Claim

Proof or give a counter example for the statement that:

\begin{align} \dim\qty(U_1+U_2+U_3) = &\dim U_1+\dim U_2+\dim U_3\\ &-\dim(U_1 \cap U_2) - \dim(U_1 \cap U_3) - \dim(U_2 \cap U_3) \\ &+\dim(U_1 \cap U_2 \cap U_3) \end{align}

Counterexample

This statement is false.

Take the following three subspaces of \(\mathbb{F}^{2}\):

\begin{align} U_1 = \qty{\mqty(a \\ 0): a \in \mathbb{F}}\\ U_2 = \qty{\mqty(0 \\ b): b \in \mathbb{F}}\\ U_3 = \qty{\mqty(c \\ c): c \in \mathbb{F}} \end{align}

Statement

Support \(W\) is finite-dimensional, and \(T \in \mathcal{L}(V,W)\). Prove that \(T\) is injective IFF \(\exists S \in \mathcal{L}(W,V)\) such that \(ST = I \in \mathcal{L}(V,V)\).

Proof

Given injectivity

Given an injective \(T \in \mathcal{L}(V,W)\), we desire that \(\exists S \in \mathcal{L}(W,V)\) such that \(ST = I \in \mathcal{L}(V,V)\).

We begin with some statements.

• Recall that, a linear map called injective when \(Tv=Tu \implies v=u\)
• Recall also that the “identity map” on \(V\) is a map \(I \in \mathcal{L}(V,V)\) such that \(Iv = v, \forall v \in V\)

Motivating \(S\)

We show that we can indeed create a function \(S\) by the injectivity of \(T\). Recall a function is a map has the property that \(v=u \implies Fv=Fu\).

Statement

Support \(W\) is finite-dimensional, and \(T \in \mathcal{L}(V,W)\). Prove that \(T\) is injective IFF \(\exists S \in \mathcal{L}(W,V)\) such that \(ST = I \in \mathcal{L}(V,V)\).

Proof

Given injectivity

Given an injective \(T \in \mathcal{L}(V,W)\), we desire that \(\exists S \in \mathcal{L}(W,V)\) such that \(ST = I \in \mathcal{L}(V,V)\).

Creating \(S\)

Define a relation \(S:range\ T\to V\) in the following manner:

\begin{equation} S(v) = a \mid Ta = v \end{equation}

Chapter 4 discussion with Lachlan

4.2

False.

The union between \(\{0\} \cup \{p \in \mathcal{P}(\mathbb{F}): deg\ p = m\}\) is not closed under addition. You can add two \(m\) degree polynomials and get something that’s not \(m\) degrees:

\begin{equation} (z^{m} + 1) - z^{m} = 1 \end{equation}

4.3

False.

The union between \(\{0\} \cup \{p \in \mathcal{P}(\mathbb{F}): deg\ p\ even\}\) is not closed also under addition, for the same reason:

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}