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This is an exercise problem given at linear algebra class. As a problem before this I was able to show that $$\det(A + uv^T) = \det A + v^T (adj A )u$$ when $A$ is order $n$ square matrix and $u,v$ are $n$-dimensional vectors.

Now I am to show this: $\det(A^2-B^2) \leq \det(A^2)$ when $A, B$ are order $n$ square matrices and $rank(B)=1$.

Moreover, using this, I am to show that $\det(A + kX) = \det(A)$ when $A$ is skew-symmetric and $X$ is a matrix with 1's at every entry.

If the problem is to show $\det(A-B)\det(A+B) \leq \det(A^2)$, it would be easy using $\det(A + uv^T) = \det A + v^T (adj A )u$. But for $\det(A^2-B^2)$, I don't know how to start.

Thank you in advance for any form of help, hint, or solution.

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    $\begingroup$ Let $B=uv'$, then $$\det (A^2 - B^2)=\det(A^2 - (uv'v) u') = \det(A^2) - (u' adj(A^2) u ) (v'v)$$ then show the two factors are both non-negative $\endgroup$
    – Yimin
    Apr 13 at 2:19
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    $\begingroup$ @Yimin Can you kindly elaborate why $B^2 = uv’vu’$? I think it should be $uv’uv’$. Isn’t $uv’vu’ = BB^T$? $\endgroup$
    – mathhello
    Apr 13 at 2:30
  • $\begingroup$ The first equality you cited needs $A$ to be nonsingular? Check $A = 0$, $u = v = 1$. $\endgroup$
    – Zhanxiong
    Apr 13 at 2:38
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    $\begingroup$ I think there is something wrong with your skew-symmetric problem in the case where $n$ is odd and $\text{rank}(A) = n-1$. E.g. with $n=3$, the ones matrix is congruent to $\mathbf e_3\mathbf e_3^T$ and congruence preserves skewness, so e.g. $C^T XC = \mathbf e_3\mathbf e_3^T$ and $C^T AC =J= \begin{bmatrix}0& -1&0 \\ 1 &0&0\\ 0&0 & 0 \end{bmatrix}$ we have $k =\det\big(J + k \mathbf e_3 \mathbf e_3^T\big) \neq \det\big(J \big) =0$ $\endgroup$ Apr 13 at 3:26
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    $\begingroup$ @hmeng I am not talking about the tag. I am talking about the statement $\det(A + kX) = \det(A)$ "when 𝐴 is skew-symmetric". I would like to hear back from OP. $\endgroup$ Apr 13 at 3:39

1 Answer 1

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As stated, the inequality is false. Let $A = \begin{pmatrix}0 & -1\\1 & 0\end{pmatrix}$ and $B = \begin{pmatrix}1 & 1\\1 & 1\end{pmatrix}$. Then $B$ has rank one, $A^2 = -I$, and $B^2 = \begin{pmatrix}2 & 2\\2 & 2\end{pmatrix}$. Now $A^2 - B^2 = \begin{pmatrix}-3 & -2\\-2 & -3\end{pmatrix}$ has determinant $3^2 - 2^2 = 5 > 1 = \det A^2$.

This is not even true when $A$ is positive definite. For if $A = \begin{pmatrix}1 & 0\\0 & 2\end{pmatrix}$ and $B = \begin{pmatrix}1 & -1\\2 & -2\end{pmatrix}$, then $$A^2 - B^2 = \begin{pmatrix}1 & 0\\0 & 4\end{pmatrix} - \begin{pmatrix}-1 & 1\\-2 & 2\end{pmatrix} = \begin{pmatrix}2 & -1\\2 & 2\end{pmatrix}$$ has determinant $2^2 - 2(-1) = 6 > 4 = \det A^2$.

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  • $\begingroup$ if $A$ is positive definite, does that hold? $\endgroup$
    – Yimin
    Apr 13 at 4:35

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