Let $A_1, A_2, ..., A_k\in M_n(\mathbb{R})$ be symmetric matrices such that $A^2_1+A^2_2+...+A^2_k=0$. Prove that $A_i=0$ for every $i=1,2,...,k$.

My attempt:

Let $B$ be a ortonormal base of $M_n(\mathbb{R})$ (this base exists becasue $M_n(\mathbb{R}$) has finite dimension), $v\in\mathbb{R}^n$ and $[T_i]_B=A_i$ for every $i=1,2,...,k$, then

$0=\left< (T^2_1+...+T^2_k)(v),v\right>=\left< T^2_1(v),v\right>+\left< T^2_2(v),v\right>+...+\left< T^2_k(v),v\right>=\left< T_1(v),T^*_1(v)\right>+\left<T_2(v),T^*_2(v) \right>+...+\left< T_k(v),T^*_k(v)\right>=\left< T_1(v),T^t_1(v)\right>+\left<T_2(v),T^t_2(v) \right>+...+\left< T_k(v),T^t_k(v)\right>=\left< T_1(v),T_1(v)\right>+\left<T_2(v),T_2(v) \right>+...+\left< T_k(v),T_k(v)\right>=||T_1(v)||^2+||T_2(v)||^2+...+||T_k(v)||^2$.

So, we need to have $||T_i(v)||^2=0$ for every $i=1,2,...,k$ ans so $T_i=0$ for every $i=1,2,...,k$ and $A_i=0$ for every $i=1,2,...,k$.

Is this correct? Thanks

  • 1
    $\begingroup$ The word symmetric is misspelled. That's for sure. $\endgroup$ Nov 10, 2022 at 18:53
  • 1
    $\begingroup$ So is orthonormal ;) $\endgroup$
    – Math Mind
    Nov 10, 2022 at 18:58
  • 2
    $\begingroup$ Looks good to me. May be you don't need that part about an orthonormal basis? After all, for all $v$ and all $i$ we have $$||A_iv||^2=(A_iv)^T(A_iv)=v^TA_i^TA_iv=v^TA_i^2v.$$ So for all $v$ we have $$\sum_i||A_iv||^2=v^T\left(\sum_iA_i^2\right)v=0.$$ And this is possible for all $v$ only when all the matrices are zero. The idea is pretty much the same though. $\endgroup$ Nov 10, 2022 at 19:07
  • 4
    $\begingroup$ note that $\text{trace}\big(A_i^2\big)=\big\Vert A_i\big\Vert_F^2$ so taking the trace of the sum gives the result $\endgroup$ Nov 10, 2022 at 19:54

1 Answer 1


It looks correct to me, although:

  1. I don't see any reason for introducing abstract linear operators $T_i$ whose expressions in a base $B$ are $A_i$, because $A_i$ are linear operators themselves. But this is a personal preference, not a mistake.
  2. The transition from $\|T_i(v)\|^2=0$ to $T_i=0$ is not clear. You may know why the former implies the latter, but you have not written this down. If I were to mark your solution, I would consider this a mistake.
  3. The transition from $T_i=0$ to $A_i=0$ could be more articulate, since you chose to deal with operators and representations. I would not consider this a mistake.
  4. It cannot see why you require an orthonormal basis. You may potentially need it to relate the two inner products (in the abstract vector space and in the space column/row vectors), but I am not even sure about that, and you are not doing anything like this.

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