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Chapter 3: Problem 7

Prove that if \(\left(v_{1}, \ldots, v_{n}\right)\) spans \(V\) and \(T \in \mathcal{L}(V, W)\) is surjective, then \(\left(T v_{1}, \ldots, T v_{n}\right)\) spans \(W\).

Short Answer

Expert verified
To prove that \((T(v_1), \ldots, T(v_n))\) spans \(W\), first note that surjectivity of \(T\) implies that for any \(w \in W\), there exists a \(v \in V\) such that \(T(v)=w\). Since \((v_1, \ldots, v_n)\) spans \(V\), any \(v \in V\) can be expressed as a linear combination of them: \(v = c_1v_1 + c_2v_2 + \cdots + c_nv_n\). Applying the linear transformation, \(T(v) = T(c_1v_1 + c_2v_2 + \cdots + c_nv_n)\), and using linearity, we get \(T(v) = c_1T(v_1) + c_2T(v_2) + \cdots + c_nT(v_n)\). Replacing \(T(v)\) with \(w\), we have \(w = c_1T(v_1) + c_2T(v_2) + \cdots + c_nT(v_n)\), showing any vector in \(W\) can be expressed as a linear combination of \((T(v_1), \ldots, T(v_n))\), which means they span \(W\).

Step by step solution

01

Define surjectivity for the linear transformation T

Surjectivity for the linear transformation \(T\) means that for any vector \(w \in W\), there exists some vector \(v \in V\) such that \(T(v) = w\).
02

Express arbitrary vector in V as a linear combination of (v_1,...,v_n)

Since \((v_1, \ldots, v_n)\) spans \(V\), any vector \(v\) in \(V\) can be expressed as a linear combination of these vectors, i.e., \( v = c_1v_1 + c_2v_2 + \cdots + c_nv_n \) for some scalars \(c_1, c_2, \ldots, c_n\).
03

Apply linear transformation T to both sides

Now we apply the linear transformation \(T\) to both sides of the equation: \( T(v) = T(c_1v_1 + c_2v_2 + \cdots + c_nv_n) \)
04

Use linearity of T

Since the transformation \(T\) is linear, we can use the linearity property to distribute it across the sum and scalar multiplications: \( T(v) = c_1T(v_1) + c_2T(v_2) + \cdots + c_nT(v_n) \)
05

Replace T(v) by w

As previously established, \(T\) is surjective, which means that \(T(v) = w\) for some \(v \in V\). We can replace \(T(v)\) by \(w\) in the previous equation: \( w = c_1T(v_1) + c_2T(v_2) + \cdots + c_nT(v_n) \)
06

Show that (T(v_1),...,T(v_n)) spans W

The equation above shows that every vector \(w\) in \(W\) can be expressed as a linear combination of the vectors \((T(v_1), \ldots, T(v_n))\). Therefore, we can conclude that \((T(v_1), \ldots, T(v_n))\) spans \(W\).

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Most popular questions from this chapter

Suppose that \(V\) and \(W\) are finite dimensional and that \(U\) is a subspace of \(V .\) Prove that there exists \(T \in \mathcal{L}(V, W)\) such that null \(T=U\) if and only if \(\operatorname{dim} U \geq \operatorname{dim} V-\operatorname{dim} W\)

Suppose that \(V\) is finite dimensional and \(T \in \mathcal{L}(V) .\) Prove that \(T\) is a scalar multiple of the identity if and only if \(S T=T S\) for every \(S \in \mathcal{L}(V)\)

Give an example of a function \(f: \mathbf{R}^{2} \rightarrow \mathbf{R}\) such that $$f(a v)=a f(v)$$ for all \(a \in \mathbf{R}\) and all \(v \in \mathbf{R}^{2}\) but \(f\) is not linear.

Prove that if \(T\) is a linear map from \(\mathbf{F}^{4}\) to \(\mathbf{F}^{2}\) such that null \(T=\left\\{\left(x_{1}, x_{2}, x_{3}, x_{4}\right) \in \mathbf{F}^{4}: x_{1}=5 x_{2} \text { and } x_{3}=7 x_{4}\right\\}\) then \(T\) is surjective

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