6
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(20 pts) Suppose \(b, c \in \mathbb{R}\). Define \(T : \mathbb{R}^3 \to \mathbb{R}^2\) by \(T(x, y, z) = (2x - 4y + 3z + b, 6x + cxyz)\). Show that \(T\) is linear if and only if \(b = c = 0\).
\(\Rightarrow\)) Since \(T\) is linear, then \(T(0,0,0)=(0,0)\Rightarrow\left(b,0\right)=\left(0,0\right)\Rightarrow b=0\)
Also \(T(1,1,1)+T(1,1,1)=(1,6+c)+(1,6+c)=\left(2,12+2c)=T(2,2,2\right)=(2,12+8c)\)
Then \(2c=8c\Rightarrow c=0\)
\(\Leftarrow\)) Assume \(b=c=0\), then \(T(x,y,z)=(2x-4y+3z,6x)\).
\(T(x_1,y_1,z_1)+T(x_2,y_2,z_3)=\left(2\left(x_1+x_2\right)-4\left(y_1+y_2\right)+3\left(z_1+z_2\right)\right)=T\left(x_1+x_2,y_1+y_2,z_1+z_2\right)~~~\checkmark\)
\(T\left(\lambda x,\lambda y,\lambda z\right)=\left(2\lambda x-4\lambda y+3\lambda z,6\lambda x\right)=\lambda T\left(x,y,z\right)~~~\checkmark\)
Thus \(T\) is a linear map
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(50 pts) Prove the following functions are linear maps. Find \(\text{Null}(T)\) and \(\text{Range}(T)\), give a basis for these subspaces, and verify the dimension theorem.
(a) \(T : \mathbb{R}^3 \to \mathbb{R}^3, \quad T(x, y, z) = (x - y + 2z, 3x + y + 4z, 5x - y + 8z).\)
\(T(x,y,z)=(x-y+2z,3x+y+4z,5x-y+8z)=\left(0,0,0\right)\)
Then we have \(\begin{cases}x-y+2z=0\\ 3x+y+4z=0\\ 5x-y+8z=0\end{cases}\Rightarrow\begin{pmatrix}1 & -1 & 2\\ 3 & 1 & 4\\ 5 & -1 & 8\end{pmatrix}\Rightarrow\begin{pmatrix}1 & 0 & \frac32\\ 0 & 1 & -\frac12\\ 0 & 0 & 0\end{pmatrix}\Rightarrow\begin{cases}x+\frac32z=0\\ y-\frac12z=0\end{cases}\)
Then \(\text{Null}\left(T\right)=\langle\left(-\frac32,\frac12,1\right)\rangle\) and the basis \(\{-\frac32,\frac12,1\}\)
\(\text{Range}\left(T\right)=\langle\left(1,3,5\right),\left(-1,1,-1\right)\rangle\) and the basis \(\{\left(1,3,5\right),\left(-1,1,-1\right)\}\)
Then \(\dim(\text{Range}(T)) = 2\) and \(\dim(\text{Null}(T)) = 1\)
Since \(\dim(\text{Range}T)+\dim(\text{Null}T)=3=\dim(V)\), then the theorem is valid.
(b) \(T : P_3(\mathbb{R}) \to P_4(\mathbb{R}), \quad T(p(x)) = (x + 1)p(x).\)
Since \(T(p(x))=\left(x+1\right)p\left(x\right)=0\Rightarrow p\left(x\right)=0\), then \(\text{Null}(T)={0}\) and basis is empty
Since the basis of \(p(x)\) is \(\{1,x,x^2,x^3\}\), then \(T(p(x))=\left(x+1\right)p\left(x\right)\), then \(\text{Range}(p(x))=\langle x+1,x^2+x,x^3+x^2,x^4+x^3\rangle\)
Then \(\dim(\text{Range}(T)) = 4\) and \(\dim(\text{Null}(T)) = 0\)
Since \(\dim(\text{Range}(T)) +\dim(\text{Null}(T)) = 4=\dim(V)\), then the theorem is valid.
(c) \(T:M_2(\mathbb{R})\to P_5(\mathbb{R}),\quad T\left(\begin{pmatrix}a & b\\ c & d\end{pmatrix}\right)\left.=\left(a-b\right)x^5+(c+d\right)x^4+(a+b)x^3+(c+d)x^2+(2b+3c)x+(7a-8b).\)
\(T\left(\begin{pmatrix}a & b\\ c & d\end{pmatrix}\right)\left.=\left(a-b\right)x^5+(c+d\right)x^4+(a+b)x^3+(c+d)x^2+(2b+3c)x+(7a-8b)=0\)
Then \(a=b,c=-d,a=-b,c=-d,2b=-3c,7a=8b\)
Then \(a=b=c=d=0\)
Thus the \(\text{Null}\left(T\right)=\left\lbrace0\right\rbrace\) and the basis is empty
\(\text{Range}(T)=\{\left(a-b\right)x^5+(c+d)x^4+(a+b)x^3+(c+d)x^2+(2b+3c)x+(7a-8b):a,b,c,d\in \R\}\)
\(=\langle x^5+x^3+7,-x^5+x^3+2x-8,x^4+x^2+3x,x^4+x^2\rangle\)
Then \(\dim(\text{Range}(T)) = 4\) and \(\dim(\text{Null}(T)) = 0\)
Since \(\dim(\text{Range}(T)) +\dim(\text{Null}(T)) = 4=\dim(V)\), then the theorem is valid.
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(30 pts) In each case define, when possible, a linear transformation \(T : \mathbb{R}^3 \to \mathbb{R}^3\) that satisfies the required conditions. When not possible, explain why not.
(a) \(\dim(\text{Range}(T)) = 2\) and \(\dim(\text{Null}(T)) = 2.\)
Impossible, since \(\dim(\text{Range}(T))\) \(+\dim(\text{Null}(T))=4\neq \dim(V)=3\)
(b) \((1, 1, 0) \in \text{Range}(T)\) and \((0, 1, 1) \in \text{Null}(T).\)
\(T\left(x,y,z\right)=\left(x,y-z,y-z-x\right)\)
(c) \((1, 1, 0) \in \text{Range}(T)\) and \((0, 1, 1), (1, 2, 1) \in \text{Null}(T).\)
Since \(-1(0,1,1)+1(1,2,1)=(1,1,0)\), then \((1,1,0)\in\text{Null}(T).\)
Contradiction
(d) \(\text{Range}(T) \subseteq \text{Null}(T).\)
\(T(x,y,z)=\left(0,0,0\right)\)
(e) \(\text{Null}(T) \subseteq \text{Range}(T).\)
\(T(x,y,z)=\left(x,y,z\right)\)