= Solution
To prove the sum is direct, suppose $\sum_{j=1}^rv_j=0$ with $v_j\in V_{\lambda_j}$. Fix $i$ and apply the polynomial in the <linear operator>
$$
P_i(T)=\prod_{j\neq i}(T-\lambda_jI).
$$
Every $v_j$ with $j\neq i$ is killed by one factor. On $v_i$, each factor is multiplication by $\lambda_i-\lambda_j$, giving
$$
0=P_i(T)\sum_jv_j=\left(\prod_{j\neq i}(\lambda_i-\lambda_j)\right)v_i.
$$
The scalar coefficient is nonzero because the <eigenvalues> are distinct. Thus $v_i=0$ for every $i$, proving
$$
\boxed{V_{\lambda_1}+\cdots+V_{\lambda_r}=V_{\lambda_1}\oplus\cdots\oplus V_{\lambda_r}.}
$$
In particular, choosing a basis in each <eigenspace> yields a <linearly independent> combined list.
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