Exercise 5.4.30

We also use the notation given in Hint of the previous exercise. By Exercise 5.4.24 we may find a basis

$$\gamma =\{v_1,v_2,\dots ,v_k\}$$

for $W$ such that ${[T_W]}_{\gamma }$ is diagonal and find a basis

$$\alpha =\{v_{k+1}+W,v_{k+2}+W,\dots ,v_n+W\}$$

for $V∕W$ such that ${[\overline{T}]}_{\alpha }$ is diagonal. For each $v+W\in V∕W$, we know that $v+W$ is an eigenvector in $V∕W$. So we may assume that

$$T(v)+W=\overline{T}(v+W)=\mathit{\lambda v}+W$$

for some scalar $\lambda$. So this means that $T(v)-\mathit{\lambda v}$ is an element in $W$. Write

$$T(v)=\mathit{\lambda v}+a_1{v}_1+a_2{v}_2+\cdots +a_k{v}_k,$$

where $v_i$’s are those elements in $\gamma$ corresponding to eigenvalues ${\lambda }_i$’s. Pick $c_i$ to be $\frac{a_i}{\lambda -{\lambda }_i}$ and set

$$v^{\prime }=v+c_1{v}_1+c_2{v}_2+\cdots +c_k{v}_k.$$

Those $c_i$’s are well-defined because $T_W$ and $\overline{T}$ has no common eigenvalue. Thus we have

$$T(v^{\prime })=T(v)+c_{1}T(v_1)+c_{2}T(v_2)+\cdots +c_{k}T(v_k)$$

$$=\mathit{\lambda v}+(a_1+c_1{\lambda }_1)v_1+(a_2+c_2{\lambda }_2)v_2+\cdots +(a_k+c_k{\lambda }_k)v_k)$$

$$=\lambda v^{\prime }$$

and $v+W=v^{\prime }+W$. By doing this process, we may assume that $v_j$ is an eigenvector in $V$ for each $v_j+W\in \alpha$. Finally we pick

$$\beta =\{v_1,v_2,\dots ,v_n\}$$

and show that it’s a basis for $V$. We have that $\gamma$ is independent and

$$\delta =\{v_{k+1},v_{k+2},\dots ,v_n\}$$

is also independent since $\eta (\delta )=\alpha$ is independent, where $\eta$ is the quotient mapping defined in Exercise 5.4.27. Finally we know that if

$$a_{k+1}{v}_{k+1}+a_{k+2}{v}_{k+2}+\cdots a_n{v}_n\in W,$$

then

$$(a_{k+1}{v}_{k+1}+a_{k+2}{v}_{k+2}+\cdots a_n{v}_n)+W=0+W$$

and so

$$a_{k+1}=a_{k+2}=\cdots =a_n=0.$$

This means that $\mathrm{span}(\delta )\cap W=\{0\}$. So $V$ is a directed sum of $W$ and $\mathrm{span}(\delta )$ and $\beta$ is a basis by Exercise 1.6.33. And thus we find a basis consisting of eigenvectors.