Exercise 5.4.23

As Hint, we prove it by induction on $k$. For $k=1$, it’s a natural statement that if $v_1$ is in $W$ then $v_1$ is in $W$. If the statement is true for $k=m-1$, consider the case for $k=m$. If we have $u=v_1+v_2+\cdots +v_m$ is in $W$, a $T$-invariant subspace, then we have

where ${\lambda }_i$ is the distinct eigenvalues. But we also have ${\lambda }_{m}u$ is in $W$. So the vector

is al so in $W$. Since those eigenvalues are distinct, we have ${\lambda }_i-{\lambda }_m$ is not zero for all $i\ne m$. So we may apply the hypothesis to

and get the result that $({\lambda }_i-{\lambda }_m)v_i$ is in $W$ and so $v_i$ is in $W$ for all $i\ne m$. Finally, we still have