# $\det(A^2-B^2) \leq \det(A^2)$ when $B$ is of rank 1

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.

• 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 Apr 13 at 2:19
• @Yimin Can you kindly elaborate why $B^2 = uv’vu’$? I think it should be $uv’uv’$. Isn’t $uv’vu’ = BB^T$? Apr 13 at 2:30
• The first equality you cited needs $A$ to be nonsingular? Check $A = 0$, $u = v = 1$. Apr 13 at 2:38
• 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 $k =\det\big(J + k \mathbf e_3 \mathbf e_3^T\big) \neq \det\big(J \big) =0$ Apr 13 at 3:26
• @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. Apr 13 at 3:39

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 $$2^2 - 2(-1) = 6 > 4 = \det A^2$$.
• if $A$ is positive definite, does that hold? Apr 13 at 4:35