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 := [ A B B ∗ C ] {\displaystyle 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 ) ≤ det ( A ) det ( C ) . {\displaystyle \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 {\displaystyle A^{-1}} and C − 1 {\displaystyle C^{-1}} are positive-definite. Let

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

We note that

D − 1 2 M D − 1 2 = [ A − 1 2 0 0 C − 1 2 ] [ A B B ∗ C ] [ A − 1 2 0 0 C − 1 2 ] = [ I p A − 1 2 B C − 1 2 C − 1 2 B ∗ A − 1 2 I q ] {\displaystyle 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 − 1 2 M D − 1 2 {\displaystyle D^{-{\frac {1}{2}}}MD^{-{\frac {1}{2}}}} , we see

det ( D − 1 2 M D − 1 2 ) ≤ ( 1 p + q t r ( D − 1 2 M D − 1 2 ) ) p + q = 1 p + q = 1. {\displaystyle \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

det ( D − 1 2 ) det ( M ) det ( D − 1 2 ) ≤ 1 ⟹ det ( M ) ≤ det ( D ) = det ( A ) det ( C ) . {\displaystyle {\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 ε > 0 {\displaystyle \varepsilon >0} , as A + ε I p {\displaystyle A+\varepsilon I_{p}} and C + ε I q {\displaystyle C+\varepsilon I_{q}} are positive-definite, we have

det ( M + ε I p + q ) ≤ det ( A + ε I p ) det ( C + ε I q ) . {\displaystyle \det(M+\varepsilon I_{p+q})\leq \det(A+\varepsilon I_{p})\det(C+\varepsilon I_{q}).}

Taking the limit as ε → 0 {\displaystyle \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 Mij, then the following inequality by Thompson is valid:

det ( M ) ≤ det ( [ det ( M i j ) ] ) {\displaystyle \det(M)\leq \det([\det(M_{ij})])}

where [det(Mij)] is the matrix whose (i,j) entry is det(Mij). In particular, if the block matrices B and C are also square matrices, then the following inequality by Everett is valid:

det ( M ) ≤ det [ det ( A ) det ( B ) det ( B ∗ ) det ( C ) ] {\displaystyle \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 ) = ∑ k = 0 n t n − k ( − 1 ) k tr ⁡ ( Λ k A ) {\displaystyle p_{A}(t)=\sum _{k=0}^{n}t^{n-k}(-1)^{k}\operatorname {tr} (\Lambda ^{k}A)}

and supposing that the blocks Mij are m x m matrices, the following inequality by Lin and Zhang is valid:

det ( M ) ≤ ( det ( [ tr ⁡ ( Λ r M i j ] ) ) ( m r ) ) m r , r = 1 , … , m {\displaystyle \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.

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