Question

The following statements refer to vectors in \({\mathbb{R}^n}\) (or \({\mathbb{R}^m}\)) with the standard inner product. Mark each statement True or False. Justify each answer.

1. The length of every vector is a positive number.
2. A vector v and its negative, \( - {\bf{v}}\), have equal lengths.
3. The distance between u and v is \(\left\| {{\bf{u}} - {\bf{v}}} \right\|\).
4. If \(r\) is any scalar, then \(\left\| {r{\bf{v}}} \right\| = r\left\| {\bf{v}} \right\|\).
5. If two vectors are orthogonal, they are linearly independent.
6. If x is orthogonal, to both u and v, then x must be orthogonal to \({\bf{u}} - {\bf{v}}\).
7. If \({\left\| {{\bf{u}} + {\bf{v}}} \right\|^2} = {\left\| {\bf{u}} \right\|^2} + {\left\| {\bf{v}} \right\|^2},\) then u and v are orthogonal.
8. If \({\left\| {{\bf{u}} - {\bf{v}}} \right\|^2} = {\left\| {\bf{u}} \right\|^2} + {\left\| {\bf{v}} \right\|^2},\) then u and v are orthogonal.
9. The orthogonal projection of y onto u is a scalar multiple of y.
10. If a vector y coincides with its orthogonal projection onto a subspace \(W\), y is in \(W\).
11. The set of all vectors in \({\mathbb{R}^n}\) orthogonal to one fixed vector is a subspace of \({\mathbb{R}^n}\).
12. If \(W\) is a subspace of \({\mathbb{R}^n}\), then \(W\) and \({W^ \bot }\) have no vectors in common.
13. If \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\) is an orthogonal set and if \({c_1},{c_2},\) and \({c_3}\) are scalars, then \(\left\{ {{c_1}{{\bf{v}}_1},{c_2}{{\bf{v}}_2},{c_3}{{\bf{v}}_3}} \right\}\) is an orthogonal set.
14. If a matrix U has orthonormal columns, then \(U{U^T} = I\).
15. A square matrix with orthogonal columns is an orthogonal matrix.
16. If a square matrix has orthonormal columns, then it also has orthonormal rows.
17. If \(W\) is a subspace, then \({\left\| {{{{\mathop{\rm proj}\nolimits} }_W}{\bf{v}}} \right\|^2} + {\left\| {{\bf{v}} - {{{\mathop{\rm proj}\nolimits} }_W}{\bf{v}}} \right\|^2} = {\left\| {\bf{v}} \right\|^2}\).
18. A least-squares solution of \(A{\mathop{\rm x}\nolimits} = b\) is the vector \(A\widehat {\bf{x}}\) in \({\mathop{\rm Col}\nolimits} A\) closest to b, so that \(\left\| {{\mathop{\rm b}\nolimits} - A\widehat {\bf{x}}} \right\| \le \left\| {{\mathop{\rm b}\nolimits} - A{\bf{x}}} \right\|\) for all x.
19. The normal equations for a least-squares solution of \(A{\mathop{\rm x}\nolimits} = b\) are given by \(\widehat {\mathop{\rm x}\nolimits} = {\left( {{A^T}A} \right)^{ - 1}}{A^T}{\mathop{\rm b}\nolimits} \).

Short answer

1. The given statement is false.
2. The given statement is true.
3. The given statement is true.
4. The given statement is false.
5. The given statement is false.
6. The given statement is true.
7. The given statement is true.
8. The given statement is true.
9. The given statement is false.
10. The given statement is true.
11. The given statement is true.
12. The given statement is false.
13. The given statement is true.
14. The given statement is false.
15. The given statement is false.
16. The given statement is true.
17. The given statement is true.
18. The given statement is false.
19. The given statement is false.

Step-by-step solution

Check whether the statement is true or false

The given statement cannot always be true because the zero vector has zero length.

Thus, the given statement (a) is false.

Check whether the statement is true or false

It is known that, \(\left\| {c{\bf{v}}} \right\| = \left| c \right|\left\| {\bf{v}} \right\|\).

For \(c = - 1\), the length is \(\left\| { - {\bf{x}}} \right\| = \left\| {\left( { - 1} \right){\bf{x}}} \right\| = \left\| { - 1} \right\|\left\| {\bf{x}} \right\| = \left\| {\bf{x}} \right\|\).

Thus, the given statement (b) is true.

