12.27

1. Suppose \(T \in \mathcal{L}(V, V)\) is such that every \(v \in V, v \neq 0\) is an eigenvector of \(T\). Prove that \(T\) is a scalar multiple of the identity operator (\(\exists \lambda \in \mathbb{F}\) such that \(T = \lambda \text{Id}\)).

   Every \(v \in V, v \neq 0\) is an eigenvector of \(T\)​\(\implies\forall v\in V,v\neq0,\exists\lambda_{v}\in\mathbb{F}\) such that \(T(v)=\lambda_{v}v\).

   So, we just need to show that for \(v, w \in V\), \(v \neq 0\), \(w \neq 0\), \(\lambda_v = \lambda_w\). (Note that for \(v = 0\), \(T(0) = 0 = \lambda \cdot 0\) for any \(\lambda \in \mathbb{F}\).)

   So, let \(v, w \in V, v, w \neq 0\). We have 2 cases:

   • \(v,w\) is linearly dependent\(\implies \exists \lambda \in \mathbb{F}\) such that \(v = c w\).
     Then \(\lambda_{v}v=T(v)=T(cw)=cT(w)=c\lambda_{w}w=\lambda_{w}(cw)=\lambda_{w}v\).

   So \(\lambda_v v = \lambda_w v\) and since \(v \neq 0\), \(\lambda_v = \lambda_w\). * \(v, w\) linearly independent: \(\lambda_{v+w}(v + w) = T(v + w) = T(v) + T(w) = \lambda_v v + \lambda_w w\).

   \(\implies (\lambda_{v+w} - \lambda_v)v + (\lambda_{v+w} - \lambda_w)w = 0\).

   So, since \(v\) and \(w\) are L.I.: \(\begin{aligned} \lambda_{v+w} - \lambda_v &= 0, \\ \lambda_{v+w} - \lambda_w &= 0. \end{aligned}\)​\(\implies \lambda_{v+w} = \lambda_v = \lambda_w\).

   So in all cases, \(\lambda_v = \lambda_w \ \forall v, w \in V, v, w \neq 0\).

   Therefore \(T(v) = \lambda v \ \forall v \in V\) for some \(\lambda \in \mathbb{F}\)​\(\implies T = \lambda \text{Id}\).

2. Suppose \(V\) is finite-dimensional and \(T \in \mathcal{L}(V, V)\) is such that every subspace \(U\) of \(V\) with \(\dim(U) = \dim(V) - 1\) is invariant under \(T\). Prove that \(T\) is a scalar multiple of the identity operator.

   Suppose by contradiction that \(T \neq \lambda \text{Id}\).

   \(\implies\) by the contrapositive of (1), \(\exists v \in V, v \neq 0\) such that \(v\) is not an eigenvector of \(T\).

   \(\implies \nexists \lambda \in \mathbb{F}\) such that \(T(v) = \lambda v\).

   So \(\{v, T(v)\}\) is L.I. \(\implies\) we extend it to a basis of \(V\): \(\{v, T(v), u_3, \dots, u_n\}\) (\(\dim(V) = n\)).

   Consider \(U = \langle v, u_3, \dots, u_n \rangle \implies \dim(U) = \dim(V) - 1\).

   \(\implies\) by hypothesis \(U\) is invariant under \(T\), but note that \(v \in U\) and \(T(v) \notin U\). Contradiction!

   (Since \(\{v, T(v), u_3, \dots, u_n\}\) is a basis, \(U = \langle v, u_3, \dots, u_n \rangle\) can't generate \(T(v)\).)

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Definition: Suppose \(T \in \mathcal{L}(V, V)\) and \(U\) is an invariant subspace under \(T\). We define:

1. The restriction operator \(T|_U : U \to U\) is defined as \(T|_U(u) = T(u)\) for \(u \in U\).

   Note that \(T|_U \in \mathcal{L}(U, U)\) since \(U\) is invariant.

2. The \(\textbf{quotient operator}\) \(T/U \in \mathcal{L}(V/U, V/U)\) is defined by \(T/U(v + U) = T(v) + U\) for \(v \in V\).

