Eigenvalues and direct sums

**Amanda1990** · May 13th 2009, 10:12 AM

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 $V_i = ker (T- a_i I)$

**NonCommAlg** · May 13th 2009, 09:15 PM

Originally Posted by Amanda1990

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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