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Chapter 4: Q20Q (page 191)

Suppose is a linearly dependent spanning set for a \(\left\{ {{{\mathop{\rm v}\nolimits} _1},...,{{\mathop{\rm v}\nolimits} _4}} \right\}\)vector space V. Show that each w in \(V\) can be expressed in more than one way as a linear combination of \({{\mathop{\rm v}\nolimits} _1},...,{{\mathop{\rm v}\nolimits} _4}\).(Hint: Let \({\mathop{\rm w}\nolimits} = {k_1}{{\mathop{\rm v}\nolimits} _1} + ... + {k_4}{{\mathop{\rm v}\nolimits} _4}\) be an arbitrary vector in \(V\). Use the linear dependence of \(\left\{ {{{\mathop{\rm v}\nolimits} _1},...,{{\mathop{\rm v}\nolimits} _4}} \right\}\) to produce another representation of w as a linear combination of \({{\mathop{\rm v}\nolimits} _1},...,{{\mathop{\rm v}\nolimits} _4}\).)

Short Answer

Expert verified

It is proved that w in \(V\) can be expressed in more than one way as a linear combination of \({{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3},\) and \({{\mathop{\rm v}\nolimits} _4}\).

Step by step solution

01

Show that each w in \(V\) can be expressed in more than one way as a linear combination

There are scalars \({k_1},{k_2},{k_3},\) and \({k_4}\) for \(w\)in \(V\) such that

\({\mathop{\rm w}\nolimits} = {k_1}{{\mathop{\rm v}\nolimits} _1} + {k_2}{{\mathop{\rm v}\nolimits} _2} + {k_3}{{\mathop{\rm v}\nolimits} _3} + {k_4}{{\mathop{\rm v}\nolimits} _4}\). 鈥�(1)

Since \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\) spans \(V\) and the set is linearly dependent, scalars \({c_1},{c_2},{c_3},\) and \({c_4}\) are not all zero such that

\(0 = {c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} + {c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\). 鈥�(2)

Add equations (1) and (2).

\(\begin{array}{c}{\mathop{\rm w}\nolimits} = {\mathop{\rm w}\nolimits} + 0\\ = \left( {{k_1} + {c_1}} \right){{\mathop{\rm v}\nolimits} _1} + \left( {{k_2} + {c_2}} \right){{\mathop{\rm v}\nolimits} _2} + \left( {{k_3} + {c_3}} \right){{\mathop{\rm v}\nolimits} _3} + \left( {{k_4} + {c_4}} \right){{\mathop{\rm v}\nolimits} _4}\end{array}\) 鈥�(3)

At least one of the weights in equation (3) is different from the corresponding weight in equation (1) since at least one of the \({c_i}\) is non-zero.

Therefore, \({\mathop{\rm w}\nolimits} \) can be expressed in more than one way as a linear combination of \({{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3},\) and \({{\mathop{\rm v}\nolimits} _4}\).

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Most popular questions from this chapter

In Exercise 5, find the coordinate vector \({\left( x \right)_{\rm B}}\) of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

5. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{5}}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{1}}\end{array}} \right)\)

(M) Let \(H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}\) and \(K = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\), where

\({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}5\\3\\8\end{array}} \right),{{\mathop{\rm v}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}1\\3\\4\end{array}} \right),{{\mathop{\rm v}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\\5\end{array}} \right),{{\mathop{\rm v}\nolimits} _4} = \left( {\begin{array}{*{20}{c}}0\\{ - 12}\\{ - 28}\end{array}} \right)\)

Then \(H\) and \(K\) are subspaces of \({\mathbb{R}^3}\). In fact, \(H\) and \(K\) are planes in \({\mathbb{R}^3}\) through the origin, and they intersect in a line through 0. Find a nonzero vector w that generates that line. (Hint: w can be written as \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2}\) and also as \({c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\). To build w, solve the equation \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} = {c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\) for the unknown \({c_j}'{\mathop{\rm s}\nolimits} \).)

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 \(T\).

Question: Determine if the matrix pairs in Exercises 19-22 are controllable.

22. (M) \(A = \left( {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\{ - 1}&{ - 13}&{ - 12.2}&{ - 1.5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\0\\0\\{ - 1}\end{array}} \right)\).

Let S be a maximal linearly independent subset of a vector space V. In other words, S has the property that if a vector not in S is adjoined to S, the new set will no longer be linearly independent. Prove that S must be a basis of V. [Hint: What if S were linearly independent but not a basis of V?]
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