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.

  • 3
    $\begingroup$ $det(A+X)det(A-X)=det((A+X)(A-X))=det(A^2-X^2)$ $\endgroup$ – 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
  • 1
    $\begingroup$ @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

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: $\det(A+X)\det(A-X) = (1-(a\vec{1}^TA^{-1}\vec{1})^2)\det(A)^2 \le \det(A)^2 = \det(A^2)$


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 $$ \text{det}(A+X)\text{det}(A-X)=\text{det}([a_1+x_1 \ldots a_n+x_n])\,\text{det}([a_1-x_1 \ldots a_n-x_n]) = \left(\text{det}([a_1 \ldots a_n])+\left(\mathop\sum_{k=1}^{n}\text{det}([a_1 \ldots a_{k-1}\,\,x_k\,\,a_{k+1}\ldots a_n])\right)\right)\times \left(\text{det}([a_1 \ldots a_n])-\left(\mathop\sum_{k=1}^{n}\text{det}([a_1 \ldots a_{k-1}\,\,x_k\,\,a_{k+1}\ldots a_n])\right)\right) \\= \text{det}([a_1 \ldots a_n])^2-\left(\mathop\sum_{k=1}^{n}\text{det}([a_1 \ldots a_{k-1}\,\,x_k\,\,a_{k+1}\ldots a_n])\right)^2\leqslant \text{det}([a_1 \ldots a_n])^2=\text{det}(A)^2=\text{det}(A^2) $$


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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