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Suppose A is a \(m \times n\) matrix with the property that for all b in \({\mathbb{R}^m}\) the equation \(A{\bf{x}} = {\bf{b}}\) has at most one solution. Use the definition of linear independence to explain why the columns of A must be linearly independent.

Short Answer

Expert verified

Theequation\(A{\bf{x}} = {\bf{b}}\)has at most one solution, and thecolumns of A are linearly independent.

Step by step solution

01

The condition for the linear independence of vectors

The vectors are said to be linearly independent if the equation \({x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} + {x_3}{{\bf{v}}_3} + ... + {x_p}{{\bf{v}}_p} = 0\) has a trivial solution, where \({{\bf{v}}_1},{{\bf{v}}_2},...,{{\bf{v}}_p}\) are the vectors.

02

The columns of A must be linearly independent

For the matrix equation\(A{\bf{x}} = {\bf{b}}\), the equation has at most one solution when\({\bf{b}} = 0\). Thus, the equation\(A{\bf{x}} = 0\)\(\left( {{\bf{x}} = 0} \right)\)has at most one solution, which means the solution is trivial. A trivial solution implies that the columns are linearly independent.

Consider\(A{\bf{x}} = 0\), where\(A = \left[ {\begin{array}{*{20}{c}}{{{\bf{a}}_1}}&{{{\bf{a}}_2}}& \cdots &{{{\bf{a}}_n}}\end{array}} \right]\), and\({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\ \vdots \\{{x_n}}\end{array}} \right]\).

Thus, it is written as shown below:

\(\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{{{\bf{a}}_1}}&{{{\bf{a}}_2}}& \cdots &{{{\bf{a}}_n}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\ \vdots \\{{x_n}}\end{array}} \right] = 0\\{x_1}{{\bf{a}}_1} + {x_2}{{\bf{a}}_2} + ... + {x_n}{{\bf{a}}_n} = 0\end{array}\)

For the equation\({x_1}{{\bf{a}}_1} + {x_2}{{\bf{a}}_2} + ... + {x_n}{{\bf{a}}_n} = 0\), the equation has\({x_1} = {x_2} = ... = {x_n} = 0\). So, the columns of A are linearly independent.

Hence, theequation\(A{\bf{x}} = {\bf{b}}\)has at most one solution, and thecolumns of A are linearly independent.

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Most popular questions from this chapter

Let \(A = \left[ {\begin{array}{*{20}{c}}1&0&{ - 4}\\0&3&{ - 2}\\{ - 2}&6&3\end{array}} \right]\) and \(b = \left[ {\begin{array}{*{20}{c}}4\\1\\{ - 4}\end{array}} \right]\). Denote the columns of \(A\) by \({{\mathop{\rm a}\nolimits} _1},{a_2},{a_3}\) and let \(W = {\mathop{\rm Span}\nolimits} \left\{ {{a_1},{a_2},{a_3}} \right\}\).

  1. Is \(b\) in \(\left\{ {{a_1},{a_2},{a_3}} \right\}\)? How many vectors are in \(\left\{ {{a_1},{a_2},{a_3}} \right\}\)?
  2. Is \(b\) in \(W\)? How many vectors are in W.
  3. Show that \({a_1}\) is in W.[Hint: Row operations are unnecessary.]

In Exercises 15 and 16, list five vectors in Span \(\left\{ {{v_1},{v_2}} \right\}\). For each vector, show the weights on \({{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _2}\) used to generate the vector and list the three entries of the vector. Do not make a sketch.

16. \({{\mathop{\rm v}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}3\\0\\2\end{array}} \right],{v_2} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\0\\3\end{array}} \right]\)


Consider two vectors v1 andv2in R3 that are not parallel.

Which vectors inlocalid="1668167992227" 3are linear combinations ofv1andv2? Describe the set of these vectors geometrically. Include a sketch in your answer.

In Exercise 1, compute \(u + v\) and \(u - 2v\).

  1. \(u = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]\), \(v = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\).

Find the elementary row operation that transforms the first matrix into the second, and then find the reverse row operation that transforms the second matrix into the first.

30.\(\left[ {\begin{array}{*{20}{c}}1&3&{ - 4}\\0&{ - 2}&6\\0&{ - 5}&9\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}1&3&{ - 4}\\0&1&{ - 3}\\0&{ - 5}&9\end{array}} \right]\)

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