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}

Warmup: 35

Suppose \(V\) is finite dimensional, \(T \in \mathcal{L}(V)\) and \(U\) is invariant under \(T\). Prove each eigenvalue of \(T / U\) is an eigenvalue of \(T\).

Now, \(\lambda\) is an eigenvalue of \(T / U\). That is:

\begin{equation} Tv + U = \lambda v + U \end{equation}

Meaning:

\begin{equation} (T-\lambda I) v \in U, \forall v \in V \end{equation}

Suppose for the sake of contradiction \(\lambda\) is not an eigenvalue of \(T\). This means no \(\lambda\) such that \(Tv = \lambda v\); specifically, that means also no \(\lambda\) such that \(T|_{u} u = \lambda u\). Now, that means \(T|_{u} - \lambda I\) is invertible given finite dimensional \(V\).

Suppose \(T \in \mathcal{L}(V)\) has a diagonal matrix \(A\) w.r.t. some basis of \(V\), and that \(\lambda \in \mathbb{F}\). Prove that \(\lambda\) appears on the diagonal of \(A\) precisely \(\dim E(\lambda, T)\) times.

Aside: “to appear on the diagonal \(n\) times”

We want to begin by giving a description for what “appearing on the diagonal” of a diagonal matrix implies.

A diagonal matrix is a special-case upper-triangular matrix, so a value being on its diagonal implies it to be an eigenvalue.

Standard Bases Back and Fourth

To map the vectors from \(B_2\) back to the standard bases, we simply have to construct the map:

\begin{equation} \mqty(2 & 1 & 2 \\ 1& 1& -1 \\ 1 & -1 & 0) \end{equation}

Each of the “standard” vectors in the new basis, when applied to this matrix, gets moved back to their original representation.

Presumably, then, moving “forward” into the new space is simply taking the inverse of this vector, which we will do separately; its inverse is:

Dot product

Calculations

Let’s calculate some dot products!

\begin{equation} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 1 \end{pmatrix} = 0 \end{equation}

\begin{equation} \begin{pmatrix} 1 \\2 \end{pmatrix} \cdot \begin{pmatrix} 2 \\1 \end{pmatrix} = 4 \end{equation}

\begin{equation} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} -1 \\1 \end{pmatrix} = 0 \end{equation}

\begin{equation} \begin{pmatrix} 1 \\1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 2 \end{pmatrix} = 4 \end{equation}

Interpretation

Geometrically, the intepretation of the dot product is the magnitude that comes from scaling the bottom projected value by the top value. This is essentially multiplying the proportion of one vector that’s parallel to the other by each other.