4.6.E: Exercise for Section 4.5

Exercise \(\PageIndex{1}\)

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \[\left[\begin{array}{r}1\\3\\-1\\1\end{array}\right],\:\left[\begin{array}{r}1\\4\\-1\\1\end{array}\right],\:\left[\begin{array}{r}1\\4\\0\\1\end{array}\right],\:\left[\begin{array}{r}1\\10\\2\\1\end{array}\right]\nonumber\]

Answer

The given vectors are linearly dependent. We express one of them as a linear combination of the others: \[ \vec{v}_4 = -2\vec{v}_1 + 3\vec{v}_2 + 1\vec{v}_3. \nonumber\] A linearly independent set that spans the same space is: \[ \left\{ \begin{bmatrix} 1 \\ 3 \\ -1 \\ 1 \end{bmatrix},\: \begin{bmatrix} 1 \\ 4 \\ -1 \\ 1 \end{bmatrix},\: \begin{bmatrix} 1 \\ 4 \\ 0 \\ 1 \end{bmatrix} \right\}. \nonumber\]

Exercise \(\PageIndex{2}\)

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \[\left[\begin{array}{r}-1\\-2\\2\\3\end{array}\right],\:\left[\begin{array}{r}-3\\-4\\3\\3\end{array}\right],\:\left[\begin{array}{r}0\\-1\\4\\3\end{array}\right],\:\left[\begin{array}{r}0\\-1\\6\\4\end{array}\right]\nonumber\]

Exercise \(\PageIndex{3}\)

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \[\left[\begin{array}{r}1\\5\\-2\\1\end{array}\right],\:\left[\begin{array}{r}1\\6\\-3\\1\end{array}\right],\:\left[\begin{array}{r}-1\\-4\\1\\-1\end{array}\right],\:\left[\begin{array}{r}1\\6\\-2\\1\end{array}\right]\nonumber\]

Exercise \(\PageIndex{4}\)

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \[\left[\begin{array}{r}1\\-1\\3\\1\end{array}\right],\:\left[\begin{array}{r}1\\6\\34\\1\end{array}\right],\:\left[\begin{array}{r}1\\0\\7\\1\end{array}\right],\:\left[\begin{array}{r}1\\0\\8\\1\end{array}\right]\nonumber\]

Answer

The given vectors are linearly dependent. We express one of them as a linear combination of the others: \[ \vec{v}_4 = -\frac{13.43}{11.43} \vec{v}_3 + \text{other terms}. \nonumber\] A linearly independent set that spans the same space is: \[ \left\{ \begin{bmatrix} 1 \\ -1 \\ 3 \\ 1 \end{bmatrix},\: \begin{bmatrix} 1 \\ 6 \\ 34 \\ 1 \end{bmatrix},\: \begin{bmatrix} 1 \\ 0 \\ 7 \\ 1 \end{bmatrix} \right\}. \nonumber\]

Exercise \(\PageIndex{5}\)

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. \[\left[\begin{array}{r}1\\3\\-1\\1\end{array}\right],\:\left[\begin{array}{r}1\\4\\-1\\1\end{array}\right],\:\left[\begin{array}{r}-3\\-10\\3\\-3\end{array}\right],\:\left[\begin{array}{r}1\\4\\0\\1\end{array}\right]\nonumber\]

Exercise \(\PageIndex{6}\)

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \[\left[\begin{array}{r}1\\3\\-3\\1\end{array}\right],\:\left[\begin{array}{r}1\\4\\-5\\1\end{array}\right],\:\left[\begin{array}{r}1\\4\\-4\\1\end{array}\right],\:\left[\begin{array}{r}1\\10\\-14\\1\end{array}\right]\nonumber\]

Exercise \(\PageIndex{7}\)

