11

Homework 11.pdf

1. (20pts) Prove or give a counterexample:

   If \(V\) is finite-dimensional and \(U\) is a subspace of \(V\) that is invariant under every \(T \in L(V, V)\), then \(U = \{0\}\) or \(U = V\).

   Proof

   Suppose \(U\subsetneq V\) and \(U\neq \{0\}\), then there exists \(v\in V,v\notin U\) and \(u\in U,u\neq 0\)​

   Then we can extend \(u\) to a basis: \(\{u,u_2,u_3,...,u_k\}\). By fundamental theorem of linear map, we can define:

   \[ T(u)=v,T(u_i)=0,\forall i\in [2,k] \]

   Since \(U\) is invariant under \(T\), then if \(u\in U,T(u)\in U\). Then \(v\in U\)

   Contradiction!

   Thus \(U = \{0\}\) or \(U = V\).

2. (20pts) Let \(T : F^n \to F^n\) be defined by

   \[ T(x_1, x_2, \dots, x_n) = (x_1, 2x_2, \dots, nx_n). \]

   1. Find all eigenvalues and the associated eigenvector spaces of \(T\).

      We need \(T(x_1,...,x_n)=\lambda(x_1,...,x_n)\). On the left hand we get \((x_1,2x_2,...,nx_n)\).

      On the right hand, we get \((\lambda x_1,...,\lambda x_n)\).

      Then \(\begin{cases}x_1=\lambda x_1\\ 2x_2=\lambda x_2\\ \ldots\\ nx_{n}=\lambda x_{n}\end{cases}\Rightarrow\begin{cases}\lambda=0,x_1=\ldots=x_{n}=0,\times\\ \lambda=i,x_1=\ldots=x_{i-1}=x_{i+1}=\ldots=x_{n}=0,x_{i}\neq0\end{cases}\)​

      Thus the eigenvalues take from \(1\) to \(n\). And the corresponding eigenvectors are from \((x_1,0,...,0,0)\) to \((0,0,...,0,x_n)\)​ 2. Find all invariant subspaces of \(T\).

      Take any \(v\in\) \(W=\langle a_1e_1,...,a_{n}e_{n}\rangle:a_{i}=\{0,1\}\), then \(T(\lambda_1a_1e_1+...+\lambda_{n}a_{n}e_{n})=\lambda_1a_1T\left(e_1\right)+\cdots+\lambda_{n}a_{n}T\left(e_{n}\right)=\left(\lambda_1a_1,2\lambda_2a_2,\ldots,n\lambda_{n}a_{n}\right)\in W\)

      Then \(\left(\lambda_1a_1,2\lambda_2a_2,\ldots,n\lambda_{n}a_{n}\right)=\left(c_1a_1,c_2a_2,\ldots,c_{n}a_{n}\right)\), then \(\begin{cases}\lambda_1=c_1\\2\lambda_2=c_2\\...\\n\lambda_n=c_n\end{cases}\).

      Thus \(W\) is invariant of \(T\) for any \(a_i\), then all invariant subspaces is \(W=\langle a_1e_1,...,a_{n}e_{n}\rangle:a_{i}=\{0,1\},\forall i\in\left\lbrack1,n\right\rbrack\)​

3. (20pts) Suppose \(V = U \oplus W\), where \(U\) and \(W\) are non-zero subspaces of \(V\). Define \(P \in L(V, V)\) by\(P(u+w)=u\quad\text{for}\quad u\in U,w\in W.\)​ Find all eigenvalues and eigenvectors of \(P\).

   Since \(P(u+w)=\lambda\left(u+w\right)\), then \(u=\lambda u+\lambda w\Rightarrow u(\lambda-1)+\lambda w=0\). Since \(U\oplus W\), then \(\lambda-1=0\wedge\lambda=0\) which is impossible at same time

   Thus there is no eigenvalues and no eigenvectors.

4. (20pts) Suppose \(T \in L(V, V)\) and \(\dim(\text{Range}(T)) = k\). Prove that \(T\) has at most \(k+1\) distinct eigenvalues.

