MAT 350: Linear Independence

Dr. Gilbert

October 9, 2025

Warm-Up Problems

Complete the following warm-up problems to re-familiarize yourself with concepts we’ll be leveraging today.

1. Determine whether the vector \(\vec{b} = \begin{bmatrix} -5\\ -2\\ -1\\ 4\end{bmatrix}\) is in \(\text{span}\left(\left\{\begin{bmatrix} 1\\ 1\\ 0\\ 2\end{bmatrix}, \begin{bmatrix} -3\\ 0\\ 1\\ 1\end{bmatrix}, \begin{bmatrix} 0\\ 1\\ 1\\ 0\end{bmatrix}\right\}\right)\)

2. Do the columns of the matrix \(\begin{bmatrix} 1 & 8 & -2 & 3\\ 0 & 0 & -1 & 1\\ 0 & 0 & 0 & 4\\ 0 & 0 & 0 & 0\end{bmatrix}\) span \(\mathbb{R}^4\)? Why or why not?

3. Describe the space spanned by the vectors \(\vec{v_1} = \begin{bmatrix} 1\\ 0\\ -1\end{bmatrix}\), \(\vec{v_2} =\begin{bmatrix} 2\\ 3\\ 0\end{bmatrix}\), and \(\vec{v_3} = \begin{bmatrix} 6\\ 15\\ 4\end{bmatrix}\).

Reminders and Today’s Goal

• Recently we’ve discussed linear combinations of vectors and spans of collections of vectors.

  • A linear combination of vectors \(\vec{v_1}, \vec{v_2}, \dots, \vec{v_k}\) is any vector that can be written as \(\vec{y} = c_1\vec{v_1} + c_2\vec{v_2} + \dots + c_k\vec{v_k}\), where \(c_1, c_2, \dots, c_k\) are scalars.
  • The span of a collection of vectors is the set of all linear combinations of those vectors.

• In our discussion on spans, we considered whether a collection of \(m\)-component vectors spanned all of \(\mathbb{R}^m\)

  • For example, does a set of three vectors from \(\mathbb{R}^3\) span all of \(\mathbb{R}^3\)? When not, we tried to determine the geometry of the space spanned by the vectors.

• It is natural now to start considering whether we can remove any vectors from a collection without changing its span.

Reminders and Today’s Goal

Goals for Today: After today’s discussion, you should be able to…

• Define the notions of linear independence and linear dependence

• Test whether a collection \(\left\{\vec{v_1},~\vec{v_2},~\cdots,~\vec{v_p}\right\}\) is linearly independent

• Explain what happens to the span if you

  • remove a vector from a linearly independent set.
  • remove a vector from a linearly dependent set.

Motivating Linear Independence / Dependence

Consider a collection of non-zero vectors \(V = \left\{\vec{v_1},~\vec{v_2},~\cdots,~\vec{v_p}\right\}\)

• If we assume that one of the vectors can be written as a linear combination of the remaining vectors in \(V\) (and – for convenience – we assume that we’ve ordered the vectors in \(V\) such that it is \(\vec{v_p}\)), then we have: \(c_1\vec{v_1} + c_2\vec{v_2} + \cdots + c_{p-1}\vec{v_{p-1}} = \vec{v_p}\)
• Because of the item above, we can write the zero-vector as a linear combination of the vectors in \(V\): \(c_1\vec{v_1} + c_2\vec{v_2} + \cdots + c_{p-1}\vec{v_{p-1}} - \vec{v_p} = \vec{0}\)
• In fact, there are infinitely many ways to write \(\vec{0}\) as a linear combination of vectors from \(V\) since: \(s\cdot c_1\vec{v_1} + s\cdot c_2\vec{v_2} + \cdots + s\cdot c_{p-1}\vec{v_{p-1}} - s\vec{v_p} = \vec{0}\) for any scalar \(s\)
• There is some redundancy here.

