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Chapter 3: Problem 1

Determine whether the following vectors are linearly independent in \(\mathbb{R}^{2}:\) (a) \(\left(\begin{array}{l}2 \\ 1\end{array}\right),\left(\begin{array}{l}3 \\\ 2\end{array}\right)\) (b) \(\left(\begin{array}{l}2 \\ 3\end{array}\right),\left(\begin{array}{l}4 \\\ 6\end{array}\right)\) (c) \(\left(\begin{array}{r}-2 \\ 1\end{array}\right),\left(\begin{array}{l}1 \\\ 3\end{array}\right),\left(\begin{array}{l}2 \\ 4\end{array}\right)\) (d) \(\left(\begin{array}{r}-1 \\ 2\end{array}\right),\left(\begin{array}{r}1 \\\ -2\end{array}\right),\left(\begin{array}{r}2 \\ -4\end{array}\right)\) (e) \(\left(\begin{array}{l}1 \\ 2\end{array}\right),\left(\begin{array}{r}-1 \\\ 1\end{array}\right)\)

Short Answer

Expert verified
The given vectors in cases (a), (b), (c), (d), and (e) are: (a) Linearly Independent (b) Linearly Dependent (c) Linearly Dependent (d) Linearly Dependent (e) Linearly Independent

Step by step solution

01

(a) Case (a): Vectors \(\begin{pmatrix}2 \\1 \end{pmatrix}\) and \(\begin{pmatrix}3 \\2 \end{pmatrix}\)

Set up the system of linear equations for the linear combination and analyze the solution: \( c_1\begin{pmatrix}2 \\1 \end{pmatrix} + c_2\begin{pmatrix}3 \\2 \end{pmatrix} = \begin{pmatrix}0 \\0 \end{pmatrix}\) This can be rewritten as the following system of linear equations: \[ \begin{cases} 2c_1 + 3c_2 = 0 \\ c_1 + 2c_2 = 0 \end{cases} \] We observe that the second equation can be rewritten as \(c_1 = -2c_2\), and substituting this into the first equation, we get \(0 = 0\). This means that we have a single unique solution to the system, which is \(c_1 = 0\) and \(c_2 = 0\). Since the only solution is the trivial one, these two vectors are linearly independent.
02

(b) Case (b): Vectors \(\begin{pmatrix}2 \\ 3\end{pmatrix}\) and \(\begin{pmatrix}4 \\ 6\end{pmatrix}\)

Create the system of linear equations and analyze the solution: \( c_1\begin{pmatrix}2 \\3 \end{pmatrix} + c_2\begin{pmatrix}4 \\6 \end{pmatrix} = \begin{pmatrix}0 \\0 \end{pmatrix}\) Rewrite the system of linear equations as: \[ \begin{cases} 2c_1 + 4c_2 = 0 \\ 3c_1 + 6c_2 = 0 \end{cases} \] Divide both sides of the first equation by 2 to get \(c_1 + 2c_2 = 0\), which is equivalent to the second equation. It means that this system of linear equations has an infinite number of solutions. In this case, one such solution is \(c_1 = 2\), and \(c_2 = -1\). Since there is at least one non-trivial solution, these two vectors are linearly dependent.
03

(c) Case (c): Vectors \(\begin{pmatrix}-2 \\ 1\end{pmatrix}\), \(\begin{pmatrix}1 \\ 3\end{pmatrix}\), and \(\begin{pmatrix}2 \\ 4\end{pmatrix}\)

Since there are three vectors in the two-dimensional space \(\mathbb{R}^2\), they must be linearly dependent. There is no need to check this case, as the dimension of the vector space is smaller than the number of vectors.
04

(d) Case (d): Vectors \(\begin{pmatrix}-1 \\ 2\end{pmatrix}\), \(\begin{pmatrix}1 \\ -2\end{pmatrix}\), and \(\begin{pmatrix}2 \\ -4\end{pmatrix}\)

Similar to case (c), there are three vectors in \(\mathbb{R}^2\), so they must be linearly dependent. There is no need to check this case, as the dimension of the vector space is smaller than the number of vectors.
05