Check whether the statement is true or false

The distance between \({\bf{u}}\) and \({\bf{v}}\), expressed as, \({\mathop{\rm dist}\nolimits} \left( {{\bf{u}},{\bf{v}}} \right)\) is equal to the length of the vector \({\bf{u}} - {\bf{v}}\). That is, \({\mathop{\rm dist}\nolimits} \left( {{\bf{u}},{\bf{v}}} \right) = \left\| {{\bf{u}} - {\bf{v}}} \right\|\).

Thus, the given statement (c) is true.

Check whether the statement is true or false 

The equation \(r\left\| {\bf{v}} \right\|\) would hold when \(r\left\| {\bf{v}} \right\|\) was replaced by \(\left| r \right|\left\| {\bf{v}} \right\|\).

Thus, the given statement (d) is false.

Check whether the statement is true or false 

Orthogonal nonzero vectors are known as the linearly independent.

Thus, the given statement (e) is false.

Check whether the statement is true or false

According to the statement, \({\bf{x}} \cdot {\bf{u}} = 0\) and \({\bf{x}} \cdot {\bf{v}} = 0\). So, \({\bf{x}} \cdot \left( {{\bf{u}} - {\bf{v}}} \right) = {\bf{x}} \cdot {\bf{u}} - {\bf{x}} \cdot {\bf{v}} = 0\).

Thus, the given statement (f) is true.

Check whether the statement is true or false

The Pythagoras theoremstates that the vectors \({\bf{u}}\) and \({\bf{v}}\) are said to beorthogonal such that if \({\left\| {{\bf{u}} + {\bf{v}}} \right\|^2} = {\left\| {\bf{u}} \right\|^2} + {\left\| {\bf{v}} \right\|^2}\).

There is an “only if” component of the Pythagoras theorem.

Thus, the given statement (g) is true.

Check whether the statement is true or false

The Pythagoras theoremstate that the vectors \({\bf{u}}\) and \({\bf{v}}\) are said to beorthogonal such that if \({\left\| {{\bf{u}} + {\bf{v}}} \right\|^2} = {\left\| {\bf{u}} \right\|^2} + {\left\| {\bf{v}} \right\|^2}\).

This is the “only if” component of the Pythagoras theorem when \({\bf{v}}\) is substituted by \( - {\bf{v}}\) , since \({\left\| { - {\bf{v}}} \right\|^2}\) would be the same as \(\left\| {\bf{v}} \right\|\).

Thus, the given statement (h) is true.

Check whether the statement is true or false 

The scalar multiple of \({\bf{u}}\), but not \({\bf{y}}\) is called the orthogonal projection of \({\bf{y}}\) onto \({\bf{u}}\) (except if y is multiple of \({\bf{u}}\)).

Thus, the given statement (i) is false.

Check whether the statement is true or false 

For any vector y, the orthogonal projection onto W is a vector in W always.

Thus, the given statement (j) is true.

Check whether the statement is true or false 

The special case in section 6.1 states that

1. A vector \({\bf{x}}\) is in \({W^ \bot }\) such that if \({\bf{x}}\) is orthogonal to any vector in a set that spans \(W\).

2. \({\mathbb{R}^n}\) has a subspace \({W^ \bot }\).

Thus, the given statement (k) is true.

Check whether the statement is true or false

There is a zero vector in both \(W\) and \({W^ \bot }\).

Thus, the given statement (l) is false.

Check whether the statement is true or false 

Recall that an orthogonal set of nonzero vectors is as \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2}} \right\}\) and \({c_1},{c_2}\) as a nonzero scalar. Then, \(\left\{ {{c_1}{{\bf{v}}_1},{c_2}{{\bf{v}}_2}} \right\}\) is an orthogonal set. This demonstrates that when the vectors in an orthogonal set are normalized, the set would still be orthogonal.

If \({{\bf{v}}_i} \cdot {{\bf{v}}_j} = 0,\) then \(\left( {{c_i}{{\bf{v}}_i}} \right) \cdot \left( {{c_j}{{\bf{v}}_j}} \right) = {c_i}{c_j}\left( {{{\bf{v}}_i} \cdot {{\bf{v}}_j}} \right) = {c_i}{c_j}\left( 0 \right) = 0\).

Thus, the given statement (m) is true.

