Show that $\det\left(xB+yB^T\right)$ is a multiple of $x+y$ where $B$ be a $3\times3$ matrix

Question

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

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

Answer 1

Let

$$C_{x,y}:=xB+yB^T \ \ \ \text{and} \ \ \ P(x,y)=\det(C_{x,y})$$

$P(x,y)$ is a polynomial (see example below) in $x$ and $y$.

Therefore, showing that $P(x,y)$ is divisible by $(x+y)$ is equivalent to show that this polynomial is zero if we replace $y$ by $-x$ in $P$.

Otherwise said, we must show that $\det(x(B-B^T))=0$ for all $x$.

It is due to the fact that $x(B-B^T)$ is an antisymmetric matrix (also called skew-symmetric) with odd order see this, thus with determinant $= 0$.

Remark : $B-B^T$ is an antisymmetric matrix because $(B-B^T)^T=B^T-B=-(B-B^T).$

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Edit : An example : let us consider

$$B=\begin{pmatrix}0&1&1\\0&1&0\\0&-1&1\end{pmatrix}$$

then $\det(C_{x,y})=(x+y)xy$

Important remark beyond the exercise itself.

In fact, if $B$ is invertible, one can factorize $P(x,y)$ in the following form :

$$P(x,y)=(x+ay)(x+y)(ax+y)\tag{*}$$

Let us explain why. First of all, two properties of polynomial $P(x,y)$. It is

• symmetric in $x,y$ : $P(x,y)=P(y,x)$

(take for that the determinant of relationship $(xB+yB^T)^T=yB+xB^T$)

• homogeneous with degree $3$ : $P(\lambda x \lambda y)=\lambda^3 P(x,y)$

(take the determinant of relationship $(\lambda x) B + (\lambda y) B^T=\lambda (xB+yB^T)$)

Now, let us place our issue in the context of characteristic polynomials. Indeed, we can write, under the condition that $B$ is invertible :

$$P(x,y)=\det(xB+yB^T)=\det[yB(\tfrac{x}{y}I+B^{-1}B^T)]=y^3 \det(B)\chi(-\tfrac{x}{y})\tag{1}$$

Where $\chi$ is the characteristic polynomial of $D:=B^{-1}B^T$.

(recall : the characteristic polynomial of a matrix $M$ is $\det(M - \lambda I)$)

Suppose that $x$ and $y$ are such that $-\tfrac{x}{y}$ is an eigenvalue of $D$ Due to the symmetry of polynomial $P$, $-\tfrac{y}{x}$ is an eigenvalue of $D$ as well. Otherwise said, if $\lambda$ is an eigenvalue, $1/\lambda$ is an eigenvalue as well. Therefore, the only possibility for the spectrum of $D$ is to be :

$$\{ a, 1, 1/a \}\tag{2}$$

where $a$ can be real or non-real.

Hence, taking $y=-x$ in (1), we have :

$$P(x,-x)= y^3 \det(B)\chi(1)=0$$

(due to (2)). This solution could qualified as more complicated than the first one, but it has a major advantage if we look for a complete factorisation of $P(x,y)$. Indeed, due to the fact that the characteristic polynomial $\chi$ can be written :

$$\chi(t)=-(t-a)(t-1)(t-\tfrac{1}{a})$$

relationship (1) gives :

$$P(x,y)=-y^3(-\tfrac{x}{y}-a)(-\tfrac{x}{y}-1)(-\tfrac{x}{y}-\tfrac{1}{a})$$ from which we deduce (*).

Final remarks : this reasoning can hold more generaly for $(2n+1) \times (2n+1)$ matrices.

This issue could be handled even more directly by using the generalized Cayley-Hamilton theorem : if $A$ and $B$ are commuting matrices and if we define :

$$f(x,y)=det(xA-yB)$$

then $f(A,B)=0$.

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