Let $A_1, A_2, \dots, A_m$ be $n\times n$ Hermitian matrices. Show that if
\[A_1^2+A_2^2+\cdots+A_m^2=\calO,\]
where $\calO$ is the $n \times n$ zero matrix, then we have $A_i=\calO$ for each $i=1,2, \dots, m$.

Recall that a complex matrix $A$ is Hermitian if the conjugate transpose of $A$ is $A$ itself.
Namely, $A$ is Hermitian if
\[\bar{A}^{\trans}=A.\]

We also use the length of a vector in the proof below.
Let $\mathbf{v}$ be $n$-dimensional complex vector. Then the length of $\mathbf{v}$ is defined to be
\[\|\mathbf{v}\|=\sqrt{\bar{\mathbf{v}}^{\trans}\mathbf{v}}.\]
The length of a complex vector $\mathbf{v}$ is a non-negative real number.

The length is also called norm or magnitude.

Proof.

Let $\mathbf{x}$ be an $n$-dimensional vector, that is, $\mathbf{x}\in \R^n$.

Then for each $i$, we have
\[\bar{\mathbf{x}}^{\trans}A_i^2\mathbf{x}=\bar{\mathbf{x}}^{\trans}\bar{A}_i^{\trans}A_i\mathbf{x}=(\overline{A_i\mathbf{x}})^{\trans}(A_i\mathbf{x})=\|A_i\mathbf{x}\|^2\geq 0.\]
Here, the first equality follows from the definition of a Hermitian matrix.

Now we compute
\begin{align*}
0&=\bar{\mathbf{x}}^{\trans}\calO \mathbf{x}=\bar{\mathbf{x}}^{\trans}(A_1^2+A_2^2+\cdots+A_m^2) \mathbf{x}\\
&=\bar{\mathbf{x}}^{\trans}A_1^2\mathbf{x}+\bar{\mathbf{x}}^{\trans}A_2^2\mathbf{x}+\cdots+\bar{\mathbf{x}}^{\trans}A_m^2 \mathbf{x}\\
&=\|A_1\mathbf{x}\|^2+\|A_2\mathbf{x}\|^2+\cdots +\|A_m\mathbf{x}\|^2.
\end{align*}

Since each length $\|A_i\mathbf{x}\|$ is a non-negative real number, this implies that we have $A_i\mathbf{x}=\mathbf{0}$ for all $\mathbf{x \in \R^n}$. Hence we must have $A_i=\calO$ for each $i=1,2,\dots, m$.

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