NUS-MATH530 5.A Problem 14

Suppose V = U \oplus W, where U and W are nonzero subspaces of V. Define P \in \mathcal{L}(V) by P(u+w) = u for u \in U, w \in W. Find all eigenvalues and eigenvectors of P.
Solutions:
\lambda = 1, v = u \in U \lambda = 0, v = w \in W For \lambda to be an eigenvalue of P, we have to have:

\begin{equation} Pv = \lambda v \end{equation}

Meaning, for WLOG v = u+w:

\begin{align} &Pv = \lambda v \\ \Rightarrow\ & P(u+w) = \lambda (u+w) \\ \Rightarrow\ & u = \lambda u + \lambda w \end{align}

Now, let’s rewrite this expression to equal to 0 to take advantage of the fact that V = U \oplus W.

\begin{align} &u = \lambda u + \lambda w \\ \Rightarrow\ & 0 = (\lambda -1) u + \lambda w \end{align}

Now, recall that a sum of subsets in a direct sum if and only if the only way to write 0 is for each of the elements of the sums to be 0. In this case, it means that:

\begin{equation} \begin{cases} (\lambda -1) u = 0 \\ \lambda w = 0 \end{cases} \end{equation}

We have two cases here: either w=0 or u=0.
Aside: why can’t u = w = 0? Suppose for the sake of contradiction let’s take u=0, w=0. Then, Pv = \lambda v, so v=u+w, and so v=0. This would make v no longer an eigenvector, by definition of eigenvector; this also makes \lambda no longer an eigenvalue. Hence, one of u or w is not 0.
w=0 We have that w=0. Replacing that in the above expression, we have that:

\begin{align} &Pv = \lambda v \\ \Rightarrow\ & P(u+w) = \lambda u + \lambda w \\ \Rightarrow\ & u = \lambda u + 0 \\ \Rightarrow\ & u = \lambda u \end{align}

From this expression, or the top one from before (\lambda -1 ) u = 0, we have that \lambda = 1.
Finally, then, we have:

\begin{align} Pv &= \lambda v \\ &= v \end{align}

Any valid solution for v is an eigenvector.
So:

\begin{align} &Pv = v \\ \Rightarrow\ & P(u+w) = v \\ \Rightarrow\ & u = v \end{align}

Hence, all u \in U is an eigenvector of P with eigenvalue 1.
u=0 We now have that u=0. So, we have that:

\begin{align} &Pv = \lambda v \\ \Rightarrow\ & P(u+w) = \lambda u + \lambda w \\ \Rightarrow\ & u = \lambda u + \lambda w \\ \Rightarrow\ & 0 = \lambda w \end{align}

From this expression, or the bottom one from \lambda w = 0, we have that \lambda = 0.
Finally, then, we have:

\begin{align} Pv &= \lambda v \\ &= 0 \end{align}

Any valid solution for v is an eigenvector.
So:

\begin{align} &Pv = 0 \\ \Rightarrow\ & P(u+w) = 0 \\ \Rightarrow\ & u = 0 \end{align}

Recall now that v = u+w, so v = 0 +w, making v = w.
Hence, all w \in W is an eigenvector of P with eigenvalue 0.