= Solution
{c}
A <projection> is a linear operator $E$ satisfying $E^2=E$; it need not be an orthogonal projection without an inner product. Pairwise annihilation gives
$$
P^2=\left(\sum_iE_i\right)^2=\sum_iE_i^2+\sum_{i\ne j}E_iE_j=\sum_iE_i=P.
$$
Thus their sum is a projection.
The identities $E_iP=E_i$ imply $T(I-P)=0$. Since $T$ is nonsingular, $P=I$. Also
$$
TE_i=\sum_j\lambda_jE_jE_i=\lambda_iE_i,
$$
so $\operatorname{im}E_i\subseteq V_i$. Conversely, if $Tx=\lambda_ix$, applying $E_j$ gives $(\lambda_j-\lambda_i)E_jx=0$. For $j\ne i$ the eigenvalues are distinct, hence $E_jx=0$. Since $P=I$, $x=E_ix$, proving $\operatorname{im}E_i=V_i$.
Every vector is $\sum_iE_ix$. If $\sum_iv_i=0$ with $v_i\in V_i$, applying $E_j$ gives $v_j=0$. Thus
$$
\boxed{V=V_1\oplus\cdots\oplus V_r.}
$$
These operators are the <spectral projections of a diagonalizable matrix>, expressed without choosing an eigenbasis.
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