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How many pivot columns must a \(7 \times 5\) matrix have if its columns are linearly independent? why?

Short Answer

Expert verified

The matrix must have five pivot columns so that the columns are linearly independent.

Step by step solution

01

Determine the pivot columns of the \(7 \times 5\) matrix

If the columns of a \(7 \times 5\) matrix are linearly independent, the matrix must have five pivot columns.

02

Describe why the \(7 \times 5\) matrix has five pivot columns

The column of matrix \(A\) islinearly independent if and only if the equation \(Ax = 0\) has only a trivial solution.

The matrix must contain five pivot columns if the columns are linearly independent.

It is because the equation \(Ax = 0\) must have only a trivial solution. Alternatively, the equation \(Ax = 0\) can have a free variable for the columns of \(A\) to be linearly dependent.

Thus, the \(7 \times 5\) matrix must have five pivot columns.

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Most popular questions from this chapter

In Exercises 3 and 4, display the following vectors using arrows

on an \(xy\)-graph: u, v, \( - {\bf{v}}\), \( - 2{\bf{v}}\), u + v , u - v, and u - 2v. Notice that u - vis the vertex of a parallelogram whose other vertices are u, 0, and \( - {\bf{v}}\).

4. u and v as in Exercise 2

Consider the problem of determining whether the following system of equations is consistent for all \({b_1},{b_2},{b_3}\):

\(\begin{aligned}{c}{\bf{2}}{x_1} - {\bf{4}}{x_2} - {\bf{2}}{x_3} = {b_1}\\ - {\bf{5}}{x_1} + {x_2} + {x_3} = {b_2}\\{\bf{7}}{x_1} - {\bf{5}}{x_2} - {\bf{3}}{x_3} = {b_3}\end{aligned}\)

  1. Define appropriate vectors, and restate the problem in terms of Span \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\). Then solve that problem.
  1. Define an appropriate matrix, and restate the problem using the phrase “columns of A.”
  1. Define an appropriate linear transformation T using the matrix in (b), and restate the problem in terms of T.

Explain why a set \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\) in \({\mathbb{R}^5}\) must be linearly independent when \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\) is linearly independent and \({{\mathop{\rm v}\nolimits} _4}\) is not in Span \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\).

Use the accompanying figure to write each vector listed in Exercises 7 and 8 as a linear combination of u and v. Is every vector in \({\mathbb{R}^2}\) a linear combination of u and v?

8.Vectors w, x, y, and z

Give a geometric description of span \(\left\{ {{v_1},{v_2}} \right\}\) for the vectors \({{\mathop{\rm v}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}8\\2\\{ - 6}\end{array}} \right]\) and \({{\mathop{\rm v}\nolimits} _2} = \left[ {\begin{array}{*{20}{c}}{12}\\3\\{ - 9}\end{array}} \right]\).

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