/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Prove that if \(\\{x, y\\}\) is ... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 7

Prove that if \(\\{x, y\\}\) is a basis for a vector space over \(C\), then so is $$ \left\\{\frac{1}{2}(x+y), \frac{1}{2 i}(x-y)\right\\} . $$

Short Answer

Expert verified
In order to prove that the set \(\{ \frac{1}{2}(x+y), \frac{1}{2 i}(x-y) \}\) is also a basis for the vector space over C, we need to show that it is linearly independent and spans the same vector space as {x, y}. We can prove linear independence by showing that the only solution for the coefficients \(c_1\) and \(c_2\) in the linear combination \(c_1(\frac{1}{2}(x+y)) + c_2(\frac{1}{2 i}(x-y)) = 0\) are \(c_1 = 0\) and \(c_2 = 0\). In addition, we can prove that the new set spans the same vector space as {x, y} by showing that for any vector v and any constants \(a_1\) and \(a_2\), there exist constants \(b_1\) and \(b_2\) such that \(v = b_1(\frac{1}{2}(x+y)) + b_2(\frac{1}{2 i}(x-y))\). Since the new set is linearly independent and spans the same vector space as {x, y}, we can conclude that it is also a basis for the vector space over C.

Step by step solution

01

Show linear independence of the new set

To show that the new set is linearly independent, we need to prove that the only way to express the zero vector as a linear combination of the new set is: $$c_1\left(\frac{1}{2}(x+y)\right) + c_2\left(\frac{1}{2i}(x-y)\right) = 0$$ with the coefficients \(c_1\) and \(c_2\) equalling zero.
02

Solve for coefficients

Let's express the linear combination as follows: \(c_1\left(\frac{1}{2}(x+y)\right) + c_2\left(\frac{1}{2i}(x-y)\right) = c_1\frac{1}{2}x + c_1\frac{1}{2}y + c_2\frac{1}{2i}x - c_2\frac{1}{2i}y = 0\) Now, rearrange the terms and group the x and y terms: \(\left(c_1\frac{1}{2} + c_2\frac{1}{2i}\right)x + \left(c_1\frac{1}{2} - c_2\frac{1}{2i}\right)y = 0\)
03

Determine coefficients to make zero vector

Since {x, y} is a basis for the vector space, it is linearly independent. This means that the only way to express the zero vector as a linear combination of {x, y} is when the coefficients are zero: \(c_1\frac{1}{2} + c_2\frac{1}{2i} = 0\) and \(c_1\frac{1}{2} - c_2\frac{1}{2i} = 0\) Now, we can solve this system of equations. From the first equation, we can write: \(c_1\frac{1}{2} = - c_2\frac{1}{2i}\) Multiply both sides by 2 to eliminate the fractions: \(c_1 = - c_2\frac{1}{i}\) Now, substitute this expression into the second equation: \((- c_2\frac{1}{i})\frac{1}{2} - c_2\frac{1}{2i} = 0\) Simplify the equation: \(-c_2\frac{1}{2i} - c_2\frac{1}{2i} = 0\) Combine the terms: \(-2c_2\frac{1}{2i} = 0\) Divide both sides by -2: \(c_2\frac{1}{i} = 0\) Now, the only way for the equation to hold true is if c_2 = 0. Substitute this back into the equation for c_1: \(c_1 = - c_2\frac{1}{i} = -0\) So, c_1 = 0 as well. Therefore, the new set is linearly independent.
04

Show that the new set spans the same vector space

Since {x, y} is a basis, we can express any vector v in the vector space as: \(v = a_1x + a_2y\) To show that the new set spans the same vector space, we must find constants \(b_1\) and \(b_2\) such that: \(v = b_1\left(\frac{1}{2}(x+y)\right) + b_2\left(\frac{1}{2i}(x-y)\right)\) Rearrange the terms and group the x and y terms: \(v = \left(b_1\frac{1}{2} + b_2\frac{1}{2i}\right)x + \left(b_1\frac{1}{2} - b_2\frac{1}{2i}\right)y\) Now, compare the coefficients: \(a_1 = b_1\frac{1}{2} + b_2\frac{1}{2i}\) and \(a_2 = b_1\frac{1}{2} - b_2\frac{1}{2i}\) We can see that given any pair of constants \(a_1\) and \(a_2\), we can find constants \(b_1\) and \(b_2\) that satisfy the equations. Therefore, the new set \(\{ \frac{1}{2}(x+y), \frac{1}{2i}(x-y) \}\) spans the same vector space as {x, y}.
05

Conclusion

We have proven that the new set \(\{ \frac{1}{2}(x+y), \frac{1}{2i}(x-y) \}\) is linearly independent and spans the same vector space as {x, y}. Therefore, we can conclude that this new set is also a basis for the vector space over C.

