If $\mathbf{A}$ is real symmetric, prove $\mathbf{A}^n=\mathbf{0}$ implies $\mathbf{A}=\mathbf{0}$ For all $\mathbf{x}$, we see that $\mathbf{A}^n\mathbf{x}=\mathbf{0}$. Therefore we can construct a linearly independent basis of eigenvectors (each with eigenvalue $0$). Following from this, the eigenvectors of $\mathbf{A}$ must also form a basis, with eigenvalues $0^{1/n}=0$. Since a square matrix is diagonalizable iff there is a basis of its eigenvectors, there exists a frame in where $\mathbf{A} = \mathrm{diag}(0,\cdots,0) = \mathbf{0}$. However $\mathbf{0}\mathbf{M}=\mathbf{0}$ for all matrices $\mathbf{M}$ so therefore $\mathbf{A} = \mathbf{0}$ in any frame.
Is this proof complete? Where have I used the fact that $\boldsymbol{A}$ is real symmetric? The question actually asks you to consider the quadratic form $Q = \mathbf{x^T}\mathbf{A}\mathbf{x}$ but how does this help? Are there any other proofs?

Notice that $\text{tr}(A^TA)=0$ implies $A=0$. Since $A$ is symmetric, then it
follows from $A^2=0$ that $A=0$.
Moreover, if $A$ is symmetric and $A^n=0$, there exists $k\in\mathbb{N}$ such that $2^k\ge n$ and then $A^{2^k}=A^{n}A^{2^k-n}=0$. By induction we obtain
$$A^{2^k}=0\Longrightarrow A^{2^{k-1}}=0\Longrightarrow\cdots\Longrightarrow A=0.$$
