Filipjack wrote: b) If $\det \left(\sum\limits_{k=1}^n A_k^2 \right) = 0$ and $A_0 \ne aA_i$ for any $a \in \mathbb{R},$ then $\sum\limits_{k=1}^n A_k^2=O_2.$ This is not the correct statement (you can easily find counterexamples with diagonal matrices.) The correct one has $A_2$ in place of $A_0$. Actually, the correct statement is (which is more or less the same, as you didn't quantify $i$) b) If $\det \left(\sum\limits_{k=1}^n A_k^2 \right) = 0$ and $A_2 \ne aA_1$ for any $a \in \mathbb{R},$ then $\sum\limits_{k=1}^n A_k^2=O_2.$

a) There are $a_k,b_k\in\mathbb{R}$ s.t. $A_k=a_kI_2+b_kA_0$. We may assume that $A_0$ is triangular, that is, $A_0=\begin{pmatrix}p&*\\0&q\end{pmatrix}$, where $p,q \in\mathbb{R}$ or $q=\overline{p}$. Thus $\sum_{k=1}^nA_k^2=\begin{pmatrix}\sum_k(a_k+pb_k)^2&*\\0&\sum_k(a_k+qb_k)^2\end{pmatrix}$ and $\det(\sum_{k=1}^nA_k^2)=(\sum_k(a_k+pb_k)^2)(\sum_k(a_k+qb_k)^2)\geq 0$. $\textbf{Remark.}$ The fact that $n\times n$ real matrices $(A_k)_k$ pairwise commute implies directly that $\det(\sum_{k=1}^nA_k^2)\geq 0$. Here the proof is much easier than in the general case.

b) Here $a_1b_2-a_2b_1\not= 0$. Assume that $\det(\sum_{k=1}^nA_k^2)=0$. $\bullet$ Case 1. $p,q\in\mathbb{R}$. For example $\sum_k(a_k+pb_k)^2=0$, that is, for every $k$, $a_k+pb_k=0$. Yet, $a_1+pb_1=a_2+pb_2=0$ is impossible. $\bullet$ Case 2. $q=\overline{p}\notin\mathbb{R}$. Then we can choose $A_0=diag(p,\overline{p})$ and, therefore $\sum_{k=1}^nA_k^2=diag(\sum_k(a_k+pb_k)^2,\sum_k(a_k+\overline{p}b_k)^2)$. Then $\sum_k(a_k+pb_k)^2=0$ AND $\sum_k(a_k+\overline{p}b_k)^2=0$ and we are done. $\square$