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I want to prove that $\det (A+X)\det (A-X) \leq \det (A^2)$ where $X $ is a matrix whose $n^2$ entries are all the same.

I tried to write down the expressions involved but that didn't help me prove the inequality.

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    $\begingroup$ $det(A+X)det(A-X)=det((A+X)(A-X))=det(A^2-X^2)$ $\endgroup$ – Jorge Fernández Hidalgo Jan 17 '16 at 20:43
  • $\begingroup$ X^2 is essentially just a matrix with the same non-negative entry. $\endgroup$ – Jorge Fernández Hidalgo Jan 17 '16 at 20:44
  • $\begingroup$ @dREaM: I, too, was thinking along those lines -- but we need that $X$ commutes with anything, which is apparently true, but how to prove it? $\endgroup$ – Eli Rose Jan 17 '16 at 20:45
  • $\begingroup$ Oh yeah, good point. $\endgroup$ – Jorge Fernández Hidalgo Jan 17 '16 at 20:45
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    $\begingroup$ @dREaM: Oops, not true. (Anything that commutes with all matrices is a scalar multiple of $I$). $\endgroup$ – Eli Rose Jan 17 '16 at 20:50
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The matrix determinant lemma states that if $A$ is an invertible $n \times n$ matrix and $u,w$ are $n \times 1$ vectors, then $\det(A+uw^T) = (1+v^TA^{-1}u)\det(A)$.

Since $X$ has all of its $n^2$ entries equal to some number $a$, we have $X = a\vec{1}\vec{1}^T$, where $\vec{1}$ is a vector of all ones.

Hence, $\det(A+X) = \det(A+a\vec{1}\vec{1}^T) = (1+a\vec{1}^TA^{-1}\vec{1})\det(A)$,

and $\det(A-X) = \det(A-a\vec{1}\vec{1}^T) = (1-a\vec{1}^TA^{-1}\vec{1})\det(A)$.

Multiply these two equations to get: $\det(A+X)\det(A-X) = (1-(a\vec{1}^TA^{-1}\vec{1})^2)\det(A)^2 \le \det(A)^2 = \det(A^2)$

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Schematically, $A = [a_1\ldots a_n]$ where $a_1$ to $a_n$ are the columns of $A$; use similar notation for $X$. By i) the multilinearity of the determinant, ii) $\text{det}C = 0$ whenever the columns of $C$ are repeated, and iii) $\text{det}(BC)=\text{det}B\,\text{det}C$, we have $$ \text{det}(A+X)\text{det}(A-X)=\text{det}([a_1+x_1 \ldots a_n+x_n])\,\text{det}([a_1-x_1 \ldots a_n-x_n]) = \left(\text{det}([a_1 \ldots a_n])+\left(\mathop\sum_{k=1}^{n}\text{det}([a_1 \ldots a_{k-1}\,\,x_k\,\,a_{k+1}\ldots a_n])\right)\right)\times \left(\text{det}([a_1 \ldots a_n])-\left(\mathop\sum_{k=1}^{n}\text{det}([a_1 \ldots a_{k-1}\,\,x_k\,\,a_{k+1}\ldots a_n])\right)\right) \\= \text{det}([a_1 \ldots a_n])^2-\left(\mathop\sum_{k=1}^{n}\text{det}([a_1 \ldots a_{k-1}\,\,x_k\,\,a_{k+1}\ldots a_n])\right)^2\leqslant \text{det}([a_1 \ldots a_n])^2=\text{det}(A)^2=\text{det}(A^2) $$

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By a change of basis, we may assume that $X$ is the matrix with $n$ at its $(1,1)$-th entry and zero elsewhere. Let $m_{11}$ be the $(1,1)$-th minor of new $A$ obtained by the aforementioned change of basis. By Laplace expansion along the first row, we see that $\det(A\pm X)=\det(A)\pm nm_{11}$. Hence the result follows.

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  • $\begingroup$ I don't think this change of basis is obvious to me.. $\endgroup$ – Patricia Jan 17 '16 at 21:11

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