$\det(A_1\cdot B_1 +A_2\cdot B_2)=0$

Question

Let $A_1, A_2\in M_n(\mathbb{R})$ two symmetric matrices s.t. $det(A_1^2+A_2^2)=0$.

Show that $det(A_1\cdot B_1 +A_2\cdot B_2)=0$ for every $B_1, B_2\in M_n(\mathbb{R})$.

My idea:

I consider the matrix C :\begin{bmatrix} A_1 & A_2 \\ B_1^t & B_2^t \end{bmatrix}

$det(C\cdot C^t)\geq 0\Rightarrow det(\begin{bmatrix} A_1^2 +A_2^2 & D \\ D^t & E\\ \end{bmatrix})\geq 0 $ where $D=A_1B_1+A_2B_2$.

I tried to expand the determinant with Laplace Rule. I am not sure if $ det(\begin{bmatrix} A_1^2 +A_2^2 & D \\ D^t & E\\ \end{bmatrix}) = -det(D\cdot D^t)$.

In this way I would get $det(D)=0$.

Answer 1

If $\det (A_1^2+A_2^2) = 0$, then $x^TA_1^2+x^TA_2^2=0$ for some $x\neq 0$ and so $\|x^TA_1\|^2 + \|x^TA_2\|^2 = 0$.

Since $x^TA_1=x^TA_2 = 0$, we see that $x^T (A_1 B_1+A_2 B_2) =0$ and so $\det (A_1 B_1+A_2 B_2) =0$

Answer 2

Let $\det(A^2_1+A^2_2)=0$ then there exists some vector $x\neq 0$ such that $(A^2_1+A^2_2)x=0$ this implies $$0=\langle (A_1^2+A^2_2)x,x\rangle=\langle A^2_1x,x\rangle+\langle A^2_2x,x\rangle=\langle A_1 x,A^T_1x\rangle+\langle A_2x,A^T_2x\rangle$$ Since $A_1, A_2$ are symmetric then $A_1=A_1^T, A_2=A^T_2$. Therefore $$0=\langle A_1 x,A^T_1x\rangle+\langle A_2x,A^T_2x\rangle=\langle A_1 x,A_1x\rangle+\langle A_2x,A_2x\rangle=||A_1x||^2+||A_2x||^2$$ This holds only if $A_1x=A_2x=0$ in which case we obtain that $\det A_1=\det A_2=0$. Now let $C:=A_1B_1+A_2B_2$ then $C^T=B_1^TA^T_1+B^T_2A^T_2=B_1^TA_1+B^T_2A_2$. This then gives $$C^T x=(B_1^TA_1+B^T_2A_2) x=B_1^TA_1x+B^T_2A_2x=B_1^T0+B^T_20=0$$ Therefore $\det C^T=0$ but $\det C=\det C^T$ for any matrix $C$. The result follows $$\det C=\det (A_1B_1+A_2B_2)=0$$