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Most popular questions from this chapter

Pendular Motion. It is well known that the motion of a pendulum is approximated by the differential equation $$ \theta^{\prime \prime}+\frac{g}{l} \theta=0, $$where \(\theta(t)\) is the angle in radians that the pendulum makes with a vertical line at time \(t\) (see Figure 2.8), interpreted so that \(\theta\) is positive if the pendulum is to the right and negative if the pendulum is to the left of the vertical line as viewed by the reader. Here \(l\) is the length of the pendulum and \(g\) is the magnitude of acceleration due to gravity. The variable \(t\) and constants \(l\) and \(g\) must be in compatible units (e.g., \(t\) in seconds, \(l\) in meters, and \(g\) in meters per second per second). (a) Express an arbitrary solution to this equation as a linear combination of two real-valued solutions. (b) Find the unique solution to the equation that satisfies the conditions $$ \theta(0)=\theta_{0}>0 \text { and } \theta^{\prime}(0)=0 \text {. } $$ (The significance of these conditions is that at time \(t=0\) the pendulum is released from a position displaced from the vertical by \(\theta_{0}\).) (c) Prove that it takes \(2 \pi \sqrt{l / g}\) units of time for the pendulum to make one circuit back and forth. (This time is called the period of the pendulum.)

Find the inverse of each of the following matrices (if it exists): \\[ A=\left[\begin{array}{ll} 7 & 4 \\ 5 & 3 \end{array}\right], \quad B=\left[\begin{array}{ll} 2 & 3 \\ 4 & 5 \end{array}\right], \quad C=\left[\begin{array}{rr} 4 & -6 \\ -2 & 3 \end{array}\right], \quad \quad D=\left[\begin{array}{ll} 5 & -2 \\ 6 & -3 \end{array}\right] \\]

Show that \(A=\left[\begin{array}{cc}\frac{1}{3}-\frac{2}{3} i & \frac{2}{3} i \\\ -\frac{2}{3} i & -\frac{1}{3}-\frac{2}{3} i\end{array}\right]\) is unitary.

For the following functions \(f\) on a vector space \(V\), determine which are linear functionals. (a) $\mathrm{V}=\mathrm{P}(R) ; \mathrm{f}(p(x))=2 p^{\prime}(0)+p^{\prime \prime}(1)$, where ' denotes differentiation (b) \(\mathrm{V}=\mathrm{R}^{2} ; \mathrm{f}(x, y)=(2 x, 4 y)\) (c) $\mathrm{V}=\mathrm{M}_{2 \times 2}(F) ; \mathrm{f}(A)=\operatorname{tr}(A)$ (d) \(\mathrm{V}=\mathrm{R}^{3} ; \mathrm{f}(x, y, z)=x^{2}+y^{2}+z^{2}\) (e) \(\mathrm{V}=\mathrm{P}(R) ; \mathrm{f}(p(x))=\int_{0}^{1} p(t) d t\) (f) \(\mathrm{V}=\mathrm{M}_{2 \times 2}(F) ; \mathrm{f}(A)=A_{11}\)

Refer to the following matrices: $$A=\left[\begin{array}{rr} 1 & 2 \\ 3 & -4 \end{array}\right], \quad B=\left[\begin{array}{rr} 5 & 0 \\ -6 & 7 \end{array}\right], \quad C=\left[\begin{array}{rrr} 1 & -3 & 4 \\ 2 & 6 & -5 \end{array}\right], \quad D=\left[\begin{array}{rrr} 3 & 7 & -1 \\ 4 & -8 & 9 \end{array}\right]$$ Find \(\quad(a) A^{T}\) (b) \(B^{T}\) \((\mathrm{c})(A B)^{T}\) (d) \(A^{T} B^{T}\). [Note that \(A^{T} B^{T} \neq(A B)^{T}\).]
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