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Chapter 4: Problem 22

Suppose the vectors \(u, v, w\) are linearly independent. Show that the vectors \(u+v, u-v\) \(u-2 v+w\) are also linearly independent.

Short Answer

Expert verified
To show that the vectors \(u+v\), \(u-v\), and \(u-2v+w\) are linearly independent, consider the linear combination \(c_1(u + v) + c_2(u - v) + c_3(u - 2v + w) = 0\). Distribute the coefficients and simplify, leading to \((c_1 + c_2 + c_3)u + (c_1 - c_2 - 2c_3)v + c_3w = 0\). Since u, v, and w are linearly independent, the coefficients of each vector must be zero: \(c_1 + c_2 + c_3 = 0\), \(c_1 - c_2 - 2c_3 = 0\), and \(c_3 = 0\). Solving these equations, we find that \(c_1 = 0\), \(c_2 = 0\), and \(c_3 = 0\). Since the only solution is the zero coefficients, the vectors \(u+v\), \(u-v\), and \(u-2v+w\) are linearly independent.

Step by step solution

01

Write down the linear combination of the given vectors

Let's consider a linear combination of the vectors u + v, u - v, and u - 2v + w that results in the zero vector: \(c_1(u + v) + c_2(u - v) + c_3(u - 2v + w) = 0\), where \(c_1\), \(c_2\), and \(c_3\) are scalar coefficients.
02

Distribute the coefficients

Now, we will distribute the coefficients and simplify the given linear combination: \(c_1u + c_1v + c_2u - c_2v + c_3u - 2c_3v + c_3w = 0\) Combine the coefficients of the vectors u, v, and w: \((c_1 + c_2 + c_3)u + (c_1 - c_2 - 2c_3)v + c_3w = 0\)
03

Use the linear independence of u, v, and w

Since the vectors u, v, and w are linearly independent, the only way the above equation can hold is if the coefficients of each vector are zero: \(c_1 + c_2 + c_3 = 0\) \(c_1 - c_2 - 2c_3 = 0\) \(c_3 = 0\)
04

Solve for the coefficients

From the third equation, we get that \(c_3 = 0\). Now we will use this result to find the values of \(c_1\) and \(c_2\): Since \(c_3 = 0\), the first and second equations become: \(c_1 + c_2 = 0\) \(c_1 - c_2 = 0\) Adding both equations, we get: \(2c_1= 0\) Which implies, \(c_1 = 0\) Now, since \(c_1 = 0\), from the first equation, we can find \(c_2\): \(0 + c_2 = 0\) Which implies, \(c_2 = 0\)
05

Interpret the results

We have found that the coefficients \(c_1 = 0\), \(c_2 = 0\), and \(c_3 = 0\) are the only possible solution to the linear combination of our resulting vectors equal to the zero vector. This means that the vectors u + v, u - v, and u - 2v + w are linearly independent since the only way to get the zero vector from their linear combination is by using all zero coefficients.

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Most popular questions from this chapter

Suppose \(W_{1}, W_{2}, \ldots, W_{r}\) are subspaces of a vector space \(V\). Show that (a) \(\operatorname{span}\left(W_{1}, W_{2}, \ldots, W_{r}\right)=W_{1}+W_{2}+\cdots+W_{r}\). (b) If \(S_{i}\) spans \(W_{i}\) for \(i=1, \ldots, r,\) then \(S_{1} \cup S_{2} \cup \ldots \cup S_{r}\) spans \(W_{1}+W_{2}+\cdots+W_{r}\).

Let \(\beta=\left\\{u_{1}, u_{2}, \ldots, u_{n}\right\\}\) be a subset of \(\mathrm{F}^{n}\) containing \(n\) distinct vectors, and let \(B\) be the matrix in \(M_{n \times n}(F)\) having \(u_{j}\) as column \(j\). Prove that \(\beta\) is a basis for \(\mathrm{F}^{n}\) if and only if \(\operatorname{det}(B) \neq 0\).

Suppose \(u, v, w\) are linearly independent vectors. Prove that \(S\) is linearly independent where (a) \(\quad S=\\{u+v-2 w, u-v-w, u+w\\}\); (b) \(S=\\{u+v-3 w, u+3 v-w, v+w\\}\).

Find a subset of \(u_{1}, u_{2}, u_{3}, u_{4}\) that gives a basis for \(W=\operatorname{span}\left(u_{i}\right)\) of \(\mathbf{R}^{5},\) where (a) \(u_{1}=(1,1,1,2,3), \quad u_{2}=(1,2,-1,-2,1), \quad u_{3}=(3,5,-1,-2,5), \quad u_{4}=(1,2,1,-1,4)\) (b) \(u_{1}=(1,-2,1,3,-1), \quad u_{2}=(-2,4,-2,-6,2), \quad u_{3}=(1,-3,1,2,1), \quad u_{4}=(3,-7,3,8,-1)\) (c) \(u_{1}=(1,0,1,0,1), \quad u_{2}=(1,1,2,1,0), \quad u_{3}=(2,1,3,1,1), \quad u_{4}=(1,2,1,1,1)\) (d) \(u_{1}=(1,0,1,1,1), \quad u_{2}=(2,1,2,0,1), \quad u_{3}=(1,1,2,3,4), \quad u_{4}=(4,2,5,4,6)\)

Find a basis for (i) the row space and (ii) the column space of each matrix \(\mathrm{M}:\) (a) \(M=\left[\begin{array}{rrrrr}0 & 0 & 3 & 1 & 4 \\ 1 & 3 & 1 & 2 & 1 \\ 3 & 9 & 4 & 5 & 2 \\ 4 & 12 & 8 & 8 & 7\end{array}\right]\), (b) \(M=\left[\begin{array}{rrrrr}1 & 2 & 1 & 0 & 1 \\ 1 & 2 & 2 & 1 & 3 \\ 3 & 6 & 5 & 2 & 7 \\ 2 & 4 & 1 & -1 & 0\end{array}\right]\).
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