Motivating Linear Independence / Dependence

Consider a collection of non-zero vectors \(V = \left\{\vec{v_1},~\vec{v_2},~\cdots,~\vec{v_p}\right\}\)

• If we assume that no vector can be written as a linear combination of the other vectors in \(V\), then the only way to write \(\vec{0}\) as a linear combination of vectors from \(V\) is to have all weights \(c_i = 0\)

  • If there was another way, then we could write \(c_1\vec{v_1} + c_2\vec{v_2} + \cdots + c_p\vec{v_p} = \vec{0}\) where at least one of the \(c_i\) is non-zero.
  • For convenience, we could assume that we’ve ordered the vectors in \(V\) such that \(c_p\), the weight \(\vec{v_p}\), is non-zero.
  • In this case, \(c_1\vec{v_1} + c_2\vec{v_2} + \cdots + c_{p-1}\vec{v_{p-1}} = -c_p\vec{v_p}\)
  • Which means that \(\frac{-c_1}{c_p}\vec{v_1} + \frac{-c_2}{c_p} + \cdots + \frac{-c_{p-1}}{c_p}\vec{v_{p-1}} = \vec{v_p}\)
  • This contradicts the assumption that \(\vec{v_p}\) is not a linear combination of the other vectors in \(V\).

Definining Linear Independence / Dependence

Definition (Linear Independence): Given a set of vectors \(V = \left\{\vec{v_1},~\vec{v_2},~\cdots,~\vec{v_p}\right\}\), if the only solution to the vector equation \(c_1\vec{v_1} + c_2\vec{v_2} + \cdots + c_p\vec{v_p} = \vec{0}\) is that \(c_1 = c_2 = \cdots = c_p = 0\), then the vectors in \(V\) are linearly independent.

Definition (Linear Dependence): Given a set of vectors \(V = \left\{\vec{v_1},~\vec{v_2},~\cdots,~\vec{v_p}\right\}\), if there exists a solution to the vector equation \(c_1\vec{v_1} + c_2\vec{v_2} + \cdots + c_p\vec{v_p} = \vec{0}\) with at least one \(c_i \neq 0\), then the vectors in \(V\) are linearly dependent.

• Linear Dependence and Vector Equations: A collection of vectors is linearly dependent if the vector equation \(x_1\vec{v_1} + x_2\vec{v_2} + \cdots + x_p\vec{v_p} = \vec{0}\) has infinitely many solutions.

Notice that reducing a linearly dependent collection of vectors to a linearly independent collection does not change its span.

Definining Linear Independence / Dependence

Definition (Linear Independence): Given a set of vectors \(V = \left\{\vec{v_1},~\vec{v_2},~\cdots,~\vec{v_p}\right\}\), if the only solution to the vector equation \(c_1\vec{v_1} + c_2\vec{v_2} + \cdots + c_p\vec{v_p} = \vec{0}\) is that \(c_1 = c_2 = \cdots = c_p = 0\), then the vectors in \(V\) are linearly independent.

Definition (Linear Dependence): Given a set of vectors \(V = \left\{\vec{v_1},~\vec{v_2},~\cdots,~\vec{v_p}\right\}\), if there exists a solution to the vector equation \(c_1\vec{v_1} + c_2\vec{v_2} + \cdots + c_p\vec{v_p} = \vec{0}\) with at least one \(c_i \neq 0\), then the vectors in \(V\) are linearly dependent.

• Linear Dependence and Vector Equations: A collection of vectors is linearly dependent if the vector equation \(x_1\vec{v_1} + x_2\vec{v_2} + \cdots + x_p\vec{v_p} = \vec{0}\) has infinitely many solutions.

Notice that reducing a linearly dependent collection of vectors to a linearly independent collection does not change its span. In some sense, we might think of linearly independent collections as being “efficient”.