   Proof

   Suppose \(T\) has at least \(k+2\) distinct eigenvalues, then \(\dim V\geq k+2\) by theorem

   Since \(\dim V=\dim\text{null}T+\dim\text{range}T\), then \(\dim \text{null}T\geq 2\)

   Then there exists at least \(\langle v_1,v_2\rangle\subseteq\text{null}T\), then \(Tv_1=0=0\cdot v_1\), \(Tv_2=0=0\cdot v_2\)

   Then \(\lambda_1=\lambda_2=0\) since \(v_1,v_2\neq 0\). Which contradicts to the distinct eigenvalues

   Thus \(T\) has at most \(k+1\) distinct eigenvalues.

5. (20pts) Suppose \(A \in M_n(F)\) and define \(T \in L(F^n, F^n)\) by \(T(v) = Av\), where the elements of \(F^n\) are thought of as \(n \times 1\) column vectors.

   1. Suppose the sum of the entries in each row of \(A\) equals 1. Prove that 1 is an eigenvalue of \(T\).

      Since \(T(v)=\lambda v=Av\), then \(\lambda Iv=Av\Rightarrow\left(A-\lambda I\right)v=0\)​

      Since \(\sum_{k=1}^{n}A_{ik}=1\) for \(i\in[1,n]\), then \(\sum_{k=1}^{n}(A-\lambda I)_{ik}=1-\lambda\sum_{k=1}^{n}I_{ik}=1-\lambda\)​

      Then \(\left(A-\lambda I\right)v=0\Rightarrow\sum_{k=1}^{n}(A-\lambda I)_{ik}v_{1k}=\left(1-\lambda\right)\left(\sum_{k=1}^{n}v_{1k}\right)=0\)

      Thus \(\lambda =1\)​ 2. Suppose the sum of the entries in each column of \(A\) equals 1. Prove that 1 is an eigenvalue of \(T\).

      Since \(T(v)=\lambda v=Av\), then \(\lambda Iv=Av\Rightarrow\left(A-\lambda I\right)v=0\)

      Since \(\sum_{k=1}^{n}A_{kj}=1\) for \(j\in[1,n]\), then \(\sum_{k=1}^{n}(A-\lambda I)_{kj}=1-\lambda\sum_{k=1}^{n}I_{kj}=1-\lambda\) (*)

      Let \(T(v)=Av=B\) where \(B\in M\)

      Then \(b_{11=}\sum_{k=1}^{n}(A-\lambda I)_{1k}v_{1k}=\left(A-\lambda I\right)_{11}v_{11}+\cdots+\left(A-\lambda I\right)_{1n}v_{1n}\)

      \(b_{11=}\sum_{k=1}^{n}(A-\lambda I)_{1k}v_{1k}=\left(A-\lambda I\right)_{11}v_{11}+\cdots+\left(A-\lambda I\right)_{1n}v_{1n}\)

      ....

      \(b_{n1=}\sum_{k=1}^{n}(A-\lambda I)_{nk}v_{kn}=\left(A-\lambda I\right)_{n1}v_{1n}+\cdots+\left(A-\lambda I\right)_{nn}v_{nn}\)

      Then we know \(b_{11}+...+b_{n1}=0\), then

      \(\left(A-\lambda I\right)_{11}v_1+\cdots+\left(A-\lambda I\right)_{1n}v_{n}+\\\left(A-\lambda I\right)_{21}v_1+\cdots+\left(A-\lambda I\right)_{2n}v_{n}+\\\ldots+\\\left(A-\lambda I\right)_{n1}v_1+\cdots+\left(A-\lambda I\right)_{nn}v_{n}=0\)

      Then we have \(v_1\left\lbrack(A-\lambda I)_{11}+\ldots+(A-\lambda I)_{n1}\right\rbrack+\cdots+v_{n}\left\lbrack(A-\lambda I)_{1n}+\ldots+(A-\lambda I)_{nn}\right\rbrack=0\)

      We use (*), then we have \((1-\lambda)(v_1+...+v_n)=0\Rightarrow \lambda=1\)

      Thus \(\lambda =1\)

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