Suppose vectors \({{\bf{b}}_{\bf{1}}}\),….\({{\bf{b}}_p}\) span a subspace W, and let \(\left\{ {{{\bf{a}}_{\bf{1}}},....,{{\bf{a}}_p}} \right\}\) be any set in W containing more than p vectors. Fill in the details of the following argument to show that \(\left\{ {{{\bf{a}}_{\bf{1}}},....,{{\bf{a}}_q}} \right\}\) must be linearly dependent. First, let \(B = \left[ {{{\bf{b}}_{\bf{1}}}\,...\,\,{{\bf{b}}_p}} \right]\) and \(A = \left[ {{{\bf{a}}_{\bf{1}}},....,{{\bf{a}}_q}} \right]\).

a. Explain why for each vector \({{\bf{a}}_j}\), there exist a vector \({{\bf{c}}_j}\) in \({\mathbb{R}^p}\) such that \({{\bf{a}}_j} = B{{\bf{c}}_j}\).

b. Let \(C = \left[ {{{\bf{c}}_{\bf{1}}}\,\,...\,\,{{\bf{c}}_q}} \right]\). Explain why there is a nonzero vector u such that \(C{\bf{u}} = {\bf{0}}\).

c. Use B and C to show that \(A{\bf{u}} = {\bf{0}}\). This shows that the columns of A are linearly dependent.

Step by step solution

01

Find the answer for section (a)

B is the basis for the subspace \(W = {\mathbb{R}^p}\). Hence, for any vector in W, the vector \({{\bf{a}}_j}\) can be expressed as shown below:

\(\begin{array}{c}{{\bf{a}}_j} = {{\bf{c}}_1}{{\bf{a}}_1} + {{\bf{c}}_2}{{\bf{a}}_2} + .... + {{\bf{c}}_p}{{\bf{a}}_p}\\ = \sum\limits_{i = 1}^p {{{\bf{c}}_i}{{\bf{a}}_i}} \\ = B{c_i}\end{array}\)

02

find the answer for section (b)

The given set Chas more vectors than set B. As C represents a linearly dependent set, the equation \(C{\bf{u}} = 0\) has a non-trivial solution.

03

Find the answer for section (c)

Matrix A can be expressed as shown below:

\(A = \left[ {{{\bf{a}}_1}\,\,...\,\,{{\bf{a}}_q}} \right] = \left[ {B{{\bf{c}}_1},\,\,....\,\,B{{\bf{c}}_q}} \right] = BC\)

Also,

\(\begin{array}{c}A{\bf{u}} = BC{\bf{u}}\\A{\bf{u}} = B\left( {C{\bf{u}}} \right)\\A{\bf{u}} = B\left( 0 \right)\\A{\bf{u}} = 0\end{array}\)

So, the columns of A are linearly dependent.

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