Homogeneous Systems

Definition (Homogeneous System/Equation): When the right-hand side / constant vector in a system or equation is the zero vector (ie. \(\vec{b} = \vec{0}\)), then that system or equation is called homogeneous

Homogeneous systems and equations are always consistent, because the zero-vector is always a solution.

\[\left\{\begin{array}{lcl} a_{11}x_1 + a_{12}x_2 + \cdots + a_{1p}x_p & = & 0\\ a_{21}x_1 + a_{22}x_2 + \cdots + a_{2p}x_p & = & 0\\ & \vdots & \\ a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mp}x_p & = & 0 \end{array}\right.\]

Has solution \(x_1 = 0,~x_2 = 0,~\cdots,~x_p = 0\)

Homogeneous Systems

Definition (Homogeneous System/Equation): When the right-hand side / constant vector in a system or equation is the zero vector (ie. \(\vec{b} = \vec{0}\)), then that system or equation is called homogeneous

Homogeneous systems and equations are always consistent, because the zero-vector is always a solution.

\[A\vec{x} = \vec{0} \text{ or } \left[\begin{array}{cccc|c} \vec{a_1} & \vec{a_2} & \cdots & \vec{a_p} & \vec{0}\end{array}\right]\]

Has solution \(\vec{x} = \vec{0}\)

Homogeneous Systems

Definition (Homogeneous System/Equation): When the right-hand side / constant vector in a system or equation is the zero vector (ie. \(\vec{b} = \vec{0}\)), then that system or equation is called homogeneous

Homogeneous systems and equations are always consistent, because the zero-vector is always a solution.

\[x_1\vec{v_1} + x_2\vec{v_2} + \cdots + x_p\vec{v_p} = \vec{0}\]

Has solution \(x_1 = 0,~x_2 = 0,~\cdots,~x_p = 0\)

Homogeneous Systems and Linear Independence

Investigating homogeneous equations provides us with a simple strategy for determining whether a collection of vectors is linearly independent or linearly dependent.

Strategy: We identify whether \(\left\{\vec{v_1},~\vec{v_2},~\cdots,~\vec{v_p}\right\}\) is linearly independent or dependent by determining whether or not the system corresponding to the augmented matrix \(\left[\begin{array}{cccc|c} v_{11} & v_{12} & \cdots & v_{1p} & 0\\ v_{21} & v_{22} & \cdots & v_{2p} & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ v_{m1} & v_{m2} & \cdots & v_{mp} & 0\\ \end{array}\right]\) has a free variable.

Homogeneous Systems and Linear Independence

Strategy: We identify whether \(\left\{\vec{v_1},~\vec{v_2},~\cdots,~\vec{v_p}\right\}\) is linearly independent or dependent by determining whether or not the system corresponding to the augmented matrix \(\left[\begin{array}{cccc|c} v_{11} & v_{12} & \cdots & v_{1p} & 0\\ v_{21} & v_{22} & \cdots & v_{2p} & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ v_{m1} & v_{m2} & \cdots & v_{mp} & 0\\ \end{array}\right]\) has a free variable.

Recall that, since \(\vec{x} = \vec{0}\) is a solution to the homogeneous system, the only way for a non-zero solution to exist is to have infinitely many solutions.

We’ve discovered previously that a system has infinitely many solutions if it has at least one free variable.

Any column of the coefficient matrix which is not a pivot column corresponds to a free variable.

Completed Example #1

Example: Determine whether the vectors \(\vec{v_1} = \begin{bmatrix} 0\\ -1\\ 2\end{bmatrix}\), \(\vec{v_2} = \begin{bmatrix} 3\\ 1\\ -1\end{bmatrix}\), and \(\vec{v_1} = \begin{bmatrix} 2\\ 0\\ 1\end{bmatrix}\) are linearly independent.

\[\begin{align}\left[\begin{array}{rrr|r} 0 & 3 & 2 & 0\\ -1 & 1 & 0 & 0\\ 2 & -1 & 1 & 0\end{array}\right]\end{align}\]

Completed Example #1

Example: Determine whether the vectors \(\vec{v_1} = \begin{bmatrix} 0\\ -1\\ 2\end{bmatrix}\), \(\vec{v_2} = \begin{bmatrix} 3\\ 1\\ -1\end{bmatrix}\), and \(\vec{v_1} = \begin{bmatrix} 2\\ 0\\ 1\end{bmatrix}\) are linearly independent.