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \[\left[\begin{array}{r}1\\0\\3\\1\end{array}\right],\:\left[\begin{array}{r}1\\1\\8\\1\end{array}\right],\:\left[\begin{array}{r}1\\7\\34\\1\end{array}\right],\:\left[\begin{array}{r}1\\1\\7\\1\end{array}\right]\nonumber\]

Exercise \(\PageIndex{8}\)

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \[\left[\begin{array}{r}1\\4\\-2\\1\end{array}\right],\:\left[\begin{array}{r}1\\5\\-3\\1\end{array}\right],\:\left[\begin{array}{r}1\\7\\-5\\1\end{array}\right],\:\left[\begin{array}{r}1\\5\\-2\\1\end{array}\right]\nonumber\]

Exercise \(\PageIndex{9}\)

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. \[\left[\begin{array}{r}1\\2\\2\\-4\end{array}\right],\:\left[\begin{array}{r}3\\4\\1\\-4\end{array}\right],\:\left[\begin{array}{r}0\\-1\\0\\4\end{array}\right],\:\left[\begin{array}{r}0\\-1\\-2\\5\end{array}\right]\nonumber\]

Answer

The given vectors are linearly independent because the only solution to the homogeneous system

\[ a \begin{bmatrix} 1 \\ 2 \\ 2 \\ -4 \end{bmatrix} + b \begin{bmatrix} 3 \\ 4 \\ 1 \\ -4 \end{bmatrix} + c \begin{bmatrix} 0 \\ -1 \\ 0 \\ 4 \end{bmatrix} + d \begin{bmatrix} 0 \\ -1 \\ -2 \\ 5 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \nonumber\]

is the trivial solution \( a = b = c = d = 0 \). Therefore, the vectors are linearly independent.

Exercise \(\PageIndex{10}\)

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \[\left[\begin{array}{r}2\\3\\1\\-3\end{array}\right],\:\left[\begin{array}{r}-5\\-6\\0\\3\end{array}\right],\:\left[\begin{array}{r}-1\\-2\\1\\3\end{array}\right],\:\left[\begin{array}{r}-1\\-2\\0\\4\end{array}\right]\nonumber\]

Exercise \(\PageIndex{11}\)

Suppose \( S = \{ \vec{v}_1, \vec{v}_2, \dots, \vec{v}_n \} \) is a linearly dependent set of vectors in \( \mathbb{R}^n \). Prove that at least one of these vectors can be written as a linear combination of the others.

Answer

We are given that the set \( S = \{ \vec{v}_1, \vec{v}_2, \dots, \vec{v}_n \} \) is \textbf{linearly dependent} in \( \mathbb{R}^n \). By definition, this means that there exist scalars \( a_1, a_2, \dots, a_n \), not all zero, such that:

\[a_1 \vec{v}_1 + a_2 \vec{v}_2 + \dots + a_n \vec{v}_n = \vec{0}. \nonumber \]

Since not all of the \( a_i \) are zero, at least one of them is nonzero. Without loss of generality, assume that \( a_k \neq 0 \) for some \( k \), meaning the term \( a_k \vec{v}_k \) is present in the equation. We can solve for \( \vec{v}_k \) in terms of the other vectors:

\[a_k \vec{v}_k = - (a_1 \vec{v}_1 + a_2 \vec{v}_2 + \dots + a_{k-1} \vec{v}_{k-1} + a_{k+1} \vec{v}_{k+1} + \dots + a_n \vec{v}_n). \nonumber\]

Since \( a_k \neq 0 \), we can divide by \( a_k \) to express \( \vec{v}_k \) as a linear combination of the other vectors:

\[\vec{v}_k = -\frac{a_1}{a_k} \vec{v}_1 - \frac{a_2}{a_k} \vec{v}_2 - \dots - \frac{a_{k-1}}{a_k} \vec{v}_{k-1} - \frac{a_{k+1}}{a_k} \vec{v}_{k+1} - \dots - \frac{a_n}{a_k} \vec{v}_n.\nonumber\]

Thus, we have shown that at least one vector in \( S \) can be written as a linear combination of the others.

