Homework 6
Yue Shi 999027873
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Give the following systems of linear equations in their matrix form and find the space of solutions by row-reducing said associated matrix.
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\(\begin{cases} \frac{1}{3}x_{1} + 2x_{2} - 6x_{3} = 0 \\ -4x_{1} + 5x_{3} = 0 \\ -3x_{1} + 6x_{2} - 13x_{3} = 0 \\ -\frac{7}{3}x_{1} + 2x_{2} - \frac{8}{3}x_{3} = 0\end{cases}\)
In matrix \(\left(\begin{array}{ccc|c}\frac{1}{3} & 2 & -6 & 0 \\-4 & 0 & 5 & 0 \\-3 & 6 & -13 & 0 \\-\frac{7}{3} & 2 & -\frac{8}{3} & 0\end{array}\right)\quad \Longleftrightarrow \quad\left(\begin{array}{ccc|c}1 & 0 & -\frac{5}{4} & 0 \\0 & 1 & -\frac{67}{24} & 0 \\0 & 0 & 0 & 0 \\0 & 0 & 0 & 0\end{array}\right)\)
Then we get \(x_1 = \frac{5}{4}x_3 \quad \text{and} \quad x_2 = \frac{67}{24}x_3\)
Thus the solution space is \(\left\langle \left( \frac{5}{4}, \frac{67}{24}, 1 \right) \right\rangle\) 2. \(\begin{cases} x_{1} - x_{2} + 2x_{3} = 1 \\ 2x_{1} + 2x_{3} = 1 \\ x_{1} - 3x_{2} + 4x_{3} = 2\end{cases}\)
In matrix \(\left( \begin{array}{ccc|c} 1 & -1 & 2 & 1 \\ 2 & 0 & 2 & 1 \\ 1 & -3 & 4 & 2 \end{array}\right)\quad \Longleftrightarrow \quad\left( \begin{array}{ccc|c} 1 & 0 & 1 & \frac{1}{2} \\ 0 & 1 & -1 & -\frac{1}{2} \\ 0 & 0 & 0 & 0 \end{array}\right)\)
Then we get \(x_1 + x_3 = \frac{1}{2} \quad \text{and} \quad x_2 - x_3 = -\frac{1}{2}\)
Then we have \(x_1 = \frac{1}{2} - x_3, \quad x_2 = x_3 - \frac{1}{2}\)Thus the solution space is \(\left\{ \left( \frac{1}{2} - a, a - \frac{1}{2}, a \right) : a \in \mathbb{R} \right\}\)
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Find the values of \(k \in \mathbb{R}\) for which the following system is:
\(\begin{cases} 2k x_{1} + 4x_{2} + 2k x_{3} = 2 \\ k x_{1} + (k + 4)x_{2} + 3x_{3} = -2 \\ -k x_{1} - 2x_{2} + x_{3} = 1 \\ (k + 2)x_{2} + (3k + 1)x_{3} = -1\end{cases}\)
- inconsistent
- consistent independent
- consistent dependent
\(\left( \begin{array}{ccc|c} 2k & 4 & 2k & 2 \\ k & k+4 & 3 & -2 \\ -k & -2 & 1 & 1 \\ 0 & k+2 & 3k+1 & -1 \end{array}\right)\quad\longrightarrow{\quad}\left( \begin{array}{ccc|c} k & 2 & -1 & -1 \\ 0 & k+2 & 4 & -1 \\ 0 & 0 & k+1 & 2 \\ 0 & 0 & 0 & 1-k \end{array}\right)\)
Inconsistent: By theorem, \(1-k \neq 0 \Rightarrow k \neq 1\).
When \(k = 1\): \(\left(\begin{array}{ccc|c}1 & 2 & -1 & -1 \\0 & 3 & 4 & -1 \\0 & 0 & 1 & 1\end{array}\right)\) has a unique solution since \(\text{rank}=3\)
Thus inconsistent: \(k\neq 1\), consistent independent: \(k=1\), consistent dependent: impossible
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Decide whether the following functions are linear maps or not. Justify.
