I am studying diagonalizability right now and I'm having trouble understanding this theorem (5.11, last page).
A linear operator T on finite-dimensional vector space V is diagonalizable if and only if V is the direct sum of the eigenspaces of T.
I know that a linear transformation $T$ is diagonalizable iff there exists subspaces $v_i,...v_k$ such that
(i) $V = V_1 \oplus ... \oplus V_k$
(ii) for every $1\leq i \leq k$, the subspace $V_i$ is T-invariant. That is, $T(v_i) \leq V_i$
(iii) $T$ (restricted on $v_i$) $= \lambda_i$ Id$_{v_i}$. That is, for each $v \in V$, we have $T(v) = \lambda_i v$
Also, $T$ is diagonalizable if there exists a basis $B$ of $V$ consisting of eigenvectors for $T$. That is, for each $v \in B$ there exists a scalar $\lambda$ such that $T(v) = \lambda v$
But how do we conclude,
$T$ is diagonalizable $\iff \oplus{V_ \lambda }$ where $\lambda\in Spec(T)$
In the theorem above?