Prove that if W is a subspace of a finite dimensional vector space V, then dim(W) ≤ dim(V).

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

We proved a theorem stating that for a linear mapping α : U → V , we havedim Kerα + dim Imα = dimU.A part of the proof went as follows. Assuming dim Kerα = m, there is a basis of Kerα consisting ofm vectors; denote them by u1, . . . , um. Since dim Imα = n, there is a basis of Imα consisting of nvectors; denote them by v1, . . . , vn. Since v1 is in Imα, there is w1 ∈ U such that α(w1) = v1, andsimilarly for all other vectors in the set v1, . . . , vn; in this way, we define n vectors w1, . . . , wn.Now consider a vector u ∈ U . Since α(u) is in Imα, we have α(u) = d1v1 + · · · + dnvn for somenumbers d1, . . . , dn. Consider a vector u′ ∈ U defined as u′ = d1w1 + · · · + dnwn.Prove that u − u′ ∈ Kerα.

We proved a theorem stating that for a linear mapping α : U → V , we havedim Kerα + dim Imα = dimU.A part of the proof went as follows. Assuming dim Kerα = m, there is a basis of Kerα consisting ofm vectors; denote them by u1, . . . , um. Since dim Imα = n, there is a basis of Imα consisting of nvectors; denote them by v1, . . . , vn. Since v1 is in Imα, there is w1 ∈ U such that α(w1) = v1, andsimilarly for all other vectors in the set v1, . . . , vn; in this way, we define n vectors w1, . . . , wn.Now consider a vector u ∈ U . Since α(u) is in Imα, we have α(u) = d1v1 + · · · + dnvn for somenumbers d1, . . . , dn. Consider a vector u′ ∈ U defined as u′ = d1w1 + · · · + dnwn.Continuing this proof, prove that (u − u′) ∈ Kerα.

Suppose {v,w} is a linearly independent set in the vector space V.Then the set {v,v+w} is also linearly independent.A. TrueB. False

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.