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Homework 8

  1. (30 pts) For the following linear maps, find \(\text{Ker}(T)\) and \(\text{Im}(T)\), give a basis for these subspaces and verify the dimension theorem.

    (a) \(T: \mathbb{R}^3 \rightarrow \mathbb{R}^5\), \(T(x, y, z) = (x - y + z, x + y + 2z, 2x + 3y - 5z, 2x - y + z, 4x + 3y - z)\).

    (b) \(T: M_2(\mathbb{R}) \rightarrow M_2(\mathbb{R})\), \(T\begin{pmatrix} x & y \\ z & w \end{pmatrix} = \begin{pmatrix} -x + 3y - 2z + w & 7x - 6y + 2z - w \\ -2x + 6y - 4z + 2w & -11x + 13y - 6z + 3w \end{pmatrix}\).

    (c) \(T: M_2(\mathbb{R}) \rightarrow P_5(\mathbb{R})\), \(T\begin{pmatrix} a & b \\ c & d \end{pmatrix} = (a - b)x^5 + (c + d)x^4 + (a + b)x^3 + (c + d)x^2 + (2b + 3c)x + (7a - 8b)\).

    (a) Take standard basis: \([T]_{E} = \begin{pmatrix} 1 & -1 & 1 \\ 1 & 1 & 2 \\ 2 & 3 & -5 \\ 2 & -1 & 1 \\ 4 & 3 & -1 \end{pmatrix}\).
    The row reduced echelon form of \([T]_E\) is \(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\).
    Thus \(\text{ker}(T) = \{0\}\). \(\text{Im}(T) = \langle (1, 1, 2, 2, 4), (-1, 1, 3, -1, 3), (1, 2, -5, 1, -1) \rangle\).
    ​\(\dim \mathbb{R}^3 = 3 = \dim \text{ker}(T) + \dim (\text{Im}(T)) = 0 + 3 = 3\).

    (b) Take standard basis \(\mathcal{E}\): \([T]_{\mathcal{E}}= \begin{pmatrix} -1 & 3 & -2 & 1 \\ 7 & -6 & 2 & -1 \\ -2 & 6 & -4 & 2 \\ -11 & 13 & -6 & 3 \end{pmatrix}\).

    The row reduced echelon form is \(\begin{pmatrix} 1 & 0 & -\frac{2}{5} & \frac{1}{5} \\ 0 & 1 & -\frac{4}{5} & \frac{2}{5} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}\).

    \(\Rightarrow x=\frac{2}{5}z-\frac{1}{5}w,y=\frac{4}{5}z-\frac{2}{5}w\), \(z=0\Rightarrow\text{ker}(T)=\left.\langle(2/5,4/5,1,0),(-1/5,-2/5,0,1).\right>\).

    \(\text{Im}(T)=\left<(-1,7,-2,-11),(3,-6,6,13)\right>\).

    Then \(\text{dim }M_{2}(\mathbb{R}) = 4 = \text{dim ker }T + \text{dim Im}(T) = 2 + 2 = 4\).

    (c) Take standard basis: \([T]_{E}= \begin{pmatrix} 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 2 & 3 & 0 \\ 7 & -8 & 0 & 0 \end{pmatrix}\)
    The row reduced echelon form is \(\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}\)
    Then \(\text{ker}(T) = \{0\}\). \(\text{Im}(T)=\langle(1,0,1,0,0,7),(-1,0,1,0,2,-8),(0,1,0,1,3,0),(0,1,0,1,0,0)\rangle\)
    Then \(\text{dim }M_{2}(\mathbb{R}) = 4 = \text{dim ker}(T ) + \text{dim Im}(T) = 0 + 4 = 4\)

  2. (40 pts) Defining linear maps.
    (a) Define a linear map \(T: \mathbb{R}^3 \to \mathbb{R}^3\) such that \((1,1,0) \in \text{Im}(T)\) and \((0,1,1), (1,2,1) \in \text{Ker}(T)\). Find \(T(x,y,z)\).
    (b) Define a linear map \(T: \mathbb{R}^3 \to \mathbb{R}^3\) such that \(\text{Im}(T)\) is spanned by \((1,0,-1)\) and \((1,2,2)\). Find \(T(x,y,z)\).
    (c) Define a linear map \(T: \mathbb{R}^3 \to M_2(\mathbb{R})\) such that \(\text{Ker}(T) = \{(x,y,z) \in \mathbb{R}^3: z = 2x = y\}\) and \(\text{Im}(T) = \left\{\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in M_2(\mathbb{R}): b = a-c, b-d = a+c \right\}\). Find \(T(x,y,z)\).
    (d) Prove there does not exist a linear map \(T: \mathbb{R}^3 \to M_2(\mathbb{R})\) such that \(\text{Ker}(T) = \{(x,y,z) \in \mathbb{R}^3: z = 2x = y\}\) and \(\text{Im}(T) = \left\{\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in M_2(\mathbb{R}): b-d = a+c \right\}\).

