Fischer's inequality
# Fischer's inequality

In mathematics, Fischer's inequality gives an upper bound for the determinant of a positive-semidefinite matrix whose entries are complex numbers in terms of the determinants of its principal diagonal blocks. Suppose A, C are respectively p×p, q×q positive-semidefinite complex matrices and B is a p×q complex matrix. Let

    $M:=\left[{\begin{matrix}A&B\\B^{*}&C\end{matrix}}\right]$

so that M is a (p+q)×(p+q) matrix.

Then Fischer's inequality states that

    $\det(M)\leq \det(A)\det(C).$

If M is positive-definite, equality is achieved in Fischer's inequality if and only if all the entries of B are 0. Inductively one may conclude that a similar inequality holds for a block decomposition of M with multiple principal diagonal blocks. Considering 1×1 blocks, a corollary is Hadamard's inequality. On the other hand, Fischer's inequality can also be proved by using Hadamard's inequality, see the proof of Theorem 7.8.5 in Horn and Johnson's Matrix Analysis.

## Proof

Assume that A and C are positive-definite. We have $A^{-1}$ and $C^{-1}$ are positive-definite. Let

    $D:=\left[{\begin{matrix}A&0\\0&C\end{matrix}}\right].$

We note that

    $D^{-{\frac {1}{2}}}MD^{-{\frac {1}{2}}}=\left[{\begin{matrix}A^{-{\frac {1}{2}}}&0\\0&C^{-{\frac {1}{2}}}\end{matrix}}\right]\left[{\begin{matrix}A&B\\B^{*}&C\end{matrix}}\right]\left[{\begin{matrix}A^{-{\frac {1}{2}}}&0\\0&C^{-{\frac {1}{2}}}\end{matrix}}\right]=\left[{\begin{matrix}I_{p}&A^{-{\frac {1}{2}}}BC^{-{\frac {1}{2}}}\\C^{-{\frac {1}{2}}}B^{*}A^{-{\frac {1}{2}}}&I_{q}\end{matrix}}\right]$

Applying the AM-GM inequality to the eigenvalues of $D^{-{\frac {1}{2}}}MD^{-{\frac {1}{2}}}$, we see

    $\det(D^{-{\frac {1}{2}}}MD^{-{\frac {1}{2}}})\leq \left({1 \over p+q}\mathrm {tr} (D^{-{\frac {1}{2}}}MD^{-{\frac {1}{2}}})\right)^{p+q}=1^{p+q}=1.$

By multiplicativity of determinant, we have

    ${\begin{aligned}\det(D^{-{\frac {1}{2}}})\det(M)\det(D^{-{\frac {1}{2}}})\leq 1\\\Longrightarrow \det(M)\leq \det(D)=\det(A)\det(C).\end{aligned}}$

In this case, equality holds if and only if M = D that is, all entries of B are 0.

For $\varepsilon >0$, as $A+\varepsilon I_{p}$ and $C+\varepsilon I_{q}$ are positive-definite, we have

    $\det(M+\varepsilon I_{p+q})\leq \det(A+\varepsilon I_{p})\det(C+\varepsilon I_{q}).$

Taking the limit as $\varepsilon \rightarrow 0$ proves the inequality. From the inequality we note that if M is invertible, then both A and C are invertible and we get the desired equality condition.

## Improvements

If M can be partitioned in square blocks $M_{ij},$ then the following inequality by Thompson is valid:

    $\det(M)\leq \det([\det(M_{ij})])$

where $[det(M_{ij})]$ is the matrix whose (i,j) entry is $det(M_{ij}).$

In particular, if the block matrices B and C are also square matrices, then the following inequality by Everett is valid:

    $\det(M)\leq \det {\begin{bmatrix}\det(A)&&\det(B)\\\det(B^{*})&&\det(C)\end{bmatrix}}$

Thompson's inequality can also be generalized by an inequality in terms of the coefficients of the characteristic polynomial of the block matrices. Expressing the characteristic polynomial of the matrix A as

    $p_{A}(t)=\sum _{k=0}^{n}t^{n-k}(-1)^{k}\operatorname {tr} (\Lambda ^{k}A)$

and supposing that the blocks $M_{ij}$ are m x m matrices, the following inequality by Lin and Zhang is valid:

    $\det(M)\leq \left({\frac {\det([\operatorname {tr} (\Lambda ^{r}M_{ij}]))}{\binom {m}{r}}}\right)^{\frac {m}{r}},\quad r=1,\ldots ,m$

Note that if r = m, then this inequality is identical to Thompson's inequality.

