rom the axioms of a linear space, prove that if −v = −w then v = w.

From the axioms of a linear space, prove that 0v = 0 for each vector v

Let v be a vector and let k be a scalar. From the axioms of a vector space, prove that if kv = 0 theneither k = 0 or v = 0.

Let V=Rn be a vector space and W = {(a1, a2,…, an) such that a2=a12. Check whether W isa subspace of V or not, with respect to usual addition and scalar multiplication.

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}.