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Chapter 4: Q4.5-31E (page 191)

Exercises 31 and 32 concern finite-dimensional vector spaces V and W and a linear transformation \(T:V \to W\).

Let H be a nonzero subspace of V, and let \(T\left( H \right)\) be the set of images of vectors in H. Then \(T\left( H \right)\) is a subspace of W, by Exercise 35 in section 4.2. Prove that \({\bf{dim}}T\left( H \right) \le {\bf{dim}}\left( H \right)\).

Short Answer

Expert verified

\(\dim T\left( H \right) \le p = \dim H\)

Step by step solution

01

Write the transformation vector in subspace H

Let the set \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},....{{\bf{v}}_p}} \right\}\) be a basis for H,i.e. \(\dim H = p\). For any \({\bf{v}}\) in the subspace H:

\(T\left( {\bf{v}} \right) = T\left( {{c_1}{{\bf{v}}_1} + .... + {c_p}{{\bf{v}}_p}} \right)\)

02

Check for statement (b)

Any vector \(T\left( {\bf{v}} \right) \in T\left( H \right)\) is a linear combination of \(T\left( {{{\bf{v}}_1}} \right)\),….,\(T\left( {{{\bf{v}}_p}} \right)\), i.e.

\(T\left( H \right) = {\rm{span}}\left\{ {T\left( {{{\bf{v}}_1}} \right),.....,T\left( {{{\bf{v}}_p}} \right)} \right\}\)

As \(T\left( {{{\bf{v}}_1}} \right)\),….,\(T\left( {{{\bf{v}}_p}} \right)\) are not linearly independent, \(\dim T\left( H \right) \le p = \dim H\).

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Most popular questions from this chapter

Suppose \({{\bf{p}}_{\bf{1}}}\), \({{\bf{p}}_{\bf{2}}}\), \({{\bf{p}}_{\bf{3}}}\), and \({{\bf{p}}_{\bf{4}}}\) are specific polynomials that span a two-dimensional subspace H of \({P_{\bf{5}}}\). Describe how one can find a basis for H by examining the four polynomials and making almost no computations.

In Exercise 7, find the coordinate vector \({\left( x \right)_{\rm B}}\) of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

7. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{1}}}\\{ - {\bf{3}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{3}}}\\{\bf{4}}\\{\bf{9}}\end{array}} \right),{b_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{2}}}\\{\bf{4}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{\bf{8}}\\{ - {\bf{9}}}\\{\bf{6}}\end{array}} \right)\)

Let H be an n-dimensional subspace of an n-dimensional vector space V. Explain why \(H = V\).

Let \({M_{2 \times 2}}\) be the vector space of all \(2 \times 2\) matrices, and define \(T:{M_{2 \times 2}} \to {M_{2 \times 2}}\) by \(T\left( A \right) = A + {A^T}\), where \(A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\).

  1. Show that \(T\)is a linear transformation.
  2. Let \(B\) be any element of \({M_{2 \times 2}}\) such that \({B^T} = B\). Find an \(A\) in \({M_{2 \times 2}}\) such that \(T\left( A \right) = B\).
  3. Show that the range of \(T\) is the set of \(B\) in \({M_{2 \times 2}}\) with the property that \({B^T} = B\).
  4. Describe the kernel of \(T\).

In Exercise 1, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

1. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{6}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{3}}\end{array}} \right)\)
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