Hadamard’s Determinant Inequality Author(s): Kenneth Lange Source: The American Mathematical Monthly, Vol. 121, No. 3 (March), pp. 258-259 Published by: Mathematical Association of America Stable URL: http://www.jstor.org/stable/10.4169/amer.math.monthly.121.03.258 . Accessed: 18/05/2014 07:56 Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at . http://www.jstor.org/page/info/about/policies/terms.jsp
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Hadamard’s Determinant Inequality Kenneth Lange Abstract. This note is devoted to a short, but elementary, proof of Hadamard’s determinant inequality.
Let X be an n × n matrix with complex entries and columns x 1 , . . . , x n . Hadamard’s determinant inequality [2] reads |det X| ≤
n Y
kx i k2 .
i=1
Horn and Johnson [1] supply at least a half-dozen proofs for this classical result. Hadamard’s inequality is obviously true if any x i = 0. If we assume otherwise and divide both sides by the right-hand side, then Hadamard’s inequality reduces to the inequality |det X| ≤ 1, subject to the Euclidean length constraints kx i k2 = 1. Equality is attained by taking X to be the identity matrix I or any unitary matrix. Recall that a unitary matrix X has orthonormal columns and consequently satisfies |det X|2 = det X ∗ det X = det(X ∗ X) = det I = 1. In view of Weierstrass’ theorem, the necessarily positive maximum of |det X| is attained on the compact set defined by the constraints. Suppose that X has columns of unit length and yields the maximum value, but two columns of X, say the first two, are not orthogonal. Let us demonstrate that |det X| can be increased by replacing the first column by the linear combination y1 = ax 1 + bx 2 for carefully chosen complex scalars a and b. The determinant of the new matrix Y satisfies det Y = det(ax 1 + bx 2 , x 2 , x 3 , . . . , x n ) = a det X. We force y1 to be a unit vector by imposing the constraint |a|2 + |b|2 + 2Re(abs) ¯ = 1,
s = x ∗1 x 2 .
The Cauchy–Schwarz inequality implies |s| ≤ 1. Furthermore, s 6 = 0 by assumption, and |s| 6= 1, for otherwise x 1 and x 2 are collinear and det X vanishes. The reader can check that we may choose 1 p 2 a 1 − |s| = . b s¯ −p 2 1 − |s| Since a > 1, the identity det Y = a det X shows that |det Y | > |det X|. This contradiction proves that X has orthonormal columns, so it is unitary and |det X| = 1. http://dx.doi.org/10.4169/amer.math.monthly.121.03.258 MSC: Primary 15A15, Secondary 15A45
258
c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121
This content downloaded from 86.157.158.68 on Sun, 18 May 2014 07:56:59 AM All use subject to JSTOR Terms and Conditions
REFERENCES 1. R. A. Horn, C. R. Johnson, Matrix Analysis. Second edition. Cambridge University Press, Cambridge, 2013. 2. J. Hadamard, R´esolution d’une question relative aux d´eterminants, Bulletin des Sciences Math 2 (1893) 240–246. Departments of Biomathematics, Human Genetics, and Statistics, University of California, Los Angeles, CA 90095 [email protected]
Mean of the Mean of the Mean . . . Define the “mean value function” of a continuous function f : [0, 1] −→ R by Z x 1 f (t) dt for 0 < x ≤ 1 T ( f )(x) = x 0 f (0) for x = 0. Then we get a linear operator T : C([0, 1]) −→ C([0, 1]) and the following property holds. Proposition. T n ( f ) tends to the constant function f (0) with respect to the maximum norm on C([0, 1]). ak Proof. For a polynomial function p(t) = ak t k + · · · + a1 t + a0 , we have T ( p)(x) = k+1 xk + a1 n · · · + 2 x + a0 . It follows that T ( p) → p(0) as n → ∞. Furthermore, it can be checked that the operator T is non-expansive, namely:
||T ( f ) − T (g)|| ≤ || f − g||. Suppose f (0) = 0. Given ε > 0, choose a polynomial p with || f − p|| < ε (by Weierstrass). Then ||T n ( f ) − T n ( p)|| ≤ || f − p|| < ε, and hence ||T n ( f )|| → 0 as n → ∞. Now, for any f ∈ C([0, 1]), T n ( f ) = T n ( f (0)) + T n f − f (0) → f (0) as n → ∞.
1
1
1
1 f
1
1
1 T 2( f )
T( f )
1 T 3( f )
—Submitted by S¸. Koc¸ak and M. Limoncu, Department of Mathematics, Anadolu University, 26470 Eskis¸ehir, Turkey http://dx.doi.org/10.4169/amer.math.monthly.121.03.259 MSC: Primary 26A18
March 2014]
NOTES
This content downloaded from 86.157.158.68 on Sun, 18 May 2014 07:56:59 AM All use subject to JSTOR Terms and Conditions
259
. JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected].