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Suppose \(T \in \mathcal{L}(V, V)\). Prove that:

1. \(T/\text{Range}(T) = 0\)

   Let \(v + \text{Range}(T) \in V / \text{Range}(T)\).

   \(\implies \left[T / \text{Range}(T)\right](v + \text{Range}(T)) = T(v) + \text{Range}(T) = 0 + \text{Range}(T) = \text{Range}(T)\).

   \(\uparrow\) is the zero element of \(V / \text{Range}(T)\).

   Therefore, \(T / \text{Range}(T) = 0\) (the zero operator).

2. \(T / \text{Null}(T)\) is injective \(\iff \text{Null}(T) \cap \text{Range}(T) = \{0\}\).

   Proof

   \(\Rightarrow)\)​ Let \(w \in \text{Null}(T) \cap \text{Range}(T)\) \(\implies w = T(v)\) for some \(v\).

   Now \(T / \text{Null}(T)\left(v + \text{Null}(T)\right) = T(v) + \text{Null}(T) = w + \text{Null}(T) = 0 + \text{Null}(T) = T / \text{Null}(T)\left(0 + \text{Null}(T)\right)\), where \(w \in \text{Null}(T)\).

   Since \(T / \text{Null}(T)\) is injective, \(v + \text{Null}(T) = 0 + \text{Null}(T)\) \(\implies v - 0 \in \text{Null}(T)\).

   Therefore \(w = T(v) = 0 \implies w = 0\).

   So \(\text{Null}(T) \cap \text{Range}(T) = \{0\}\).

   \(\Leftarrow)\) We need to see \(T / \text{Null}(T)\) is injective.

   It is enough to see that \(\text{Null}(T/\text{Null}(T))=\left\lbrace0+\text{Null}(T)\right\rbrace\).

   Let \(v + \text{Null}(T) \in \text{Null}(T / \text{Null}(T))\).

   \(\implies 0 + \text{Null}(T) = T / \text{Null}(T)(v + \text{Null}(T)) = T(v) + \text{Null}(T)\).

   \(\implies T(v)-0\in\text{Null}(T)\), but then \(T(v) \in \text{Null}(T) \cap \text{Range}(T) = \{0\}\) (by hypothesis).

   \(\implies T(v) = 0 \implies v \in \text{Null}(T)\).

   Therefore, \(v + \text{Null}(T) = 0 + \text{Null}(T)\).

   \(\implies \text{Null}(T / \text{Null}(T)) = \{0 + \text{Null}(T)\}\).

   \(\implies T / \text{Null}(T)\) is injective.

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

   Proof

   Let \(\lambda \in \mathbb{F}\) be an eigenvalue of \(T / U\).

   \(\implies \exists v + U \in V / U\), non-zero, such that \(T(v) + U = T / U(v + U) = \lambda(v + U)\) (by definition).

   \(\implies T(v) + U = \lambda v + U \implies T(v) - \lambda v \in U \implies (T - \lambda \text{Id})(v) \in U\).

   We have two cases:

   1. If \(\lambda\) is an eigenvalue of \(T|_U\), \(\lambda\) is an eigenvalue of \(T\) (any eigenvector is in \(U\)).

   2. If \(\lambda\) is not an eigenvalue of \(T|_U\), remember theorem in lecture

      Therefore, since \(\lambda\) is not an eigenvalue of \(T|_U\),

      \(\implies (T|_U - \lambda \text{Id})|_U\) is invertible, in particular, it is surjective.

      \(\implies\)​\(\exists u\in U\) such that \((T|_{U}-\lambda\text{Id})(u)=T(v)-\lambda v\).

      \(\implies T(u)-\lambda u=T(v)-\lambda v\implies T(u-v)=\lambda(u-v)\).

      So \(\lambda\) is an eigenvalue of \(T\) with eigenvector \(u-v\neq0\) (since \(u \neq v\)).

      Therefore, in all cases, \(\lambda\) is an eigenvalue of \(T\).

      \(\implies\) Every eigenvalue of \(T / U\) is an eigenvalue of \(T\).

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