\[\begin{align}\left[\begin{array}{rrr|r} 0 & 3 & 2 & 0\\ -1 & 1 & 0 & 0\\ 2 & -1 & 1 & 0\end{array}\right] &\substack{R_1 \leftrightarrow R_2\\ \longrightarrow} \left[\begin{array}{rrr|r} -1 & 1 & 0 & 0\\ 0 & 3 & 2 & 0\\ 2 & -1 & 1 & 0\end{array}\right]\end{align}\]

Completed Example #1

Example: Determine whether the vectors \(\vec{v_1} = \begin{bmatrix} 0\\ -1\\ 2\end{bmatrix}\), \(\vec{v_2} = \begin{bmatrix} 3\\ 1\\ -1\end{bmatrix}\), and \(\vec{v_1} = \begin{bmatrix} 2\\ 0\\ 1\end{bmatrix}\) are linearly independent.

\[\begin{align}\left[\begin{array}{rrr|r} 0 & 3 & 2 & 0\\ -1 & 1 & 0 & 0\\ 2 & -1 & 1 & 0\end{array}\right] &\substack{R_1 \leftrightarrow R_2\\ \longrightarrow} \left[\begin{array}{rrr|r} -1 & 1 & 0 & 0\\ 0 & 3 & 2 & 0\\ 2 & -1 & 1 & 0\end{array}\right]\\ &\substack{R_1 \leftarrow \left(-1\right)R_1\\ \longrightarrow} \left[\begin{array}{rrr|r} 1 & -1 & 0 & 0\\ 0 & 3 & 2 & 0\\ 2 & -1 & 1 & 0\end{array}\right]\end{align}\]

Completed Example #1

Example: Determine whether the vectors \(\vec{v_1} = \begin{bmatrix} 0\\ -1\\ 2\end{bmatrix}\), \(\vec{v_2} = \begin{bmatrix} 3\\ 1\\ -1\end{bmatrix}\), and \(\vec{v_1} = \begin{bmatrix} 2\\ 0\\ 1\end{bmatrix}\) are linearly independent.

\[\begin{align}\left[\begin{array}{rrr|r} 0 & 3 & 2 & 0\\ -1 & 1 & 0 & 0\\ 2 & -1 & 1 & 0\end{array}\right] &\substack{R_1 \leftrightarrow R_2\\ \longrightarrow} \left[\begin{array}{rrr|r} -1 & 1 & 0 & 0\\ 0 & 3 & 2 & 0\\ 2 & -1 & 1 & 0\end{array}\right]\\ &\substack{R_1 \leftarrow \left(-1\right)R_1\\ \longrightarrow} \left[\begin{array}{rrr|r} 1 & -1 & 0 & 0\\ 0 & 3 & 2 & 0\\ 2 & -1 & 1 & 0\end{array}\right]\\ &\substack{R_3 \leftarrow R_3 + \left(-2\right)R_1\\ \longrightarrow} \left[\begin{array}{rrr|r} 1 & -1 & 0 & 0\\ 0 & 3 & 2 & 0\\ 0 & 1 & 1 & 0\end{array}\right]\end{align}\]

Completed Example #1

Example: Determine whether the vectors \(\vec{v_1} = \begin{bmatrix} 0\\ -1\\ 2\end{bmatrix}\), \(\vec{v_2} = \begin{bmatrix} 3\\ 1\\ -1\end{bmatrix}\), and \(\vec{v_1} = \begin{bmatrix} 2\\ 0\\ 1\end{bmatrix}\) are linearly independent.

\[\begin{align}\left[\begin{array}{rrr|r} 0 & 3 & 2 & 0\\ -1 & 1 & 0 & 0\\ 2 & -1 & 1 & 0\end{array}\right] &\sim \left[\begin{array}{rrr|r} 1 & -1 & 0 & 0\\ 0 & 3 & 2 & 0\\ 0 & 1 & 1 & 0\end{array}\right]\end{align}\]

