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Euge

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- #1

Euge

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- #2

julian

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$$

\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.

- #3

julian

Gold Member

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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.

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.

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

$$

\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.

Last edited:

- #4

julian

Gold Member

- 736

- 225

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

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]##.

$$

\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]##.

Last edited:

- #5

Euge

Gold Member

MHB

POTW Director

- 1,980

- 1,332

$$

\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.

- #6

Office_Shredder

Staff Emeritus

Science Advisor

Gold Member

- 5,469

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Wow, that's a ridiculously slick proof.

- #7

Euge

Gold Member

MHB

POTW Director

- 1,980

- 1,332

Thanks @Office_Shredder!

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