(e) Case (e): Vectors \(\begin{pmatrix}1 \\ 2\end{pmatrix}\) and \(\begin{pmatrix}-1 \\ 1\end{pmatrix}\)

Set up the system of linear equations for the linear combination and analyze the solution: \( c_1\begin{pmatrix}1 \\2 \end{pmatrix} + c_2\begin{pmatrix}-1 \\1 \end{pmatrix} = \begin{pmatrix}0 \\0 \end{pmatrix}\) This system of linear equations can be rewritten as: \[ \begin{cases} c_1 - c_2 = 0 \\ 2c_1 + c_2 = 0 \end{cases} \] From the first equation, we have \(c_1 = c_2\). Substituting this into the second equation, we get \(0 = 0\). This means that we have a single unique solution to the system, which is \(c_1 = 0\) and \(c_2 = 0\). Since the only solution is the trivial one, these two vectors are linearly independent.

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Most popular questions from this chapter

Is it possible to find a pair of two-dimensional subspaces \(U\) and \(V\) of \(\mathbb{R}^{3}\) such that \(U \cap V=\\{0\\} ?\) Prove your answer. Give a geometrical interpretation of your conclusion. [Hint: Let \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}\right\\}\) and \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) be bases for \(U\) and \(V,\) respectively. Show that \(\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{v}_{1}, \mathbf{v}_{2}\) are linearly dependent.

Determine the null space of each of the following matrices: $$\text { (a) }\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right]$$ $$\text { (b) }\left[\begin{array}{rrrr} 1 & 2 & -3 & -1 \\ -2 & -4 & 6 & 3 \end{array}\right]$$ $$\text { (c) }\left[\begin{array}{rrr} 1 & 3 & -4 \\ 2 & -1 & -1 \\ -1 & -3 & 4 \end{array}\right]$$ $$\text { (d) }\left[\begin{array}{rrrr} 1 & 1 & -1 & 2 \\ 2 & 2 & -3 & 1 \\ -1 & -1 & 0 & -5 \end{array}\right]$$

Show that the element 0 in a vector space is unique.

Let \(A\) be a \(4 \times 5\) matrix and let \(U\) be the reduced row echelon form of \(A\). If \\[ \begin{array}{c} \mathbf{a}_{1}=\left(\begin{array}{r} 2 \\ 1 \\ -3 \\ -2 \end{array}\right), \quad \mathbf{a}_{2}=\left(\begin{array}{rr} -1 \\ 2 \\ 3 \\ 1 \end{array}\right) \\ U=\left(\begin{array}{rrrrr} 1 & 0 & 2 & 0 & -1 \\ 0 & 1 & 3 & 0 & -2 \\ 0 & 0 & 0 & 1 & 5 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right) \end{array} \\] (a) find a basis for \(N(A)\) (b) given that \(\mathbf{x}_{0}\) is a solution of \(A \mathbf{x}=\mathbf{b},\) where \\[ \mathbf{b}=\left(\begin{array}{l} 0 \\ 5 \\ 3 \\ 4 \end{array}\right) \quad \text { and } \quad \mathbf{x}_{0}=\left(\begin{array}{l} 3 \\ 2 \\ 0 \\ 2 \\ 0 \end{array}\right) \\] (i) find all solutions to the system. (ii) determine the remaining column vectors of \(A\).

Let \(S, T,\) and \(U\) be subspaces of a vector space \(V\) We can form new subspaces by using the operations of \(\cap\) and \(+\) defined in Exercises 20 and \(22 .\) When we do arithmetic with numbers, we know that the operation of multiplication distributes over the operation of addition in the sense that \\[ a(b+c)=a b+a c \\] It is natural to ask whether similar distributive laws hold for the two operations with subspaces. (a) Does the intersection operation for subspaces distribute over the addition operation? That is does \\[ S \cap(T+U)=(S \cap T)+(S \cap U) \\] (b) Does the addition operation for subspaces disdoes tribute over the intersection operation? That is, \\[ S+(T \cap U)=(S+T) \cap(S+U) \\]
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