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Exercise 32 reveal an important connection between linear independence and linear transformations and provide practice using the definition of linear dependence. Let V and W be vector spaces, let \(T:V \to W\) be a linear transformation, and let \(\left\{ {{v_{\bf{1}}},...,{v_p}} \right\}\) be a subset of V.

32. Suppose that T is a one-to-one transformation, so that an equation \(T\left( u \right) = T\left( v \right)\) always implies \(u = v\). Show that if the set of images \(\left\{ {T\left( {{v_{\bf{1}}}} \right),...,T\left( {{v_p}} \right)} \right\}\) is linearly dependent, then \(\left\{ {{v_{\bf{1}}},...,{v_p}} \right\}\) is linearly dependent. This fact shows that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set (because in this case the set of images cannot be linearly dependent).

Step by step solution

01

Write the given statement

Suppose \(\left\{ {T\left( {{v_1}} \right),...,T\left( {{v_p}} \right)} \right\}\) is linearly dependent.

02

Use the definition of linear dependence

There exist scalars \({c_1},...,{c_p}\) that are not all zeros, such that

\({c_1}T\left( {{v_1}} \right) + ... + {c_p}T\left( {{v_p}} \right) = 0\).

Since T is a linear transformation,

\(\begin{array}{c}T\left( {{c_1}{v_1} + ... + {c_p}{v_p}} \right) = 0\\ = T\left( 0 \right).\end{array}\)

Since T is one-to-one, \({c_1}{v_1} + ... + {c_p}{v_p} = 0\).

Since not all scalars are zeros.

03

Draw a conclusion

Thus, \(\left\{ {{v_1},...,{v_p}} \right\}\) is linearly dependent.

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