To show that the span of vectors \( u \), \( v \), and \( w \) is equal to \( \mathbb{R}^2 \), we need to demonstrate that any arbitrary vector in \( \mathbb{R}^2 \) can be expressed as a linear combination of \( u \), \( v \), and \( w \). Let's consider an arbitrary vector \( x \) in \( \mathbb{R}^2 \): \[ x = \begin{pmatrix} a \\ b \end{pmatrix} \] We need to find scalars \( \alpha \), \( \beta \), and \( \gamma \) such that: \[ x = \alpha u + \beta v + \gamma w \] Given vectors: \[ u = \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \quad v = \begin{pmatrix} 2 \\ -1 \end{pmatrix}, \quad w = \begin{pmatrix} -5 \\ 1 \end{pmatrix} \] Substituting the given vectors into the equation, we get: \[ \begin{pmatrix} a \\ b \end{pmatrix} = \alpha \begin{pmatrix} 1 \\ 1 \end{pmatrix} + \beta \begin{pmatrix} 2 \\ -1 \end{pmatrix} + \gamma \begin{pmatrix} -5 \\ 1 \end{pmatrix} \] This leads to the following system of equations: \[ \alpha + 2\beta - 5\gamma = a \] \[ \alpha - \beta + \gamma = b \] Since we have two equations and three unknowns, we can express one of the unknowns, say \( \gamma \), in terms of \( \alpha \) and \( \beta \), and then solve for \( \alpha \) and \( \beta \) to express any vector \( x \) as a linear combination of \( u \), \( v \), and \( w \). However, we have already established that \( w \) can be written as a linear combination of \( u \) and \( v \) (from the previous answer, \( w = -1u - 2v \)). This means that any linear combination of \( u \), \( v \), and \( w \) can be reduced to a linear combination of just \( u \) and \( v \). Since \( u \) and \( v \) are linearly independent, they span \( \mathbb{R}^2 \), and thus the span of \( u \), \( v \), and \( w \) is also \( \mathbb{R}^2 \). In conclusion, the span of \( u \), \( v \), and \( w \) is \( \mathbb{R}^2 \) because \( u \) and \( v \) are sufficient to express any vector in \( \mathbb{R}^2 \), and \( w \) does not add any new dimension to the span.

Let w1, w2, w3, u and v be vectors in Rn. Suppose the vectors u and vare in Span{w1, w2, w3}. Show that u + v is also in Span{w1, w2, w3}.

Suppose {v1,v2,U3, v4} is a linearly dependent spanning set for a vector space V. Show that each w in V can be expressed in more than one way)as a linear combination of v1,...,v4.

To write the vector \( w \) as a linear combination of vectors \( u \) and \( v \), we need to find scalars \( a \) and \( b \) such that \( w = au + bv \). 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 solve the following system of equations for \( a \) and \( b \): \[ a(1) + b(2) = -5 \] \[ a(1) + b(-1) = 1 \] This can be written as two simultaneous equations: \[ a + 2b = -5 \] \[ a - b = 1 \] Solving for \( a \) and \( b \), we can subtract the second equation from the first to eliminate \( a \): \[ (a + 2b) - (a - b) = -5 - 1 \] \[ 3b = -6 \] \[ b = -2 \] Now we can substitute \( b \) into the second equation to find \( a \): \[ a - (-2) = 1 \] \[ a + 2 = 1 \] \[ a = -1 \] Therefore, the vector \( w \) can be written as a linear combination of \( u \) and \( v \) as follows: \[ w = -1u - 2v \] \[ w = -1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} - 2 \begin{pmatrix} 2 \\ -1 \end{pmatrix} \] \[ w = \begin{pmatrix} -1 \\ -1 \end{pmatrix} + \begin{pmatrix} -4 \\ 2 \end{pmatrix} \] \[ w = \begin{pmatrix} -5 \\ 1 \end{pmatrix} \] This confirms that \( w \) is indeed a linear combination of \( u \) and \( v \) with the coefficients \( a = -1 \) and \( b = -2 \).

To determine if the vectors \( u \), \( v \), and \( w \) are linearly dependent, we need to check if there exists a non-trivial linear combination of these vectors that equals the zero vector. In other words, we need to find scalars \( a \), \( b \), and \( c \) (not all zero) such that: \[ a \cdot u + b \cdot v + c \cdot w = 0 \] 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 have already found that \( w \) can be written as a linear combination of \( u \) and \( v \) with specific scalars \( a = -1 \) and \( b = -2 \): \[ w = -1 \cdot u - 2 \cdot v \] This equation shows that there is a non-trivial linear combination of \( u \), \( v \), and \( w \) that equals the zero vector: \[ -1 \cdot u - 2 \cdot v + 1 \cdot w = 0 \] Since we have found a non-trivial solution where the scalars are not all zero, the vectors \( u \), \( v \), and \( w \) are linearly dependent. The fact that \( w \) can be expressed as a linear combination of \( u \) and \( v \) is sufficient to establish their linear dependence.

Let be U = Span{[3,0,1]} a subspace of W=Span{[1,0,0],[1,0,1]}, and the space V the orthogonal complement of U. Answer these two following questions:1. Give a basis v=[ , , ] for V,2. Give dim V = .(If your result is not an integer, then please give your result with 2 decimals (rounded - not cut-off; use a dot as a decimal separator))