

and so the second derivative is

$$
f'' (t) = - \sum_{i=1}^n \frac{(1- \nu_i)^2}{(\nu_i + t (1- \nu_i))^2} < 0 .
$$

As $f(t)$ is twice-differentiable and $f''(t)$ is non-positive the function is concave. Combining this with $f(0) = f(1) = 0$, we have

$$
f (t) = f ((1-t) \cdot 0 + t \cdot 1) \geq (1-t) f(0) + t f(1) = 0
$$

for $t \in [0,1]$.

 

[Euge]

Spoiler

For every $t\in [0,1]$, $tA + (1 - t)B$ is real positive definite. Hence
$$
\frac{\pi^{n/2}}{\sqrt{\det(tA + (1 - t)B)}} = \int_{\mathbf{R}^n} e^{-\mathbf{x}\cdot (tA + (1 - t)B)\mathbf{x}}\, d\mathbf{x} = \int_{\mathbf{R}^n} (e^{-\mathbf{x}\cdot A\mathbf{x}})^t (e^{-\mathbf{x}\cdot B\mathbf{x}})^{1-t}\, d\mathbf{x}
$$
By Hölder's inequality applied with conjugate exponents $1/t$ and $1/(1-t)$, the latter integral is dominated by
$$
\left(\int_{\mathbf{R}^n} e^{-\mathbf{x}\cdot A\mathbf{x}}\, d\mathbf{x}\right)^t \left(\int_{\mathbf{R}^n} e^{-\mathbf{x}\cdot B\mathbf{x}}\, d\mathbf{x}\right)^{1-t}
$$
The latter expression equals
$$
\left(\frac{\pi^{n/2}}{\sqrt{\det A}}\right)^t \left(\frac{\pi^{n/2}}{\sqrt{\det B}}\right)^{1-t} = \frac{\pi^{n/2}}{\sqrt{(\det A)^t (\det B)^{1-t}}}
$$
Thus
$$
\frac{\pi^{n/2}}{\sqrt{\det(tA + (1 - t)B)}} \le \frac{\pi^{n/2}}{\sqrt{(\det A)^t (\det B)^{1-t}}}
$$
which is equivalent to the desired inequality.

 

[Office_Shredder]

Wow, that's a ridiculously slick proof.

 

[Euge]

Thanks @Office_Shredder!