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7

Homework 7.pdf

  1. (30 pts) Consider the linear map given by \(T : P_2(\mathbb{R}) \to P_1(\mathbb{R})\), \(T(ax^2 + bx + c) = (a + b)x + 2c - a\).
    (a) Let \(C = \{x, x^2, 1\}\) and \(C' = \{1, x\}\). Find the matrix of \(T\) with respect to \(C\) and \(C'\).

    Since \(T(x)=x,T\left(x^2\right)=x-1,T\left(1\right)=2\), then \(T(x)=0\cdot1+1\cdot x,T(x^2)=-1\cdot1+1\cdot x,T(1)=2\cdot1+0\cdot x\)​

    Thus the matrix is \(\begin{pmatrix}0 & -1 & 2\\ 1 & 1 & 0\end{pmatrix}\)

    (b) Let \(\mathcal{B} = \{x^2 + 1, x^2 + x + 1, x^2\}\) and \(\mathcal{B}' = \{1, x - 2\}\). Find the matrix of \(T\) with respect to \(\mathcal{B}\) and \(\mathcal{B}'\).

    Since \(T(x^2+1)=x+1,T\left(x^2+x+1\right)=2x+1,T\left(x^2\right)=x-1\), then \(T(x^2+1)=3\cdot1+1\cdot\left(x-2),T(x^2+x+1\right)=5\cdot1+2\cdot\left(x-2),T(x^2\right)=1\cdot1+1\cdot\left(x-2\right)\)

    Thus the matrix is \(\begin{pmatrix}3 & 5 & 1\\ 1 & 2 & 1\end{pmatrix}\)
    (c) Give a basis for \(\text{Null}(T)\) and \(\text{Range}(T)\).

    Since \(T_{CC'}=\begin{pmatrix}0 & -1 & 2\\ 1 & 1 & 0\end{pmatrix}\), then \(T_{CC^{\prime}}=\begin{pmatrix}0 & -1 & 2\\ 1 & 1 & 0\end{pmatrix}=\begin{pmatrix}0 & 1 & -2\\ 1 & 0 & 2\end{pmatrix}=\begin{pmatrix}1 & 0 & 2\\ 0 & 1 & -2\end{pmatrix}\)​

    Thus we have \(\begin{pmatrix}1 & 0 & 2\\ 0 & 1 & -2\end{pmatrix}\cdot\begin{pmatrix}b\\ a\\ c\end{pmatrix}\Rightarrow\begin{cases}b+2c=0\\a-2c=0\end{cases}\)​

    Thus the basis of \(\text{Null}T\) is \(\{2x^2 -2x+x\}\)​

    The basis of \(\text{Range}T\) is \(\{x,x-1\}\)​

  1. (30 pts) Let \(T : V \to W\) be a linear map. Prove the following:
    (a) If \(T(v) = 0\) for all \(v \in V\), then for any bases \(\mathcal{B}_V\) and \(\mathcal{B}_W\) of \(V\) and \(W\) respectively, the matrix of \(T\) with respect to \(\mathcal{B}_V\) and \(\mathcal{B}_W\) is the zero matrix.

    Suppose the basis of \(V\) is \(\{v_1,...,v_n\}\) and the basis of \(W\) is \(\{w_1,\ldots,w_{m}\}\)

    Then \(T(v_i)=0\cdot w_1+...+0\cdot w_m\) for any \(i\in[1,n]\)

    Thus each column of matrix is \(0\), then the matrix is zero matrix.

    (b) If \(\text{Null}(T)\) is non-trivial, then there exists a basis \(\mathcal{B}_V\) of \(V\) such that for any basis \(\mathcal{B}_W\) of \(W\), the matrix of \(T\) with respect to \(\mathcal{B}_V\) and \(\mathcal{B}_W\) has at least one zero column. Moreover, one can take \(\mathcal{B}_V\) in such a way as to have \(\dim(\text{Null}(T))\) zero columns.

    Suppose the basis of \(\text{Null}(T)\) is \(\{v_1,\ldots,v_{k}\}\), then we can extend it to the basis of \(V\) which is \(\{v_1,\ldots,v_{k},v_{k+1},\ldots,v_{n}\}\)​

    Then \(T(v_{i})=0=0\cdot w_1+...+0\cdot w_{m},\forall i\in\left\lbrack1,k\right\rbrack\) since \(w_1,...,w_m\) is the basis

    Then take the coordinates of \(T(v_{i})\) we have at least one zero column.

    Since \(i\in[1,k]\), then we have \(k\) zero columns equal to \(\dim(\text{Null}T)=k\)
    (c) There exist bases \(\mathcal{B}_V\) and \(\mathcal{B}_W\) of \(V\) and \(W\) respectively such that the matrix of \(T\) with respect to \(\mathcal{B}_V\) and \(\mathcal{B}_W\) is \(\begin{pmatrix}I_{m} & 0\\ 0 & 0\end{pmatrix}\) where \(I_m\) is the \(m \times m\) identity matrix and \(m = \dim(\text{Range}(T))\).

