[proofplan]
Stack the images $W_1 = A_1 Z$ and $W_2 = A_2 Z$ into a single Gaussian vector $W = BZ$ and read off independence from its covariance. The orthogonality hypothesis $A_1 A_2 = 0$ forces the cross-covariance block to vanish, and for a joint normal this is equivalent to independence of the two blocks. Each quadratic form $Z^\top A_i Z$ factors as $W_i^\top W_i$ because $A_i$ is a symmetric projection, so independence of $W_1$ and $W_2$ transfers to the quadratic forms.
[/proofplan]
[step:Stack the linear images $A_1 Z$ and $A_2 Z$ into a joint Gaussian vector]
Define the block matrix
\begin{align*}
B: \mathbb{R}^n &\to \mathbb{R}^{2n} \\
z &\mapsto \begin{pmatrix} A_1 \\ A_2 \end{pmatrix} z,
\end{align*}
so that $BZ = (W_1^\top, W_2^\top)^\top$ where $W_i := A_i Z \in \mathbb{R}^n$. Since $Z \sim N_n(\mathbf{0}, \sigma^2 I_n)$ and $B$ is a deterministic linear map, the stacked vector $BZ$ is multivariate normal with mean $B \cdot \mathbf{0} = \mathbf{0}$ and covariance
\begin{align*}
\operatorname{Cov}(BZ) = B\,(\sigma^2 I_n)\,B^\top = \sigma^2 B B^\top.
[/step]
[step:Compute the covariance blocks and use $A_1 A_2 = 0$ to kill the off-diagonal]
Writing $BB^\top$ in $2 \times 2$ block form and using the symmetry of each $A_i$ (so $A_i^\top = A_i$):
\begin{align*}
B B^\top = \begin{pmatrix} A_1 \\ A_2 \end{pmatrix} \begin{pmatrix} A_1^\top & A_2^\top \end{pmatrix} = \begin{pmatrix} A_1 A_1^\top & A_1 A_2^\top \\ A_2 A_1^\top & A_2 A_2^\top \end{pmatrix} = \begin{pmatrix} A_1^2 & A_1 A_2 \\ A_2 A_1 & A_2^2 \end{pmatrix}.
\end{align*}
Idempotence gives $A_i^2 = A_i$, so the diagonal blocks reduce to $A_1$ and $A_2$. The off-diagonal block $A_1 A_2$ vanishes by the hypothesis $A_1 A_2 = 0$, and taking transposes (using $A_1^\top = A_1$ and $A_2^\top = A_2$) also yields $A_2 A_1 = (A_1 A_2)^\top = 0$. Therefore
\begin{align*}
\operatorname{Cov}(BZ) = \sigma^2 \begin{pmatrix} A_1 & 0 \\ 0 & A_2 \end{pmatrix}.
\end{align*}
[guided]
We want to show that the two quadratic forms are independent, and the cleanest route to independence for Gaussians is through the covariance matrix: for a *jointly normal* vector, independence of two blocks is equivalent to the vanishing of the cross-covariance block. We therefore assemble $W_1 = A_1 Z$ and $W_2 = A_2 Z$ into a single joint Gaussian and compute the covariance blocks.
Writing $BB^\top$ in block form and using $A_i^\top = A_i$:
\begin{align*}
B B^\top = \begin{pmatrix} A_1 A_1^\top & A_1 A_2^\top \\ A_2 A_1^\top & A_2 A_2^\top \end{pmatrix} = \begin{pmatrix} A_1^2 & A_1 A_2 \\ A_2 A_1 & A_2^2 \end{pmatrix}.
\end{align*}
The diagonal blocks: idempotence was assumed precisely so that $A_i^2 = A_i$ — this is what makes $A_i$ a *projection*, and it will be needed again when we factor the quadratic forms in the next step. The off-diagonal block is $A_1 A_2$, which equals $\mathbf{0}$ by the orthogonality hypothesis. Its transpose is $A_2 A_1 = (A_1 A_2)^\top = \mathbf{0}^\top = \mathbf{0}$, so the lower-left block also vanishes. This is where the orthogonality hypothesis $A_1 A_2 = 0$ is consumed: without it, the two quadratic forms would generally be correlated and the argument would fail.
The covariance of $BZ$ is therefore block-diagonal:
\begin{align*}
\operatorname{Cov}(BZ) = \sigma^2 \begin{pmatrix} A_1 & 0 \\ 0 & A_2 \end{pmatrix}.
\end{align*}
[/guided]
[/step]
[step:Deduce independence of $W_1$ and $W_2$ from the block-diagonal covariance]
Since $BZ = (W_1^\top, W_2^\top)^\top$ is a centered multivariate Gaussian vector in $\mathbb{R}^{2n}$ with covariance matrix block-diagonal as computed above, the sub-vectors $W_1$ and $W_2$ are uncorrelated. For jointly Gaussian random vectors, zero cross-covariance is equivalent to independence — a standard fact about the multivariate normal. Therefore $W_1$ and $W_2$ are independent $\mathbb{R}^n$-valued random vectors.
[guided]
We have just shown that the joint distribution of $(W_1, W_2)$ is multivariate normal with block-diagonal covariance. We now extract independence.
Why is zero cross-covariance enough? In general, uncorrelated random variables need **not** be independent — one can construct counterexamples even for bounded random variables. The Gaussian case is special: if $(W_1, W_2)$ is jointly normal, then the joint characteristic function factorises,
\begin{align*}
\mathbb{E}\!\left[\exp(is_1^\top W_1 + is_2^\top W_2)\right] = \exp\!\left(-\tfrac{1}{2}(s_1^\top \Sigma_1 s_1 + s_2^\top \Sigma_2 s_2)\right)
\end{align*}
precisely because the cross-covariance block vanishes. This factorisation is the product of the marginal characteristic functions, which by inversion gives independence. The joint-normality hypothesis is essential: marginal normality of $W_1$ and $W_2$ alone would not suffice. It holds here because each $W_i$ is a linear image of the Gaussian vector $Z$, and the stacked vector $BZ$ is also a linear image of $Z$, hence jointly Gaussian.
We conclude that $W_1$ and $W_2$ are independent random vectors in $\mathbb{R}^n$.
[/guided]
[/step]
[step:Factor each quadratic form $Z^\top A_i Z$ as $W_i^\top W_i$]
For $i \in \{1, 2\}$, using symmetry $A_i^\top = A_i$ and idempotence $A_i^2 = A_i$:
\begin{align*}
W_i^\top W_i = (A_i Z)^\top (A_i Z) = Z^\top A_i^\top A_i Z = Z^\top A_i^2 Z = Z^\top A_i Z.
\end{align*}
Hence $Z^\top A_1 Z$ is a (deterministic) function of $W_1$ alone, and $Z^\top A_2 Z$ is a function of $W_2$ alone.
[/step]
[step:Transfer independence from $(W_1, W_2)$ to the two quadratic forms]
Measurable functions of independent random vectors are independent. Since $W_1$ and $W_2$ are independent by Step 3, and since $Z^\top A_i Z = W_i^\top W_i$ is a continuous (hence Borel-measurable) function $\mathbb{R}^n \to \mathbb{R}$ of $W_i$ by Step 4, we conclude
\begin{align*}
Z^\top A_1 Z \text{ and } Z^\top A_2 Z \text{ are independent.}
\end{align*}
This completes the proof.
[/step]
