Please help in leading further in: If $\vert A\vert+\vert B\vert =0,$ then find the value of $\vert A+B\vert$.

Question

There are two square matrices $A$ and $B$ of same order such that $A^2=I$ and $B^2=I,$Where $I$ is a unit matrix.If $\vert A\vert+\vert B\vert =0,$ then find the value of $\vert A+B\vert ,$here $\vert A\vert$ denotes the determinant of matrix $A$

Solution:Let $A$ and $B$ are square matrices of order $n$.

Since $A^2=I$ then Cayley-Hamilton theorem implies that the characteristic polynomial of last equation is

$\lambda_A^2-1=0\implies \lambda_A \in \{+1,-1\}\implies tr(A)\in\{-2,0,2\}\tag{1}$.

Since $B^2=I$ then Cayley-Hamilton theorem implies that the characteristic polynomial of last equation is

$\lambda_B^2-1=0\implies \lambda_B \in \{+1,-1\}\implies tr(B)\in\{-2,0,2\}\tag{2}$.

Using equations $(1)$ and equation $(2)$ we get $$tr(A+B)=tr(A)+tr(B)\in \{-4,-2,0,2,4\}$$

Now making use of $\vert A\vert+\vert B\vert=0$:If $\vert A\vert=1\implies \vert B\vert=-1$,this will lead to two cases

Case1: Both Eigenvalues of $A$ are $1$ $\implies$ Both Eigenvalues of $B$ are $-1$.So $tr(A)=n\implies tr(B)=-n$

Case2: Both Eigenvalues of $A$ are $-1$ $\implies$ Both Eigenvalues of $B$ are $1$.So $tr(A)=-n\implies tr(B)=n$

Both Cases leads us to the conclusion $tr(A+B)=tr(A)+tr(B)=0$

Now $$(A+B)^2=(A+B)(A+B)=A^2+B^2+AB+BA$$ $$\implies (A+B)^2=I+I+AB+BA$$ $$\implies tr(A+B)^2=tr(I)+tr(I)+tr(AB)+tr(BA)$$ $$\implies tr(A+B)^2=n+n+2tr(AB)$$since $tr(AB)=tr(BA)$

Now we know that $$\vert P\vert=\frac{1}{2}[(tr(P))^2-tr(P^2)]$$

Putting $P=A+B$ in above equation we get

$$\vert A+B\vert=\frac{1}{2}[(tr(A+B))^2-tr((A+B)^2)]$$.

Since $tr(A+B)=0$,so we get $$\vert A+B\vert=\frac{1}{2}[-tr((A+B)^2)]$$.

$$\vert A+B\vert=\frac{1}{2}[-n-n-2tr(AB)]$$.

What to do next??

Is there anyother method other than the above method??

Answer 1

From $A^2=B^2=I$ and $\det(A)+\det(B)=0$, we have $\det(A)=-\det(B)=\pm1$ and $(A+B)=A(B+A)B$. Taking determinants on both sides of the latter equality, we obtain $\det(A+B)=-\det(A+B)$. Therefore $\det(A+B)=0$ when the characteristic of the underlying field is not $2$.

Answer 2

It is unclear what field you are working over; I assume you are considering matrices with real entries. The approach for complex entries is similar.

Hint:

• If $A^2 = I$, then $\Bbb R^n = \ker(A - I) \oplus \ker(A + I)$. That is, $\dim \ker(A - I) + \dim \ker(A + I) = n$, and $\ker(A - I) \cap \ker(A + I) = \{0\}$.
• $|A| = (-1)^{\dim \ker (A + I)}$
• We can deduce that either $\ker(A - I) \cap \ker(B + I)$ or $\ker(A + I) \cap \ker(B - I)$ contains a non-zero element. For a vector $x$ from either intersection, we have $(A + B)x = 0$.