Question

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–Schmidt 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 \(R = {Q^T}A\).

Short answer

a. False, because this statement is true when \(c \ne 0\).

b. True, because \({\rm{span}}\left\{ {{x_1}, \ldots ,{x_p}} \right\} = {\rm{span}}\left\{ {{v_1}, \ldots ,{v_p}} \right\}\).

c. True, by using the definition of \(QR\) factorization of a matrix.

Step-by-step solution

\(QR\) factorization of a Matrix

A matrix which has order \(m \times n\) can be written as the multiplication of a upper triangular matrix \(R\) and a matrix \(Q\) which is formed by applying Gram–Schmidt orthogonalization process to the \({\rm{col}}\left( A \right)\).

The matrix \(R\) can be found by the formula \({Q^T}A = R\).

Checking whether the given statements are true of false

a.

This is false. Because this statement is true when \(c \ne 0\).

If \(c = 0\), then we will have \(\left\{ {{v_1},{v_2},0} \right\}\). We know that any set containing zero vector is not a linearly independent set.

b.

The given statement is true. From the Theorem 11, since \(\left\{ {{x_1}, \ldots ,{x_p}} \right\}\)gives the set of orthogonal vectors \(\left\{ {{v_1}, \ldots ,{v_p}} \right\}\), thus we have

\({\rm{span}}\left\{ {{x_1}, \ldots ,{x_p}} \right\} = {\rm{span}}\left\{ {{v_1}, \ldots ,{v_p}} \right\}\).

c.

The statement is true. From the definition of \(QR\) factorization of a matrix it is true.

We have \({Q^T}Q = I\), since the columns of the matrix \(Q\) are orthonormal.

\(\begin{aligned}{}A &= QR\\{Q^T}A &= {Q^T}QR\\{Q^T}A &= IR\\{Q^T}A &= R\end{aligned}\)

Thus, the statement is true.