Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

How to prove the following inequality: Let X and Y be n*m matrices with real entries. Prove that det(XY^T)^2\leq det(XX``^T)det(YY^T)

share|improve this question
add comment

3 Answers

If you replace determinants by traces, then this inequality is just Cauchy-Schwartz for the inner product $(X,Y)=\mathop{\mathrm{tr}}(XY^T)$ on the space of matrices. Now, we just have to recall that determinants are traces: for a $n\times n$-matrix $A$ we have $\det(A)=\mathop{\mathrm{tr}}(\Lambda^n(A))$, where $\Lambda^n(A)$ is the $n$-th exterior power of the linear operator $A$.

(Now that I took time to look at answers above I believe this is more or less precisely Wadim Zudilin's proof made a tiny bit more invariant.)

share|improve this answer
1  
The answers are not above any more! :-) –  Wadim Zudilin May 25 '10 at 11:19
    
True - I keep forgetting about the built-in nonlinearity :-) –  Vladimir Dotsenko May 25 '10 at 12:46
add comment

If $n>m$ all the determinants are zero.

The Cauchy-Binet formula allows you to write the determinant of $XY^t$ as the sum $$ \sum_{I\subset\lbrace1,2,\dots,m\rbrace}\det(X_IY_I^t) =\sum_{I\subset\lbrace1,2,\dots,m\rbrace}\det(X_I)\det(Y_I^t) $$ where the sum is over all $n$-element subsets of $\lbrace1,2,\dots,m\rbrace$. Here $X_I$ and $Y_I$ are quadratic $n\times n$ matrices with columns of $X$ and $Y$ from the set $I$. The similar formulas are available for $\det(XX^t)$ and $\det(YY^t)$, so the wanted inequality reduces to verification of $$ \left(\sum_{I}x_Iy_I\right)^2\le \sum_{I}x_I^2\sum_Iy_I^2. $$ This a known inequality.

share|improve this answer
    
Yep, I think this is the easiest answer. Thank you very much! –  Marine May 28 '10 at 4:12
    
You are welcome. –  Wadim Zudilin May 28 '10 at 6:27
add comment

Denotes vectors-rows of $X$ by $x_1$, $\dots$, $x_n$, of $Y$ by $y_1$, $\dots$, $y_n$, all $x_i$, $y_i$ belong to $\mathbb{R}^m$. Let $y_i=z_i+w_i$, where $z_i$ belongs to linear span of $x_1$,$\dots$,$x_n$, and $w_i$ is orthogonal to this linear span. If we replace $y_i$ to $z_i$, the left hand side does not change, while the right hand side may only decrease ($YY^t=ZZ^t+WW^t$ with natural notations, and all matrices are non-negative definite, so all eigenvalues of $YY^t$ are at least the same as those of $ZZ^t$, hence $\det(YY^t)\geq \det(ZZ^t)$.) But if all $x_i$, $y_i$ belong to the same subspace of dimension $n$, then equality in inicial inequality occurs, as we may think that $m=n$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.