Completed Example #1

Example: Determine whether the vectors \(\vec{v_1} = \begin{bmatrix} 0\\ -1\\ 2\end{bmatrix}\), \(\vec{v_2} = \begin{bmatrix} 3\\ 1\\ -1\end{bmatrix}\), and \(\vec{v_1} = \begin{bmatrix} 2\\ 0\\ 1\end{bmatrix}\) are linearly independent.

\[\begin{align}\left[\begin{array}{rrr|r} 0 & 3 & 2 & 0\\ -1 & 1 & 0 & 0\\ 2 & -1 & 1 & 0\end{array}\right] &\sim \left[\begin{array}{rrr|r} 1 & -1 & 0 & 0\\ 0 & 3 & 2 & 0\\ 0 & 1 & 1 & 0\end{array}\right]\\ &\substack{R_2\leftrightarrow R_3\\ \longrightarrow} \left[\begin{array}{rrr|r} 1 & -1 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 3 & 2 & 0 \end{array}\right]\end{align}\]

Completed Example #1

Example: Determine whether the vectors \(\vec{v_1} = \begin{bmatrix} 0\\ -1\\ 2\end{bmatrix}\), \(\vec{v_2} = \begin{bmatrix} 3\\ 1\\ -1\end{bmatrix}\), and \(\vec{v_1} = \begin{bmatrix} 2\\ 0\\ 1\end{bmatrix}\) are linearly independent.

\[\begin{align}\left[\begin{array}{rrr|r} 0 & 3 & 2 & 0\\ -1 & 1 & 0 & 0\\ 2 & -1 & 1 & 0\end{array}\right] &\sim \left[\begin{array}{rrr|r} 1 & -1 & 0 & 0\\ 0 & 3 & 2 & 0\\ 0 & 1 & 1 & 0\end{array}\right]\\ &\substack{R_2\leftrightarrow R_3\\ \longrightarrow} \left[\begin{array}{rrr|r} 1 & -1 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 3 & 2 & 0 \end{array}\right]\\ &\substack{R_3\leftarrow R_3 + \left(-3\right)R_2\\ \longrightarrow} \left[\begin{array}{rrr|r} 1 & -1 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 0 & -1 & 0 \end{array}\right]\end{align}\]

Completed Example #1

Example: Determine whether the vectors \(\vec{v_1} = \begin{bmatrix} 0\\ -1\\ 2\end{bmatrix}\), \(\vec{v_2} = \begin{bmatrix} 3\\ 1\\ -1\end{bmatrix}\), and \(\vec{v_1} = \begin{bmatrix} 2\\ 0\\ 1\end{bmatrix}\) are linearly independent.

\[\begin{align}\left[\begin{array}{rrr|r} 0 & 3 & 2 & 0\\ -1 & 1 & 0 & 0\\ 2 & -1 & 1 & 0\end{array}\right] &\sim \left[\begin{array}{rrr|r} 1 & -1 & 0 & 0\\ 0 & 3 & 2 & 0\\ 0 & 1 & 1 & 0\end{array}\right]\\ &\substack{R_2\leftrightarrow R_3\\ \longrightarrow} \left[\begin{array}{rrr|r} 1 & -1 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 3 & 2 & 0 \end{array}\right]\\ &\substack{R_3\leftarrow R_3 + \left(-3\right)R_2\\ \longrightarrow} \left[\begin{array}{rrr|r} \boxed{~1~} & -1 & 0 & 0\\ 0 & \boxed{~1~} & 1 & 0\\ 0 & 0 & \boxed{~-1~} & 0 \end{array}\right]\end{align}\]

Completed Example #1

Example: Determine whether the vectors \(\vec{v_1} = \begin{bmatrix} 0\\ -1\\ 2\end{bmatrix}\), \(\vec{v_2} = \begin{bmatrix} 3\\ 1\\ -1\end{bmatrix}\), and \(\vec{v_1} = \begin{bmatrix} 2\\ 0\\ 1\end{bmatrix}\) are linearly independent.

\[\begin{align}\left[\begin{array}{rrr|r} 0 & 3 & 2 & 0\\ -1 & 1 & 0 & 0\\ 2 & -1 & 1 & 0\end{array}\right] &\sim \left[\begin{array}{rrr|r} \boxed{~1~} & -1 & 0 & 0\\ 0 & \boxed{~1~} & 1 & 0\\ 0 & 0 & \boxed{~-1~} & 0 \end{array}\right]\end{align}\]

• There is a pivot in every column corresponding to a variable in the REF augmented matrix.
• There are no free variables.
• There is a unique solution to the homogeneous equation.
• The vectors \(\vec{v_1}\), \(\vec{v_2}\), and \(\vec{v_3}\) are linearly independent!

