Distinct Eigenvalues Imply Eigenspaces are Linearly Indepedent

Theorem

• Suppose $T\in \mathcal{L}(V)$
• let $c_1,\dots ,c_n$ be distinct Eigenvalue for $T$
• With $W_i=ker(T-c_{i}I),j=1,\dots ,k$
• Then, the set $\{W_1,\dots ,W_n\}$ is Linearly Independent
• If $T$ is Diagonalizable, then $V=W_1\oplus \cdots \oplus W_n$