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Math Help - Eigenvalues and direct sums

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    Eigenvalues and direct sums

    Let T be a nonsingular linear transformation on a finite dimensional complex vector space having distinct eigenvalues a_1, a_2, ... , a_r (possibly with multiplicies greater than 1). Suppose T = a_1E_1 + ... + a_r E_r where all the E_i are projections on V such that E_i E_j = 0 whenever i is not equal to j.

    Show V = V_1 \oplus V_2 \oplus ... \oplus V_r where V_i = ker (T- a_i I)
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    Quote Originally Posted by Amanda1990 View Post
    Let T be a nonsingular linear transformation on a finite dimensional complex vector space having distinct eigenvalues a_1, a_2, ... , a_r (possibly with multiplicies greater than 1). Suppose T = a_1E_1 + ... + a_r E_r where all the E_i are projections on V such that E_i E_j = 0 whenever i is not equal to j.

    Show V = V_1 \oplus V_2 \oplus ... \oplus V_r where V_i = ker (T- a_i I)
    let v \in V. since T is onto, there exists u \in V such that v=T(u)=\sum_{i=1}^ra_iE_i(u) and thus for all j: \ E_j(v)=\sum_{i=1}^r a_iE_jE_i(u)=a_jE_j(u). hence v=\sum_{i=1}^rE_i(v). \ \ \ \ (1)

    so by(1): \sum_{i=1}^rE_i(V)=V. this sum is clearly direct because if \sum_{i=1}^r E_i(v_i)=0, then 0=E_j \left(\sum_{i=1}^r E_i(v_i) \right)=E_j(v_j). so we only need to show that E_i(V)=V_i for all i:

    for any i and v \in V we have: TE_i(v)=a_iE_i(v). thus E_i(V) \subseteq V_i. conversely, if v \in V_i, then using (1) we have \sum_{j=1}^ra_iE_j(v)=a_iv=T(v)=\sum_{j=1}^r a_jE_j(v). therefore:

    \sum_{j=1}^r(a_j-a_i)E_j(v)=0 and so E_k(v)=0, for all k \neq i. thus by (1): v=E_i(v) and therefore V_i \subseteq E_i(V). Q.E.D.
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