Exercise \(\PageIndex{12}\)

Let \( B = \{ \vec{b}_1, \vec{b}_2, \dots, \vec{b}_n \} \) be a linearly independent set. Prove that every vector in \(\text{span}(B)\) can be written uniquely as a linear combination of the vectors in \(B\).

Exercise \(\PageIndex{13}\)

Prove that any set containing the zero vector is a linearly dependent set.

Answer

Let \( S = \{ \vec{v}_1, \vec{v}_2, \dots, \vec{v}_k, \vec{0} \} \) be a set of vectors in \( \mathbb{R}^n \) that contains the zero vector \( \vec{0} \). By the definition of linear dependence, we need to check if there exists a nontrivial linear combination of the vectors in \( S \) that equals the zero vector: \[ a_1 \vec{v}_1 + a_2 \vec{v}_2 + \dots + a_k \vec{v}_k + a_{k+1} \vec{0} = \vec{0} \nonumber\] for some scalars \( a_1, a_2, \dots, a_k, a_{k+1} \), not all zero. Consider setting \( a_{k+1} = 1 \) and all other coefficients \( a_1, a_2, \dots, a_k \) equal to zero: \[ 0 \cdot \vec{v}_1 + 0 \cdot \vec{v}_2 + \dots + 0 \cdot \vec{v}_k + 1 \cdot \vec{0} = \vec{0}. \nonumber\] Since this is a nontrivial linear combination (not all coefficients are zero), the set \( S \) is linearly dependent.

Exercise \(\PageIndex{14}\)

Suppose \( \vec{v}_1, \vec{v}_2, \vec{v}_3 \) are linearly independent vectors in \( \mathbb{R}^3 \). Prove that \( \vec{v}_1 + \vec{v}_2, \vec{v}_2 + \vec{v}_3, \vec{v}_3 + \vec{v}_1 \) are also linearly independent

Exercise \(\PageIndex{15}\)

Let \( A \) be an \( n \times n \) matrix. Prove that the column vectors of \( A \) are linearly independent if and only if \( \det(A) \neq 0 \).

Hint

Since this is an "if and only if" proof, you must prove each direction.

Let \( A \) be an \( n \times n \) matrix with column vectors \( \vec{a}_1, \vec{a}_2, \dots, \vec{a}_n \). We need to prove that the column vectors of \( A \) are linearly independent if and only if \( \det(A) \neq 0 \).

(⇐) Suppose \( \det(A) \neq 0 \). [We must show that the column vectors of \( A \) are linearly independent.] Consider the homogeneous system of equations: \[ A \vec x = \vec 0 \nonumber\] where \( \vec x = \langle x_1, x_2, \dots, x_n \rangle^T \) is an \( n \times 1 \) vector of unknowns. (What does that fact that \(\det(A) = 0\) say about the solution to \(A \vec x = \vec 0\)? Finish the argument.)

(⇒) Suppose the column vectors of \( A \) are linearly independent. [We must show that \( \det(A) \neq 0 \).] If the column vectors of \( A \) are linearly independent, then the only solution to the homogeneous system \( A \vec x = \vec 0 \) is the trivial solution \( \vec x= \vec 0 \) since each column of \(A\) is a pivot column. (Finish the proof from here.)

Exercise \(\PageIndex{16}\)

Suppose the set \( B = \{ \vec{b}_1, \vec{b}_2, \dots, \vec{b}_n \} \) of vectors spans \( \mathbb{R}^n \). Suppose \( \vec{v} \) is any vector in \( \mathbb{R}^n \). Prove that \( B \cup \{ \vec{v} \} \) is a linearly dependent set

Exercise \(\PageIndex{17}\)

Suppose \( S = \{ \vec{v}_1, \vec{v}_2, \dots, \vec{v}_k \} \) is a linearly independent set of vectors. Prove that any subset of \( S \) is also linearly independent