Check whether the statement is true or false 

Theorem 10states that when anorthonormal basisfor a subspace \(W\) of \({\mathbb{R}^n}\) is \(\left\{ {{{\bf{u}}_1}, \ldots ,{{\bf{u}}_p}} \right\}\) then;

\({{\mathop{\rm proj}\nolimits} _W}{\bf{y}} = \left( {{\bf{y}}.{{\bf{u}}_1}} \right){{\bf{u}}_1} + \left( {{\bf{y}}.{{\bf{u}}_2}} \right){{\bf{u}}_2} + \cdots + \left( {{\bf{y}}.{{\bf{u}}_p}} \right){{\bf{u}}_p}\) …. (4)

If \(U = \left[ {\begin{array}{*{20}{c}}{{{\bf{u}}_1}}&{{{\bf{u}}_2}}& \cdots &{{{\bf{u}}_p}}\end{array}} \right]\) then;

\({{\mathop{\rm proj}\nolimits} _W}{\bf{y}} = U{U^T}{\bf{y}}\)for every y in \({\mathbb{R}^n}\) … (5)

The given statement holds for a square matrix according to theorem 10.

Thus, the given statement (n) is false.

Check whether the statement is true or false

Asquare invertible matrix\(U\)is an orthogonal matrix as such \({U^{ - 1}} = {U^T}\). Such a square matrix contains orthonormal columns. The square matrix with orthonormal columns is clearly orthogonal.

Thus, the given statement (o) is false.

Check whether the statement is true or false

Recallthat \(U\) as a square matrix with orthonormal columns. Then, \(U\) is invertible.

Recallthat \(U\) as an \(n \times n\) orthogonal matrix. Then, the rows \(U\) produce an orthonormal basis of \({\mathbb{R}^n}\).

It is observed that when \(U\) contains orthonormal columns, then \({U^T}U = I\). When \(U\) is a square matrix then the Invertible Matrix Theorem demonstrates that the columns of \({U^T}\) are orthonormal; In other words, the rows of \(U\) are orthonormal.

Thus, the given statement (p) is true.

Check whether the statement is true or false 

subspace of \({\mathbb{R}^n}\). Then each\({\bf{y}}\) in \({\mathbb{R}^n}\) can be expressed uniquely in the form\({\bf{y}} = \widehat {\bf{y}} + {\bf{z}}\).

With \(\widehat {\bf{y}}\) is in \(W\) and \({\bf{z}}\) is in \({W^ \bot }\). In particular, when \(\left\{ {{{\mathop{\rm u}\nolimits} _1}, \ldots ,{{\mathop{\rm u}\nolimits} _p}} \right\}\) is anorthogonal basis of \(W\), then \(\widehat {\bf{y}} = \frac{{{\mathop{\rm y}\nolimits} \cdot {{\mathop{\rm u}\nolimits} _1}}}{{{{\mathop{\rm u}\nolimits} _1} \cdot {{\mathop{\rm u}\nolimits} _1}}}{{\mathop{\rm u}\nolimits} _1} + \ldots + \frac{{{\mathop{\rm y}\nolimits} \cdot {{\mathop{\rm u}\nolimits} _p}}}{{{{\mathop{\rm u}\nolimits} _p} \cdot {{\mathop{\rm u}\nolimits} _p}}}{{\mathop{\rm u}\nolimits} _p}\), and \({\bf{z}} = {\bf{y}} - \widehat {\bf{y}}\).

The vectors \({{\mathop{\rm proj}\nolimits} _W}{\bf{v}}\) and \({\bf{v}} - {{\mathop{\rm proj}\nolimits} _W}{\bf{v}}\) are orthogonal, according to the Orthogonal Decomposition Theorem. Therefore, the Pythagoras theorem leads to the given equality.

Thus, the given statement (q) is true.

Check whether the statement is true or false 

The vector \(\widehat {\bf{x}}\) (not \(A\widehat {\bf{x}}\)) is called the least-square solution. Therefore the closest point to \({\bf{b}}\) in \({\mathop{\rm Col}\nolimits} A\) is \(A\widehat {\bf{x}}\).

Thus, the given statement (r) is false.

Check whether the statement is true or false 

The equation \(\widehat {\bf{x}} = {\left( {{A^T}A} \right)^{ - 1}}{A^T}{\bf{b}}\) represents the solution of the normal equation but not the matrix form of the normal equations. Moreover, this equation is true if \({A^T}A\) is invertible.

Thus, the given statement (s) is false.