---
title: "NUS-MATH530 5.A Problem 14"
source: https://www.jemoka.com/posts/kbhnus_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} &amp;Pv = \lambda v \\ \Rightarrow\ &amp; P(u+w) = \lambda (u+w) \\ \Rightarrow\ &amp; u = \lambda u + \lambda w \end{align}
Now, let&rsquo;s rewrite this expression to equal to \(0\) to take advantage of the fact that \(V = U \oplus W\).
\begin{align} &amp;u = \lambda u + \lambda w \\ \Rightarrow\ &amp; 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&rsquo;t \(u = w = 0\)? Suppose for the sake of contradiction let&rsquo;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} &amp;Pv = \lambda v \\ \Rightarrow\ &amp; P(u+w) = \lambda u + \lambda w \\ \Rightarrow\ &amp; u = \lambda u + 0 \\ \Rightarrow\ &amp; 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 &amp;= \lambda v \\ &amp;= v \end{align}
Any valid solution for \(v\) is an eigenvector.
So:
\begin{align} &amp;Pv = v \\ \Rightarrow\ &amp; P(u+w) = v \\ \Rightarrow\ &amp; 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} &amp;Pv = \lambda v \\ \Rightarrow\ &amp; P(u+w) = \lambda u + \lambda w \\ \Rightarrow\ &amp; u = \lambda u + \lambda w \\ \Rightarrow\ &amp; 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 &amp;= \lambda v \\ &amp;= 0 \end{align}
Any valid solution for \(v\) is an eigenvector.
So:
\begin{align} &amp;Pv = 0 \\ \Rightarrow\ &amp; P(u+w) = 0 \\ \Rightarrow\ &amp; 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 \(0\).
