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12.27

  1. Suppose that \(T \in \mathcal{L}(V, V)\) and \(U_1, \ldots, U_n\) are subspaces of \(V\) invariant under \(T\).
    Prove that \(U_1 + \cdots + U_n\) is invariant under \(T\).

    Proof

    Take \(u\in (U_1 + \cdots + U_n)\), then \(u=u_1+u_2+...+u_n\). Then \(T(u)=T(u_1+...+u_{n})=T(u_1)+...+T(u_{n})\in U_1+\cdots+U_{n}\) since \(U_1, \ldots, U_n\) are subspaces of \(V\) invariant under \(T\)​

  2. Let \(T \in \mathcal{L}(\mathbb{F}^n, \mathbb{F}^n)\) be defined by \(T(x_1, \ldots, x_n) = (x_1 + \cdots + x_n, \ldots, x_1 + \cdots + x_n)\).
    Find all eigenvalues and the associated eigenvector spaces of \(T\).

    Solution

    Since \(T(x_1, \ldots, x_n) = (x_1 + \cdots + x_n, \ldots, x_1 + \cdots + x_n)\) and \(\lambda(x_1,\ldots,x_{n})=(\lambda x_1,\ldots,\lambda x_{n})\)​

    Then we need \(\begin{cases}x_1+...+x_{n}=\lambda x_1\\ ...\\ x_1+...+x_{n}=\lambda x_{n}\end{cases}\Rightarrow\begin{cases}\lambda=0,x_1+\ldots+x_{n}=0\\ \lambda=n,x_1=\ldots=x_{n}\end{cases}\)​

  3. Suppose \(T \in \mathcal{L}(V, V)\) and there exist non-zero vectors \(v\) and \(w\) in \(V\) such that \(T(v) = 3w\) and \(T(w) = 3v\).
    Prove that \(3\) or \(-3\) is an eigenvalue of \(T\).

    Since \(\forall v\in V,T(v)=3w=\lambda v\) and \(\forall v\in V,T(w)=3v=\lambda w\), then \(\lambda=\frac{3w}{v}=\frac{3v}{w}\Rightarrow\frac{w}{v}=\pm1\Rightarrow\lambda=\pm3\)​

  4. Suppose \(V\) is finite-dimensional and \(S, T \in \mathcal{L}(V, V)\). Prove that \(ST\) and \(TS\) have the same eigenvalues.

    For \(ST\), take any \(S\left(v\right)\in V\), we have \(ST(S\left(v\right))=S(T(S\left(v\right)))=\lambda S\left(v\right)\Rightarrow T\left(S\left(v\right)\right)=S^{-1}\left(\lambda S\left(v\right)\right)\Rightarrow T\left(S\left(v\right)\right)=\lambda v\)​

  5. Suppose \(V\) is finite-dimensional and \(\{v_1, \ldots, v_n\}\) is a list of vectors in \(V\).
    Prove that \(\{v_1, \ldots, v_n\}\) is linearly independent if and only if there exists \(T \in \mathcal{L}(V, V)\) such that
    ​\(\{v_1, \ldots, v_n\}\) are eigenvectors of \(T\) corresponding to distinct eigenvalues.

    Proof

    \(\Rightarrow\)) We have \(\{v_1, \ldots, v_n\}\) is linearly independent, N.T.P. the existence such \(T\)

    Then we need to define such \(T\). Since we have a linearly independent list, then we can extend it to the basis.

    Then we have a basis \(\{v_1,\ldots,v_{n},v_{n+1},\ldots,v_{m}\}\). Then by fundamental theorem of linearly algebra, there exists \(T(v_{i})=iv_{i},i\in\left\lbrack1,\ldots m\right\rbrack\)​

    \(\Leftarrow)\) Directly by theorem in lecture

  6. Give an example of a vector space \(V\), an operator \(T \in \mathcal{L}(V, V)\), and a subspace \(U\) of \(V\) that is invariant under \(T\) such that \(T / U\) has an eigenvalue that is not an eigenvalue of \(T\).

    Contradiction to the tutorial?