Completed Example #2

Example: Determine whether the vectors \(\vec{v_1} = \begin{bmatrix} 1\\ 0\\ 2\\ 0\end{bmatrix}\), \(\vec{v_2} = \begin{bmatrix} 0\\ 1\\ 1\\ -2\end{bmatrix}\), and \(\vec{v_1} = \begin{bmatrix} 2\\ -5\\ -1\\ 10\end{bmatrix}\) are linearly independent.

\[\begin{align} \left[\begin{array}{rrr|r} 1 & 0 & 2 & 0\\ 0 & 1 & -5 & 0\\ 2 & 1 & -1 & 0\\ 0 & -2 & 10 & 0\end{array}\right] \end{align}\]

Completed Example #2

Example: Determine whether the vectors \(\vec{v_1} = \begin{bmatrix} 1\\ 0\\ 2\\ 0\end{bmatrix}\), \(\vec{v_2} = \begin{bmatrix} 0\\ 1\\ 1\\ -2\end{bmatrix}\), and \(\vec{v_1} = \begin{bmatrix} 2\\ -5\\ -1\\ 10\end{bmatrix}\) are linearly independent.

\[\begin{align} \left[\begin{array}{rrr|r} 1 & 0 & 2 & 0\\ 0 & 1 & -5 & 0\\ 2 & 1 & -1 & 0\\ 0 & -2 & 10 & 0\end{array}\right] &\substack{R_3\leftarrow R_3 + \left(-2\right)R_1\\ \longrightarrow} \left[\begin{array}{rrr|r} 1 & 0 & 2 & 0\\ 0 & 1 & -5 & 0\\ 0 & 1 & -5 & 0\\ 0 & -2 & 10 & 0\end{array}\right] \end{align}\]

Completed Example #2

Example: Determine whether the vectors \(\vec{v_1} = \begin{bmatrix} 1\\ 0\\ 2\\ 0\end{bmatrix}\), \(\vec{v_2} = \begin{bmatrix} 0\\ 1\\ 1\\ -2\end{bmatrix}\), and \(\vec{v_1} = \begin{bmatrix} 2\\ -5\\ -1\\ 10\end{bmatrix}\) are linearly independent.

\[\begin{align} \left[\begin{array}{rrr|r} 1 & 0 & 2 & 0\\ 0 & 1 & -5 & 0\\ 2 & 1 & -1 & 0\\ 0 & -2 & 10 & 0\end{array}\right] &\substack{R_3\leftarrow R_3 + \left(-2\right)R_1\\ \longrightarrow} \left[\begin{array}{rrr|r} 1 & 0 & 2 & 0\\ 0 & 1 & -5 & 0\\ 0 & 1 & -5 & 0\\ 0 & -2 & 10 & 0\end{array}\right]\\ &\substack{R_3 \leftarrow R_3 + \left(-1\right)R_2\\ R_4 \leftarrow R_4 + 2R_2\\ \longrightarrow} \left[\begin{array}{rrr|r} 1 & 0 & 2 & 0\\ 0 & 1 & -5 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{array}\right] \end{align}\]

Completed Example #2

Example: Determine whether the vectors \(\vec{v_1} = \begin{bmatrix} 1\\ 0\\ 2\\ 0\end{bmatrix}\), \(\vec{v_2} = \begin{bmatrix} 0\\ 1\\ 1\\ -2\end{bmatrix}\), and \(\vec{v_1} = \begin{bmatrix} 2\\ -5\\ -1\\ 10\end{bmatrix}\) are linearly independent.

