Let $A=(a_{ij})_{n\times n}$ such $a_{ij}>0$ and $\det(A)>0$.
Defining the matrix $B:=(a_{ij}^{\frac{1}{n}})$, show that $\det(B)>0?$.
This problem is from my friend, and I have considered sometimes, but I can't. Thank you
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2Your friend has fun problems. But this one is only true for $n=1$ and $n=2$ I think. – Julien Nov 22 '13 at 14:19
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4You need to change friends: they keep on tricking you. – DonAntonio Nov 22 '13 at 14:20
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@julien: With $n$, you seem to mean the (row) size of the matrix, I suppose. – ccorn Nov 22 '13 at 14:23
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@ccorn: That is in the question all right. – Marc van Leeuwen Nov 22 '13 at 14:23
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@MarcvanLeeuwen: Oh yes, thanks. I overlooked that. Well for $2\times2$ matrices, an arbitrary positive exponent would work. For matrix size $3\times3$, there exist sign discrepancies with exponent $\frac{1}{2}$ as well. – ccorn Nov 22 '13 at 14:33
3 Answers
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The statement is false. Counterexample: $$ B=\pmatrix{2&2&2\\ 3&4&2\\ 3&1&4},\ A=B\circ B\circ B=\pmatrix{8&8&8\\ 27&64&8\\ 27&1&64},\ \det(B)=-2,\ \det(A)=7000. $$
Edit. However, if $B$ is positive definite, the converse of your statement is true. More specifically, suppose $B$ is positive definite (whether it is entrywise positive is unimportant) and $A$ is the $n$-fold Hadamard product of $B$ (i.e. $A=(b_{ij}^n)$). Then $\det(A)>0$. This is because for positive semidefinite matrices $X$ and $Y$, we have $\det(X\circ Y)\ge\det(X)\det(Y)$.
user1551
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This is wrong. $$ \begin{pmatrix}27&8&1\\1&1&\epsilon\\8&\epsilon&1\end{pmatrix} $$ where $\epsilon>0$ is very small.
Marc van Leeuwen
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It is false when $n\ge 3$. Counter-example: for $x>0$ small, let
$$A(x)=\begin{pmatrix}8 & (1+3x)^3 & 1\\ 1 & (1+x)^3 & 1 \\ x^3 & 1& 1\end{pmatrix}.$$ Then
$$B(x)=\begin{pmatrix}2 & 1+3x & 1\\ 1 & 1+x & 1 \\ x & 1& 1\end{pmatrix}.$$ Direct calculation shows that when $x>0$ is small enough, $$\det A(x)=15x+O(x^2)>0,\quad \det B(x)=-x+O(x^2)<0.$$
Hu Zhengtang
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