Show that $\det(xA+yB+zI_{n})=\det(yA+xB+zI_{n})$

Question

We have $A$ and $B$ $(n×n)$ matrices with complex entries. We know that $A-B=AB-BA$. Show that $$\det(xA+yB+zI_{n})=\det(yA+xB+zI_{n})$$ for every $x,y,z$ complex numbers with $x+y≠0$.
We can see that $\operatorname{Tr}(A)=\operatorname{Tr}(B)$. I tried to suppose $A$ or $B$ are invertible and try changing $A-B=AB-BA$ somehow, but I dont know if that helps. We can also see that $$\operatorname{Tr}(xA+yB)=\operatorname{Tr}(yA+xB)=(x+y)\operatorname{Tr}(A).$$ The conclusion looks like we need to show that $f(x,y)=f(y,x)$. Maybe some calculus and we can use the continuity of polynomial functions? I also minded calculating the determinants using polynomial forms.

Answer 1

I’m making this an answer to flesh out the approach hinted by the post author and that I completed.

Write $A=X+Y$, $B=X-Y$, then $2(YX-XY)=AB-BA=2Y$ so $Y=YX-XY$.

Moreover, for any $x,y,z \in \mathbb{C}$, $xA+yB+zI_n=(x+y)X+(x-y)Y+zI_n$. Thus the problem is solved if we can prove that $\det(aX+bY+cI_n)$ is an even function of $b$.

Note that it’s enough to show, by continuity of the determinant, that for any $a \neq 0$ and any $c$ such that $X’=X+c/a I_n$ is invertible, that the function $b \longmapsto \det(aX+bY+cI_n)$ is even.

But for such $a,c$, $\det(aX+bY+cI_n)=\det(aX’+bY)=\det(X’)\det(aI_n+b(X’)^{-1}Y)$. This is an even function of $b$ (constant, in fact) as long as $Z=(X’)^{-1}Y$ is nilpotent.

Now, note that $YX’-X’Y=Y$: this implies (formally) that $ZX’-X’Z=Z$.

It is then standard to show that $Z$ is nilpotent. For instance, you can show by induction on $k$ that $Z^kX’-X’Z^k=kZ^k$: in particular, if $Z$ is not nilpotent, every $k \geq 1$ is an eigenvalue of the endomorphism $T \longmapsto TX’-X’T$, which is impossible since we’re in finite dimension.

Answer 2

From the given condition, we get $N=[N,B]$ where $N=A-B$. Therefore $[N,B]$ is nilpotent, by Jacobson's lemma. Hence $N$ is nilpotent, because it is equal to $[N,B]$. (One may also use prove this result without Jacobson's lemma. See Aphelli's comment below.) Now rewrite $N=[N,B]$ as $$ NB=(B+I)N. $$ It follows that $NB\ker(N)=(B+I)N\ker(N)=0$. That is, $B\ker(N)\subseteq\ker(N)$. We may then prove by mathematical induction that $B\ker(N^k)\subseteq\ker(N^k)$ for every $k\ge1$: $$ \begin{aligned} N^{k+1}B\ker(N^{k+1}) &=N^kNB\ker(N^{k+1})\\ &=N^k(B+I)N\ker(N^{k+1})\\ &\subseteq N^k(B+I)\ker(N^k)\\ &\subseteq N^kB\ker(N^k)+N^k\ker(N^k)\\ &\subseteq N^k\ker(N^k)+N^k\ker(N^k)\\ &=0. \end{aligned} $$ Let $m$ be the index of nilpotence of $N$. Then $\ker(N)\subseteq\ker(N^2)\subseteq\cdots\subseteq\ker(N^{m-1})=\mathbb C^n$. Let $n_k=\operatorname{nullity}(N^k)$. Extend a basis $\{v_1,v_2,\ldots,v_{n_1}\}$ of $\ker(N)$ to a basis of $\{v_1,v_2,\ldots,v_{n_2}\}$ of $\ker(N^2)$ and so on until we obtain a basis $\mathscr B=\{v_1,v_2,\ldots,v_n\}$ of $\mathbb C^n$ such that $\{v_1,v_2,\ldots,v_{n_k}\}$ is a basis of $\ker(N^k)$ for each $k\in\{1,2,\ldots,m-1\}$.

If we change the basis from the standard basis to $\mathscr B$, $B$ becomes a block triangular matrix and $N$ becomes a strictly block-upper triangular matrix. So, if we further triangularise each diagonal sub-block of $B$ (which leaves the corresponding diagonal sub-block of $N$ unchanged because it is zero), then $B$ and $N$ become simultaneously triangularised. Therefore $\det(sB+tN+zI)$ does not depend on $t$. In particular, we have $\det((x+y)B+xN+zI)=\det((x+y)B+yN+zI)$, i.e., $\det(xA+yB+zI)=\det(yA+xB+zI)$.