Let's solve the given homework question step by step: (a) Write the vector $ w $ as a linear combination of $ u $ and $ v $. Given vectors: \[ u = \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \quad v = \begin{pmatrix} 2 \\ -1 \end{pmatrix}, \quad w = \begin{pmatrix} -5 \\ 1 \end{pmatrix} \] We want to find scalars $ a $ and $ b $ such that $ w = au + bv $. This gives us the following system of equations: \[ a(1) + b(2) = -5 \] \[ a(1) + b(-1) = 1 \] Solving this system, we get: \[ a + 2b = -5 \] \[ a - b = 1 \] Adding the two equations, we get: \[ 2a + b = -4 \] \[ a = -\frac{5}{2} \] Substituting $ a $ back into the second equation: \[ -\frac{5}{2} - b = 1 \] \[ b = -\frac{7}{2} \] So, $ w $ can be written as a linear combination of $ u $ and $ v $ with $ a = -\frac{5}{2} $ and $ b = -\frac{7}{2} $: \[ w = -\frac{5}{2}u - \frac{7}{2}v \] (b) Are the vectors $ u $, $ v $, and $ w $ linearly independent? Vectors are linearly independent if the only solution to the equation $ \alpha u + \beta v + \gamma w = 0 $ is $ \alpha = \beta = \gamma = 0 $. Since we were able to express $ w $ as a linear combination of $ u $ and $ v $, this means that there exists a non-trivial solution to the equation $ \alpha u + \beta v + \gamma w = 0 $ (specifically, $ \alpha = -\frac{5}{2} $, $ \beta = -\frac{7}{2} $, and $ \gamma = 1 $). Therefore, $ u $, $ v $, and $ w $ are not linearly independent. (c) Show that span($ \{u, v, w\} $) = $ \mathbb{R}^2 $ by showing that an arbitrary vector in $ \mathbb{R}^2 $ can always be expressed as a linear combination of $ u $, $ v $, $ w $. Since $ u $ and $ v $ are linearly independent (you can verify this by checking that the matrix formed by $ u $ and $ v $ has a non-zero determinant), any vector in $ \mathbb{R}^2 $ can be expressed as a linear combination of $ u $ and $ v $. Therefore, the span of $ \{u, v\} $ is $ \mathbb{R}^2 $, and since $ w $ is also a linear combination of $ u $ and $ v $, the span of $ \{u, v, w\} $ is still $ \mathbb{R}^2 $. (d) What is the dimension of the vector space span($ \{u, v, w\} $)? The dimension of a vector space is the number of vectors in its basis. Since $ u $ and $ v $ are linearly independent and span $ \mathbb{R}^2 $, they form a basis for $ \mathbb{R}^2 $. The vector $ w $ does not add any new dimension because it is a linear combination of $ u $ and $ v $. Therefore, the dimension of the vector space span($ \{u, v, w\} $) is 2, which is the same as the dimension of $ \mathbb{R}^2 $.

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Let's solve the given homework question step by step: (a) Write the vector $w$ as a linear combination of $u$ and $v$ . Given vectors: $u = \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \quad v = \begin{pmatrix} 2 \\ -1 \end{pmatrix}, \quad w = \begin{pmatrix} -5 \\ 1 \end{pmatrix}$ We want to find scalars $a$ and $b$ such that $w = au + bv$ . This gives us the following system of equations: $a(1) + b(2) = -5$ $a(1) + b(-1) = 1$ Solving this system, we get: $a + 2b = -5$ $a - b = 1$ Adding the two equations, we get: $2a + b = -4$ $a = -\frac{5}{2}$ Substituting $a$ back into the second equation: $-\frac{5}{2} - b = 1$ $b = -\frac{7}{2}$ So, $w$ can be written as a linear combination of $u$ and $v$ with $a = -\frac{5}{2}$ and $b = -\frac{7}{2}$ : $w = -\frac{5}{2}u - \frac{7}{2}v$ (b) Are the vectors $u$ , $v$ , and $w$ linearly independent? Vectors are linearly independent if the only solution to the equation $\alpha u + \beta v + \gamma w = 0$ is $\alpha = \beta = \gamma = 0$ . Since we were able to express $w$ as a linear combination of $u$ and $v$ , this means that there exists a non-trivial solution to the equation $\alpha u + \beta v + \gamma w = 0$ (specifically, $\alpha = -\frac{5}{2}$ , $\beta = -\frac{7}{2}$ , and $\gamma = 1$ ). Therefore, $u$ , $v$ , and $w$ are not linearly independent. (c) Show that span( $\{u, v, w\}$ ) = $\mathbb{R}^2$ by showing that an arbitrary vector in $\mathbb{R}^2$ can always be expressed as a linear combination of $u$ , $v$ , $w$ . Since $u$ and $v$ are linearly independent (you can verify this by checking that the matrix formed by $u$ and $v$ has a non-zero determinant), any vector in $\mathbb{R}^2$ can be expressed as a linear combination of $u$ and $v$ . Therefore, the span of $\{u, v\}$ is $\mathbb{R}^2$ , and since $w$ is also a linear combination of $u$ and $v$ , the span of $\{u, v, w\}$ is still $\mathbb{R}^2$ . (d) What is the dimension of the vector space span( $\{u, v, w\}$ )? The dimension of a vector space is the number of vectors in its basis. Since $u$ and $v$ are linearly independent and span $\mathbb{R}^2$ , they form a basis for $\mathbb{R}^2$ . The vector $w$ does not add any new dimension because it is a linear combination of $u$ and $v$ . Therefore, the dimension of the vector space span( $\{u, v, w\}$ ) is 2, which is the same as the dimension of $\mathbb{R}^2$ .

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