\[\begin{align} \left[\begin{array}{rrr|r} 1 & 0 & 2 & 0\\ 0 & 1 & -5 & 0\\ 2 & 1 & -1 & 0\\ 0 & -2 & 10 & 0\end{array}\right] &\substack{R_3\leftarrow R_3 + \left(-2\right)R_1\\ \longrightarrow} \left[\begin{array}{rrr|r} 1 & 0 & 2 & 0\\ 0 & 1 & -5 & 0\\ 0 & 1 & -5 & 0\\ 0 & -2 & 10 & 0\end{array}\right]\\ &\substack{R_3 \leftarrow R_3 + \left(-1\right)R_2\\ R_4 \leftarrow R_4 + 2R_2\\ \longrightarrow} \left[\begin{array}{rrr|r} \boxed{~1~} & 0 & 2 & 0\\ 0 & \boxed{~1~} & -5 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{array}\right] \end{align}\]

Completed Example #2

Example: Determine whether the vectors \(\vec{v_1} = \begin{bmatrix} 1\\ 0\\ 2\\ 0\end{bmatrix}\), \(\vec{v_2} = \begin{bmatrix} 0\\ 1\\ 1\\ -2\end{bmatrix}\), and \(\vec{v_1} = \begin{bmatrix} 2\\ -5\\ -1\\ 10\end{bmatrix}\) are linearly independent.

\[\begin{align} \left[\begin{array}{rrr|r} 1 & 0 & 2 & 0\\ 0 & 1 & -5 & 0\\ 2 & 1 & -1 & 0\\ 0 & -2 & 10 & 0\end{array}\right] &\sim \left[\begin{array}{rrr|r} \boxed{~1~} & 0 & 2 & 0\\ 0 & \boxed{~1~} & -5 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{array}\right] \end{align}\]

• The third column has no pivot.
• The variable \(x_3\) is a free variable.
• There are infinitely many solutions to the homogeneous equation.
• The vectors \(\vec{v_1}\), \(\vec{v_2}\), and \(\vec{v_3}\) are not linearly independent!

Linear Independence / Dependence: Special Cases

Although they align with our observations and strategy above, it is worth pointing out a couple of special cases.

1. Any collection containing a single non-zero vector \(\left\{\vec{v_1}\right\}\) is a linearly independent set.
2. Any collection including the zero vector \(\vec{0}\) as an element is linearly dependent.

Summary

• A collection of vectors \(\left\{\vec{v_1},~\vec{v_2},~\cdots,~\vec{v_p}\right\}\) is linearly dependent if at least one of the vectors in the set can be written as a linear combination of the other vectors in the set. Otherwise, the collection is linearly independent.

• We can use several of the tools/investigations we’ve already encountered to determine whether collections \(\left\{\vec{v_1},~\vec{v_2},~\cdots,~\vec{v_p}\right\}\) are linearly independent.

  • If the corresponding homogeneous system has a unique solution, then the collection is linearly independent, otherwise the collection is linearly dependent.
  • The collection of vectors is linearly independent if the matrix \(\begin{bmatrix} \vec{v_1} & \vec{v_2} & \cdots & \vec{v_p}\end{bmatrix}\) has a pivot in every column. Otherwise, the collection is linearly dependent.

Summary (Cont’d)

Below are a few observations that we can make. Some of them were mentioned earlier in these slides, but others are mentioned here for the first time.

• A collection containing a single, non-zero vector is linearly independent.

• Any collection containing the zero vector \(\vec{0}\) is linearly dependent.

• Any collection of vectors from \(\mathbb{R}^m\) that contains more than \(m\) vectors is linearly dependent.

  • Note. The converse is not necessarily true.

Strategy: The most reliable (broadly applicable) method for determining whether a collection of vectors \(V = \left\{\vec{v_1},~\vec{v_2},~\cdots,~\vec{v_p}\right\}\) is linearly independent is to determine whether the matrix \(\begin{bmatrix} \vec{v_1} & \vec{v_2} & \cdots & \vec{v_p}\end{bmatrix}\) has a pivot in every column.

