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Chapter 4: Problem 65

Let \(S=\\{\mathbf{u}, \mathbf{v}\\}\) be a linearly independent set. Prove that the set \(\\{\mathbf{u}+\mathbf{v}, \mathbf{u}-\mathbf{v}\\}\) is linearly independent.

Short Answer

Expert verified
The vectors in the set \(\{\mathbf{u}+\mathbf{v}, \mathbf{u}-\mathbf{v}\}\) are linearly independent.

Step by step solution

01

Understanding Linear Independence

In linear algebra, a set of vectors is said to be linearly independent if no vector in the set can be represented as a linear combination of the others. That is, no vector within the set is a scalar multiple of another. So in our original set, \(S=\{\mathbf{u}, \mathbf{v}\}\), \(\mathbf{u}\) and \(\mathbf{v}\) are not scalar multiples of each other.
02

Applying Linear Combination of Vectors

Given that the new set is \(\{\mathbf{u}+\mathbf{v}, \mathbf{u}-\mathbf{v}\}\), we start by assuming that the vectors in this set are not linearly independent, i.e., there exist scalars a and b such that a*(\(\mathbf{u}+\mathbf{v}\)) + b*(\(\mathbf{u}-\mathbf{v}\)) = 0.
03

Simplify Expression

Simplify the above equation to get (a + b)\(\mathbf{u}\) + (a - b)\(\mathbf{v}\) = 0. Now, because \(\mathbf{u}\) and \(\mathbf{v}\) are linearly independent, for this equation to hold, both (a+b) and (a-b) must be equal to 0.
04

Solving for Scalars

Solving the system of equations gives us a = b = 0. Therefore, the only way our original assumption (step 2) can hold is if the scalars are zero.
05

Conclusion

Because the only solution is a = b = 0, the vectors in the set \(\{\mathbf{u}+\mathbf{v}, \mathbf{u}-\mathbf{v}\}\) are also linearly independent.

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Most popular questions from this chapter

Use Theorem 4.21 to (a) find the transition matrix from \(B\) to \(B^{\prime},\) (b) find the transition matrix from \(B^{\prime}\) to \(B\) (c) verify that the two transition matrices are inverses of each other, and (d) find \([\mathbf{x}]_{a}\) when provided with \([\mathbf{x}]_{B^{*}}\) \(B=\\{(1,3),(-2,-2)\\}, \quad B^{\prime}=\\{(-12,0),(-4,4)\\},\) \([\mathbf{x}]_{B^{\prime}}=\left[\begin{array}{r}-1 \\ 3\end{array}\right]\)

Use a graphing utility or computer software program with matrix capabilities to find the transition matrix from \(B\) to \(B^{\prime}\) $$\begin{array}{l}B=\\{(1,2,4),(-1,2,0),(2,4,0)\\}, \\ B^{\prime}=\\{(0,2,1),(-2,1,0),(1,1,1)\\}\end{array}$$

Prove that \(y=C_{1} \cos a x+C_{2} \sin a x\) is the general solution of \(y^{\prime \prime}+a^{2} y=0, a \neq 0\)

Prove that a rotation of \(\theta=\pi / 4\) will eliminate the \(x y\) -term from the equation $$a x^{2}+b x y+a y^{2}+d x+e y+f=0$$

Is the scalar multiple of a solution of a nonhomogeneous linear differential equation also a solution? Explain your answer.
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