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Chapter 6: Q21E (page 331)

Let \(A\)be an\(m \times n\)matrix whose columns are linearly independent.

  1. Use Exercise 19 to show that \({A^T}A\) is an invertible matrix.
  2. Explain why\(A\)must have at least as many rows as columns.
  3. Determine the rank of \(A\).

Short Answer

Expert verified
  1. It is proved that \({A^T}A\) is an invertible matrix.
  2. Because \(n\) linearly independent columns of \(A\) belongs to \({\mathbb{R}^m}\).
  3. The rank of \(A\) is \(n\).

Step by step solution

01

Rank of a matrix

(a).

Given that \(A\) is an \(m \times n\) matrix whose columns are linearly independent.

So, the above statement implies that \(\dim \left( {{\rm{Col}}\,A} \right) = n\) and \({\rm{rank}}\left( A \right) = n\). Now, it is known that if \(A\) is an \(m \times n\) matrix then there exist a vector \(x\) in \({\mathbb{R}^n}\) satisfies \(Ax = 0\) if and only if \({A^T}Ax = 0\). So, \(\dim \left( {{\rm{Nul}}\left( {{A^T}A} \right)} \right) = 0\).

This implies that the rank of matrix \({A^T}A\) is \(n\). Since, \({A^T}A\) is \(n \times n\) matrix and \({\rm{rank}}\left( {{A^T}A} \right) = n\). So, \({A^T}A\) is invertible.

02

Proof of part (b)

The reason is that the columns of \(A\) form a linearly independent set of vectors in \({\mathbb{R}^m}\). Therefore, the number of rows cannot be less than the number of columns.

Hence, the number of rows is at least the number of columns,

03

Rank of matrix A

(c).

It is known that the dimensions of column space of \(A\)and the rank of \(A\) are equal. Since, the number of columns that are linearly independent is \(n\).

Therefore, the rank of \(A\) is equal to \(n\).

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Most popular questions from this chapter

Find a \(QR\) factorization of the matrix in Exercise 11.

Determine which pairs of vectors in Exercises 15-18 are orthogonal.

15. \({\mathop{\rm a}\nolimits} = \left( {\begin{aligned}{*{20}{c}}8\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm b}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\{ - 3}\end{aligned}} \right)\)

Question: In Exercises 1 and 2, you may assume that\(\left\{ {{{\bf{u}}_{\bf{1}}},...,{{\bf{u}}_{\bf{4}}}} \right\}\)is an orthogonal basis for\({\mathbb{R}^{\bf{4}}}\).

2.\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{2}}}\\{\bf{1}}\\{ - {\bf{1}}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{3}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\\{ - {\bf{1}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{4}}} = \left[ {\begin{aligned}{ - {\bf{1}}}\\{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\end{aligned}} \right]\),\({\bf{x}} = \left[ {\begin{aligned}{\bf{4}}\\{\bf{5}}\\{ - {\bf{3}}}\\{\bf{3}}\end{aligned}} \right]\)

Write v as the sum of two vectors, one in\({\bf{Span}}\left\{ {{{\bf{u}}_1}} \right\}\)and the other in\({\bf{Span}}\left\{ {{{\bf{u}}_2},{{\bf{u}}_3},{{\bf{u}}_{\bf{4}}}} \right\}\).

In Exercises 17 and 18, all vectors and subspaces are in \({\mathbb{R}^n}\). Mark each statement True or False. Justify each answer.

17. a.If \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\) is an orthogonal basis for\(W\), then multiplying

\({v_3}\)by a scalar \(c\) gives a new orthogonal basis \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},c{{\bf{v}}_3}} \right\}\).

b. The Gram鈥揝chmidt process produces from a linearly independent

set \(\left\{ {{{\bf{x}}_1}, \ldots ,{{\bf{x}}_p}} \right\}\)an orthogonal set \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) with the property that for each \(k\), the vectors \({{\bf{v}}_1}, \ldots ,{{\bf{v}}_k}\) span the same subspace as that spanned by \({{\bf{x}}_1}, \ldots ,{{\bf{x}}_k}\).

c. If \(A = QR\), where \(Q\) has orthonormal columns, then \(R = {Q^T}A\).

Find the distance between \({\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{10}\\{ - 3}\end{aligned}} \right)\) and \({\mathop{\rm y}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\{ - 5}\end{aligned}} \right)\).
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