    (a) Let \(\text{ker}T = \langle (0,1,1), (1,2,1) \rangle\), \(\text{Im}T = \langle (1,1,0) \rangle\).
    Let \(B = \{(0,1,1) = v_1, (1,2,1) = v_2, (1,0,0) = v_3\}\).
    Define: \(T(0,1,1) = 0 = T(1,2,1)\). \(T(1,0,0) = (1,1,0)\).
    ​\((x,y,z) = a_1 v_1 + a_2 v_2 + a_3 v_3\)

    Then \(\begin{cases}x=a_2+a_3 \\y=a_1+2a_2 \\z=a_1+a_2\end{cases}\)
    Then \(T(x,y,z) = a_1 T(v_1) + a_2 T(v_2) + a_3 T(v_3)\) \(= (2z-y) 0 + (y-z) 0 + (x-y+z) (1,1,0)\)
    ​\(= (x-y+z, x-y+z, 0)\)
    (b) Define \(T(1,0,0)=(1,0,-1), T(0,1,0)=(1,2,2), T(0,0,1)=(0,0,0)\)
    ​\(T(x,y,z) = xT(1,0,0) + yT(0,1,0) + zT(0,0,1) = x(1,0,-1) + y(1,2,2)\)
    ​\(=(x+y, 2y, 2y-x)\)
    (c) \(\ker T = \lang(1,2,2)\rang\), From ImT, we get \(d=-2c, b=a-c\)
    Then \(\text{Im} T = \lang (1,1,0,0), (0,-1,1,-2)\rang\)
    Define \(T(1,2,2) = (0,0,0,0), T(e_2)=(1,1,0,0), T(e_3)=(0,-1,1,-2)\)
    Then \((x,y,z) = a_1 (1,2,2) + a_2 e_2 + a_3 e_3\)

    Then \(\begin{cases} x=a_1 \\ y=2a_1+a_2 \\ z=2a_1+a_3 \end{cases} \implies \begin{cases} a_1=x \\ a_2=y-2x \\ a_3=z-2x \end{cases}\)
    Then \(T(x,y,z) = a_1 T(e_1) + a_2 T(e_2) + a_3 T(e_3)\) \(= (y-2x) (1,1,0,0) + (z-2x) (0,-1,1,-2) = (y-2x, y-z, z-2x, -2z+4x)\).
    (d) From above we know dimker\(T=1\), Im\((T) = \left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in M_2(\mathbb{R}): b=a+c+d \right\}\).
    Then dimIm \((T) = 3\). Then dim Im \(T\) + dim ker \(T = 4\neq3=\) dim \(\mathbb{R}^3\).
    Thus doesn't exist.

  3. (30 pts) Suppose \(V\) and \(W\) are finite-dimensional and that \(U\) is a subspace of \(V\). Prove that there exists \(T \in \text{Hom}_F(V,W)\) such that \(\text{Ker}(T) = U\) if and only if \(\dim(U) \ge \dim(V) - \dim(W)\).

    \(\Rightarrow)\) By dimension theorem, \(\dim V = \dim \text{ker}(T) + \dim(\text{Im}(T))\)
    And we know \(\dim W \ge \dim(\text{Im}(T))\). Then \(\dim V \le \dim \text{ker} T + \dim W\)
    ​\(\Rightarrow \dim V \le \dim U + \dim W \Rightarrow \dim U \ge \dim V - \dim W\)

    \(\Leftarrow\)) Let \(B_{V} = \{u_{1}, \dots, u_{l}, v_{1}, \dots, v_{k}\}\) s.t. \(\dim U = l\), \(\dim V = k+l\). and \(B_U = \{u_1, \dots, u_l\}\), \(\dim W=m\), \(B_W = \{w_1, \dots, w_m\}\)
    Then since \(\dim U \ge \dim V - \dim W\), then \(l \ge k+l-m \Rightarrow k \le m\).
    Define \(T(u_i) = 0\) for \(\forall i \in [1,l]\). \(T(v_i) = w_i\) for \(\forall i \in [1,k]\). Then by theorem, this is a unique linear map.

    Then we need to check: \(U = \text{ker}(T)\).

    \(\subseteq\)) Take \(u \in U: u = a_1 u_1 + \dots + a_l u_l\), then \(T(u) = a_1 T(u_1) + \dots + a_l T(u_l) = 0\). Then \(u \in \ker T\).

    \(\supseteq\)) Take \(u \in \ker T: T(u) = 0\). Since \(\ker T \subseteq V\), then \(T(u)=0=T(a_{1}v_{1}+\dots+a_{k}v_{k}+a_{k+1}u_{1}+\dots+a_{k+l}u_{l})=a_{1}T(v_{1} )+\dots+a_{k}T(v_{k})+a_{k+1}T(u_{1})+\dots+a_{k+l}T(u_{l})\).
    ​\(= a_1 w_1 + \dots + a_k w_k\). Since \(\{w_1, \dots, w_k\}\) are basis, then \(a_1 = \dots = a_k = 0\).
    Thus \(u = a_{k+1}u_{1} + \dots + a_{k+l} u_{l}\). Thus \(u \in U\).
    Then \(\ker T = U\). Thus exist such linear map.

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