Fischer's Inequality
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mathematics Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantities and their changes. These topics are represented in modern mathematics ...
, Fischer's inequality gives an upper bound for the
determinant In mathematics, the determinant is a scalar value that is a function of the entries of a square matrix. It characterizes some properties of the matrix and the linear map represented by the matrix. In particular, the determinant is nonzero if and ...
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 A & B \\ B^* & C \end\right/math> so that ''M'' is a (''p''+''q'')×(''p''+''q'') matrix. Then Fischer's inequality states that : \det (M) \le \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 In mathematics, Hadamard's inequality (also known as Hadamard's theorem on determinants) is a result first published by Jacques Hadamard in 1893.Maz'ya & Shaposhnikova It is a bound on the determinant of a matrix whose entries are complex numbe ...
. 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^ and C^ are positive-definite. Let :D := \left begin A & 0 \\ 0 & C \end\right We note that :D^ M D^ = \left begin A^ & 0 \\ 0 & C^ \end\right\left begin A & B \\ B^* & C \end\right\left begin A^ & 0 \\ 0 & C^ \end\right= \left begin I_ & A^ BC^ \\ C^B^*A^ & I_\end\right/math> Applying the AM-GM inequality to the eigenvalues of D^ M D^, we see :\det (D^ M D^) \le \left( \mathrm (D^ M D^) \right)^ = 1^ = 1. By multiplicativity of
determinant In mathematics, the determinant is a scalar value that is a function of the entries of a square matrix. It characterizes some properties of the matrix and the linear map represented by the matrix. In particular, the determinant is nonzero if and ...
, we have : \begin \det(D^ ) \det(M) \det(D^ ) \le 1 \\ \Longrightarrow \det(M) \le \det(D) = \det(A) \det(C). \end 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_) \le \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 ''Mij'', then the following inequality by Thompson is valid: : \det(M) \le \det( det(M_) where et(''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) \le \det \begin \det(A) && \det(B) \\ \det(B^*) && \det(C) \end Thompson's inequality can also be generalized by an inequality in terms of the coefficients of the
characteristic polynomial In linear algebra, the characteristic polynomial of a square matrix is a polynomial which is invariant under matrix similarity and has the eigenvalues as roots. It has the determinant and the trace of the matrix among its coefficients. The chara ...
of the block matrices. Expressing the characteristic polynomial of the matrix ''A'' as : p_A (t) = \sum_^n t^ (-1)^k \operatorname(\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) \le \left(\frac \right)^,\quad r=1, \ldots, m Note that if ''r'' = ''m'', then this inequality is identical to Thompson's inequality.


See also

*
Hadamard's inequality In mathematics, Hadamard's inequality (also known as Hadamard's theorem on determinants) is a result first published by Jacques Hadamard in 1893.Maz'ya & Shaposhnikova It is a bound on the determinant of a matrix whose entries are complex numbe ...


Notes


References

* . * . {{DEFAULTSORT:Fischer's Inequality Inequalities Determinants