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\(T : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}, \quad T(x, y) = (\sin(x), y)\)
\(T(x_1, y_1) + T(x_2, y_2) = (\sin(x_1) + \sin(x_2), y_1 + y_2)\) \(\neq T(x_1 + x_2, y_1 + y_2)\) since \(\sin(x_1) + \sin(x_2) \neq \sin(x_1 + x_2)\)
Thus, it is not linear. 2. \(T : P_{2}(\mathbb{R}) \rightarrow M_{2}(\mathbb{R}), \quad T(ax^{2} + bx + c) =\begin{pmatrix} a & b + c \\ b + c & a\end{pmatrix}\)\(T(a_1 x^2 + b_1 x + c_1) + T(a_2 x^2 + b_2 x + c_2) = \begin{pmatrix} a_1 + a_2 & b_1 + b_2 + c_1 + c_2 \\ b_1 + b_2 + c_1 + c_2 & a_1 + a_2 \end{pmatrix}\)
\(= T((a_1 + a_2) x^2 + (b_1 + b_2) x + (c_1 + c_2))\)
\(\lambda T(ax^2 + bx + c) = \lambda \begin{pmatrix} a & b + c \\ b + c & a \end{pmatrix} = \begin{pmatrix} \lambda a & \lambda b + \lambda c \\ \lambda b + \lambda c & \lambda a \end{pmatrix} = T(\lambda ax^2 + \lambda bx + \lambda c)\)
Thus, T is a linear map. 3. \(T : M_{2}(\mathbb{R}) \rightarrow \mathbb{R}^{3}, \quad T\left(\begin{pmatrix} a & b \\ c & d\end{pmatrix}\right) = (a + b, cd, 2a + b - 3c + d)\)\(T \begin{pmatrix} a_{1} & b_{1} \\ c_{1} & d_{1} \end{pmatrix}+T \begin{pmatrix} a_{2} & b_{2} \\ c_{2} & d_{2} \end{pmatrix}=(a_{1}+b_{1}+a_{2}+b_{2},c_{1}d_{1}+c_{2}d_{2},2a_{1}+b_{1}-3c_{1}+ d_{1}+2a_{2}+b_{2}-3c_{2}+d_{2})\)
\(\neq T \begin{pmatrix} a_{1}+a_{2} & b_{1}+b_{2} \\ c_{1}+c_{2} & d_{1}+d_{2} \end{pmatrix}=(a_{1}+a_{2}+b_{1}+b_{2},\left(c_{1}+c_{2})\left(d_{1}+d_{2}\right) ,\ldots\right.)\) since \((c_1 + c_2)(d_1 + d_2) \neq c_1 d_1 + c_2 d_2\)
Thus, this is not a linear map. 4. \(\Phi : \text{Hom}_F(F, V) \rightarrow V, \quad \Phi(T) = T(1)\), where \(V\) is any \(F\)-vector space.\(\Phi(T_1) + \Phi(T_2) = T_1(1) + T_2(1) = (T_1 + T_2)(1)\) \(= \Phi(T_1 + T_2)\) since \(T_1, T_2 \in Hom(F, V)\) is linear
\(\lambda \Phi(T) = \lambda T(1) = \Phi(\lambda T)\)
Thus, \(\Phi\) is linear.
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Consider the vectors in \(\mathbb{R}^2\) given by
\(\alpha_1 = (1, -1), \alpha_2 = (2, -1), \alpha_3 = (-3, 2)\)
\(\beta_1 = (1, 0), \beta_2 = (0, 1), \beta_3 = (1, 1)\)
Is it possible to define a linear map \(T : \mathbb{R}^2 \rightarrow \mathbb{R}^2\) such that \(T(\alpha_i) = \beta_i\) for \(i = 1, 2, 3\)?
If so, describe \(T(x, y)\).Solution
By fundamental theorem of linear map. \(\exists ! T: T(\alpha_i) = \beta_i\), \(i = 1, 2\).
Let \(\alpha_1, \alpha_2\) be basis of \(\mathbb{R}^2\). Then \(T(x, y) = T(c\alpha_1 + d\alpha_2) \Rightarrow \begin{cases} x = c + 2d \\ y = -c - d \end{cases}\)
Thus \(d = x + y\), \(c = -x - 2y\)Then \(T(x, y) = (-x - 2y) T(\alpha_1) + (x + y) T(\alpha_2)\) \(= (-x - 2y) (1, 0) + (x + y) (0, 1)\)
\(= (-x - 2y, x + y)\)
Then evaluate \(\alpha_3\) : \(T(-3, 2) = (-(-3) - 2(2), -3 + 2) = (3 - 4, -1) = (-1, -1)\) \(\neq (1, 1)\)
Thus, \(\nexists T\).