.
Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access to The American Mathematical Monthly.
http://www.jstor.org
This content downloaded from 86.157.158.68 on Sun, 18 May 2014 07:56:59 AM All use subject to JSTOR Terms and Conditions
Hadamard’s Determinant Inequality Kenneth Lange Abstract. This note is devoted to a short, but elementary, proof of Hadamard’s determinant inequality.
Let X be an n × n matrix with complex entries and columns x 1 , . . . , x n . Hadamard’s determinant inequality [2] reads |det X| ≤
n Y
kx i k2 .
i=1
Horn and Johnson [1] supply at least a half-dozen proofs for this classical result. Hadamard’s inequality is obviously true if any x i = 0. If we assume otherwise and divide both sides by the right-hand side, then Hadamard’s inequality reduces to the inequality |det X| ≤ 1, subject to the Euclidean length constraints kx i k2 = 1. Equality is attained by taking X to be the identity matrix I or any unitary matrix. Recall that a unitary matrix X has orthonormal columns and consequently satisfies |det X|2 = det X ∗ det X = det(X ∗ X) = det I = 1. In view of Weierstrass’ theorem, the necessarily positive maximum of |det X| is attained on the compact set defined by the constraints. Suppose that X has columns of unit length and yields the maximum value, but two columns of X, say the first two, are not orthogonal. Let us demonstrate that |det X| can be increased by replacing the first column by the linear combination y1 = ax 1 + bx 2 for carefully chosen complex scalars a and b. The determinant of the new matrix Y satisfies det Y = det(ax 1 + bx 2 , x 2 , x 3 , . . . , x n ) = a det X. We force y1 to be a unit vector by imposing the constraint |a|2 + |b|2 + 2Re(abs) ¯ = 1,
s = x ∗1 x 2 .
The Cauchy–Schwarz inequality implies |s| ≤ 1. Furthermore, s 6 = 0 by assumption, and |s| 6= 1, for otherwise x 1 and x 2 are collinear and det X vanishes. The reader can check that we may choose 1 p 2 a 1 − |s| = . b s¯ −p 2 1 − |s| Since a > 1, the identity det Y = a det X shows that |det Y | > |det X|. This contradiction proves that X has orthonormal columns, so it is unitary and |det X| = 1. http://dx.doi.org/10.4169/amer.math.monthly.121.03.258 MSC: Primary 15A15, Secondary 15A45
258
c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121
This content downloaded from 86.157.158.68 on Sun, 18 May 2014 07:56:59 AM All use subject to JSTOR Terms and Conditions
REFERENCES 1. R. A. Horn, C. R. Johnson, Matrix Analysis. Second edition. Cambridge University Press, Cambridge, 2013. 2. J. Hadamard, R´esolution d’une question relative aux d´eterminants, Bulletin des Sciences Math 2 (1893) 240–246. Departments of Biomathematics, Human Genetics, and Statistics, University of California, Los Angeles, CA 90095 [email protected]
Mean of the Mean of the Mean . . . Define the “mean value function” of a continuous function f : [0, 1] −→ R by Z x 1 f (t) dt for 0 < x ≤ 1 T ( f )(x) = x 0 f (0) for x = 0. Then we get a linear operator T : C([0, 1]) −→ C([0, 1]) and the following property holds. Proposition. T n ( f ) tends to the constant function f (0) with respect to the maximum norm on C([0, 1]). ak Proof. For a polynomial function p(t) = ak t k + · · · + a1 t + a0 , we have T ( p)(x) = k+1 xk + a1 n · · · + 2 x + a0 . It follows that T ( p) → p(0) as n → ∞. Furthermore, it can be checked that the operator T is non-expansive, namely:
||T ( f ) − T (g)|| ≤ || f − g||. Suppose f (0) = 0. Given ε > 0, choose a polynomial p with || f − p|| < ε (by Weierstrass). Then ||T n ( f ) − T n ( p)|| ≤ || f − p|| < ε, and hence ||T n ( f )|| → 0 as n → ∞. Now, for any f ∈ C([0, 1]), T n ( f ) = T n ( f (0)) + T n f − f (0) → f (0) as n → ∞.
1
1
1
1 f
1
1
1 T 2( f )
T( f )
1 T 3( f )
—Submitted by S¸. Koc¸ak and M. Limoncu, Department of Mathematics, Anadolu University, 26470 Eskis¸ehir, Turkey http://dx.doi.org/10.4169/amer.math.monthly.121.03.259 MSC: Primary 26A18
March 2014]
NOTES
This content downloaded from 86.157.158.68 on Sun, 18 May 2014 07:56:59 AM All use subject to JSTOR Terms and Conditions
259
