Question

Let \({\bf{u}}\) be an eigenvector of \(A\) corresponding to an eigenvalue \(\lambda \), and let \(H\) be the line in \({\mathbb{R}^{\bf{n}}}\) through \({\bf{u}}\) and the origin.

1. Explain why \(H\) is invariant under \(A\) in the sense that \(A{\bf{x}}\) is in \(H\) whenever \({\bf{x}}\) is in \(H\).
2. Let \(K\) be a one-dimensional subspace of \({\mathbb{R}^{\bf{n}}}\) that is invariant under \(A\). Explain why \(K\) contains an eigenvector of \(A\).

Short answer

1. Because if we take \({\bf{x}}\)in\(H\)then\({\bf{x}} = c{\bf{u}}\)where \(c\) is any scalar and \(A{\bf{x}} = \left( {c\lambda } \right){\bf{u}}\) which shows that \(A{\bf{x}} \in H\).
2. If \(k\) is invariant under \(A\), then \(A{\rm{x}}\) is in \(k\) and hence \(A{\rm{x}}\) is a multiple of \({\rm{x}}\). Thus, \({\rm{x}}\) is an eigenvector of \(A\).

Step-by-step solution

Step 1: Explain why \(H\) is invariant under \(A\) in the sense that \(A{\bf{x}}\) is in \(H\) whenever \({\bf{x}}\) is in \(H\)

The set \(H\) is given by \(H = \left\{ {{\bf{x}}:{\bf{x}} = \left( {1 - t} \right){\bf{0}} + t{\bf{u}},t \in \mathbb{R}} \right\}\) then we have,

\(H = \left\{ {{\bf{x}}:{\bf{x}} = t{\bf{u}},t \in \mathbb{R}} \right\}\)

Since \(\lambda \) is an eigenvalue of \(A\), \(A{\bf{u}} = \lambda {\bf{u}}\).

Assume \({\rm{x}}\) is in \(H\). Then \({\bf{x}} = c{\bf{u}}\).

Therefore,

\(\begin{aligned}{c}A{\bf{x}} &= A\left( {c{\bf{u}}} \right)\\ &= c\left( {A{\bf{u}}} \right)\\ &= c\left( {\lambda {\bf{u}}} \right)\\ &= \left( {c\lambda } \right){\bf{u}}\end{aligned}\)

Thus, \(A{\bf{x}} \in H\).

Hence if we take \({\bf{x}}\) in \(H\) then \({\bf{x}} = c{\bf{u}}\) where \(c\) is any scalar and \(A{\bf{x}} = \left( {c\lambda } \right){\bf{u}}\) which shows that \(A{\bf{x}} \in H\).

Step 2: Explain why \(K\) contains an eigenvector of \(A\)

Now take a nonzero vector \({\rm{x}}\) in \(K\). Since \(K\) is a one-dimensional subspace of\({R^n}\)then\(K\) is a set of all scalar multiples of \({\rm{x}}\).

Also, since \(K\) is invariant under \(A\), \(A{\rm{x}}\) is in \(K\).

This implies that \(A{\rm{x}}\) is multiple of \({\rm{x}}\), so \(A{\bf{x}} = \lambda {\bf{x}}\).

Thus, \({\rm{x}}\) is an eigenvector of \(A\).

Hence if \(k\) is invariant under \(A\), then \(A{\rm{x}}\) is in \(k\) and hence \(A{\rm{x}}\) is a multiple of \({\rm{x}}\). Thus, \({\rm{x}}\) is an eigenvector of \(A\).