    Assume \(m = \dim(\text{Range}(T))\), then consider a linearly independent list \(\{v_1,...,v_m\}\) since \(\dim V\geq m\) by dimension theorem

    Since \(v=a_1v_1+...+a_mv_m\) is unique way, then \(T(v)=T\left(a_1v_1+...+a_{m}v_{m}\right)=a_1T\left(v_1\right)+\cdots+a_{m}T\left(v_{m}\right)\) which is unique.

    Then we have a linear independent list \(\{T(v_1),\ldots,T(v_{m})\}\) and the length is \(m\), then it is a basis of \(\text{Range}T\)

    Also we can extend the list to the \(\{v_1,...,v_{m},v_{m+1},\ldots,v_n\}\) which is the basis of \(V\)

    And the basis of \(W\) is \(\{w_1,...,w_{m},w_{m+1},\ldots,w_{p}\}\)​

    Then we can define \(T(v_{i})=\begin{cases}w_{i},\forall i\in\left\lbrack1,m\right\rbrack\\ 0,\forall i\in\left\lbrack m+1,p\right\rbrack\end{cases}\)

    Thus \(T(v_{i})=\begin{cases}0\cdot w_1+\cdots+1\cdot w_{i}+0\cdot w_{i+1}+\cdots0\cdot w_{m},\forall i\in\left\lbrack1,m\right\rbrack\\ 0,\forall i\in\left\lbrack m+1,p\right\rbrack\end{cases}\)

    Thus the first \(m\) columns and rows is identity matrix, and the rest are \(0\)

    Then we have the \(\begin{pmatrix}I_{m} & 0\\ 0 & 0\end{pmatrix}\)

  1. (20 pts) Suppose \(V\) is finite-dimensional and \(S, T, U \in \mathcal{L}(V, V)\) and \(STU = I\). Show that \(T\) is invertible and that \(T^{-1} = US\).

    Since \(STU=I\) and \(S, T, U \in \mathcal{L}(V, V)\)​, then \(S^{-1}STU=S^{-1}\Rightarrow TU=S^{-1}\Rightarrow TUU^{-1}=S^{-1}U^{-1}\Rightarrow T=S^{-1}U^{-1}\)

    Then \(S^{-1}U^{-1}\in\mathcal{L}(V,V)\) and \(US\in\mathcal{L}(V,V)\)​

    Then let's check \((US)\circ T=US\circ\left(S^{-1}U^{-1}\right)=US\circ\left(US\right)^{-1}=I\)​

    \(T\circ(US)=\left(S^{-1}U^{-1}\right)\circ\left(US\right)=\left(US\right)^{-1}\circ\left(US\right)=I\)

    Thus \(T\) is invertible and \(T^{-1} = US\).

  1. (20 pts) Prove that every linear map from \(M_{n \times 1}(\mathbb{F})\) to \(M_{m \times 1}(\mathbb{F})\) is given by a matrix multiplication. In other words, prove that if \(T \in \mathcal{L}(M_{n \times 1}(\mathbb{F}), M_{m \times 1}(\mathbb{F}))\), then there exists an \(m \times n\) matrix \(A\) such that \(T(x) = Ax\) for every \(x \in M_{n \times 1}(\mathbb{F})\).

    Proof

    Since \(T:M_{n\times1}(\mathbb{F})\rightarrow\) \(M_{m \times 1}(\mathbb{F})\) is linear, then let the basis of \(M_{n\times1}(\mathbb{F})\) is \(B=\{e_1,e_2,...,e_{n}\}\) and the basis of \(M_{m\times1}(\mathbb{F})\) is \(B^{\prime}=\{f_1,f_2,...,f_{m}\}\)

    Then we know \(T(e_{i})=\sum_{k=1}^{m}a_{ki}f_{k}\), then we have a matrix corresponding to \(T\) : \(A=\begin{pmatrix}a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix}\)

    If we have \(x \in M_{n \times 1}(\mathbb{F})\), then \(x=\sum_{j=1}^{n}b_{j}e_{j}=\begin{pmatrix}b_1\\ \vdots\\ b_{n}\end{pmatrix}\)

    Then \(T(x)=T\left(\sum_{j=1}^{n}b_{j}e_{j}\right)=\sum_{j=1}^{n}b_{j}T\left(e_{j}\right)=\sum_{j=1}^{n}b_{j}\sum_{k=1}^{m}a_{kj}f_{k}=\sum_{j=1}^{n}\sum_{k=1}^{m}a_{kj}b_{j}f_{k}\)​

    Then \(Ax=\begin{pmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \vdots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn}\end{pmatrix}\begin{pmatrix}b_1\\ \vdots\\ b_{n}\end{pmatrix}=\begin{pmatrix}\sum_{l=1}^{n}a_{1l}b_1\\ \sum_{l=2}^{n}a_{2l}b_2\\ \vdots\\ \sum_{l=m}^{n}a_{ml}b_{m}\end{pmatrix}\in M_{m\times1}(\mathbb{F})\)

    Thus such \(A\) exists such that \(T(x)=Ax\)

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