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Let $B$ be a $3\times3$ matrix, and let $f(x,y)= \det\left(xB+yB^T\right)$. Show that $\det\left(xB+yB^T\right)$ is a multiple of $x+y$, where $x,y$ are any real numbers and $B$ is any $3 \times 3$ matrix with real entries.

I can only think a tedious method to show this question by assuming a determinant to do it. Can anyone give me a idea or method to finish it easily?

Lê Thành Đạt
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Anderson
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1 Answers1

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Let $f(x,y) = \det(xB + yB^T)$. Then $f(x, -x) = \det(xB - xB^T) = x^3 \det(B- B^T)$

The matrix $A= B-B^T$ is an odd-size skew-symmetric $(A^T = -A)$ matrix. Since $\det(A^T) = \det(A)$ and $\det(-A) = (-1)^3 \det(A) = -\det(A)$, it follows that $\det(A) = 0$.

Added: $f(x,y) \in \mathbb{R}[x,y] = (\mathbb{R}[x])[y]$ is a polynomial, hence $f(x,y) = (x+y)g(x,y) + h(x)$. Therefore $x+y$ divides $f(x,y)$ in $\mathbb{R}[x,y]$ if and only if $h(x) = 0$, and the latter is equivalent to $f(x, -x) = 0$.

Catalin Zara
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    Since it's not obvious how you approached this, you should say explicitly that for a general polynomial $p(x,y)$ on two variables, $(x + y) \mid p(x,y)$ if and only if $p(x,-x) = 0$. – Ben Grossmann Mar 16 '20 at 16:28
  • Or perhaps the less complicated version is this: saying that a function $f$ divides a function $g$ really means that if $f = 0$, then $g = 0$. – Ben Grossmann Mar 16 '20 at 16:30
  • @Omnomnomnom I am a bit confused, it is necessary, but not really sufficient in general. – Severin Schraven Mar 16 '20 at 16:36
  • @SeverinSchraven I agree - multiplicities can play a tricky role sometimes ... For example $f(x) = x^2$ and $g(x) = x$. – Catalin Zara Mar 16 '20 at 16:40
  • That was a bit of a careless statement, then. Thanks for the correction – Ben Grossmann Mar 16 '20 at 17:28
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    This is exactly the same question (https://math.stackexchange.com/q/3582940) you have posted 3 hours before this one. Ii is a bad way of using this site. I had given you an answer (very similar to the answer by @Catalin Zara), The least you should have done is mentionning the previous question. – Jean Marie Mar 17 '20 at 08:27
  • @Omnomnomnom I have added an edit to my previous answer to the initial question (math.stackexchange.com/q/3582940) generalizing this factorization (therefore providing a way to manage possible multiplicities) – Jean Marie Mar 17 '20 at 19:16