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Thread: 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 $\displaystyle a_1, a_2, ... , a_r $ (possibly with multiplicies greater than 1). Suppose $\displaystyle T = a_1E_1 + ... + a_r E_r$ where all the $\displaystyle E_i$ are projections on V such that $\displaystyle E_i E_j = 0$ whenever i is not equal to j.

    Show $\displaystyle V = V_1 \oplus V_2 \oplus ... \oplus V_r$ where $\displaystyle V_i = ker (T- a_i I)$
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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 $\displaystyle a_1, a_2, ... , a_r $ (possibly with multiplicies greater than 1). Suppose $\displaystyle T = a_1E_1 + ... + a_r E_r$ where all the $\displaystyle E_i$ are projections on V such that $\displaystyle E_i E_j = 0$ whenever i is not equal to j.

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

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

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

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