• If the answer is yes, then the vectors in \(V\) are linearly independent.
• If the answer is no, then the vectors in \(V\) are linearly dependent.

Examples to Try #1

Example: Determine whether the collections of vectors below are linearly independent.

1. \(\left\{\left[\begin{array}{r} 1\\ -2\\ 0\\ 0\\ 3\end{array}\right]\right\}\)

2. \(\left\{\left[\begin{array}{r} 1\\ 2\end{array}\right], \left[\begin{array}{r} -1\\ 1\end{array}\right], \left[\begin{array}{r} 3\\ 3\end{array}\right]\right\}\)

3. \(\left\{\left[\begin{array}{r} 1\\ 0\\ 3\end{array}\right], \left[\begin{array}{r} 0\\ 0\\ 0\end{array}\right]\right\}\)

4. \(\left\{\left[\begin{array}{r} 1\\ 1\\ 0\end{array}\right], \left[\begin{array}{r} 0\\ -2\\ 1\end{array}\right], \left[\begin{array}{r} -1\\ 0\\ 1\end{array}\right]\right\}\)

Examples to Try #2

Example: Determine the value(s) for \(h\) which make \(\vec{v_3} \in \text{span}\left(\left\{\vec{v_1},~\vec{v_2}\right\}\right)\) where \(\vec{v_1} = \left[\begin{array}{r} 1\\ -3\\ 5\end{array}\right]\), \(\vec{v_2} = \left[\begin{array}{r} -3\\ 9\\ 15\end{array}\right]\), and \(\vec{v_3} = \left[\begin{array}{r}2\\ -5\\ h\end{array}\right]\). For which values of \(h\) is \(\left\{\vec{v_1},~\vec{v_2},~\vec{v_3}\right\}\) linearly independent?

Examples to Try #3

Example: Determine the value(s) of \(h\) which make \(\left[\begin{array}{r} 1\\ -2\\ 4\end{array}\right]\), \(\left[\begin{array}{r} -3\\ 7\\ 6\end{array}\right]\), and \(\left[\begin{array}{r} 2\\ 1\\ h\end{array}\right]\) linearly dependent.

Examples to Try #4

Example: What are the possible row echelon forms of the matrices in the following scenarios:

1. \(A\) is a \(2\times 2\) matrix with linearly dependent columns.
2. \(A\) is a \(3\times 3\) matrix with linearly independent columns.
3. \(A\) is a \(4\times 2\) matrix \(\left[\begin{array}{rr} \vec{a_1} & \vec{a_2}\end{array}\right]\) with \(\vec{a_2}\) not a scalar multiple of \(\vec{a_1}\).
4. \(A\) is a \(4\times 3\) matrix \(\left[\begin{array}{rrr} \vec{a_1} & \vec{a_2} & \vec{a_3}\end{array}\right]\) such that \(\left\{\vec{a_1},~\vec{a_2}\right\}\) is linearly independent and \(\vec{a_3}\) is not in \(\text{span}\left(\left\{\vec{a_1}, \vec{a_2}\right\}\right)\).

Examples to Try #5

Example: How many pivot columns must a \(6\times 4\) matrix have if its columns span \(\mathbb{R}^4\)? Why?

Examples to Try #6

Example: Construct \(3\times 2\) matrices \(A\) and \(B\) such that \(A\vec{x} = \vec{0}\) has a non-trivial solution but \(B\vec{x} = \vec{0}\) has only the trivial solution.

Exit Ticket Task

Navigate to our MAT350 Exit Ticket Form, answer the questions, and complete the task below.

Note. Today’s discussion is listed as 11. Linear Independence

Task: Determine whether the columns of the matrix \(A = \begin{bmatrix} 1 & -3 & 2 & 5\\ 0 & 3 & 9 & -2\\ 0 & 0 & -4 & 0\end{bmatrix}\) are linearly independent.

Homework

\[\Huge{\text{Finish Homework 5}}\] \[\Huge{\text{on MyOpenMath}}\]

Next Time…

\(\Huge{\text{Matrix Transformations}}\)