Find a basis {u1, u2, u3} for R^3 such that P is the change-of-coordinates matrix from {u1, u2, u3} to the basis {v1, v2, v3}. [Hint: What do the columns of P represent?]
Understand the Problem
The question is asking to find a basis for R^3 such that a given matrix P represents the change-of-coordinates from one basis to another. This requires understanding linear transformations and basis representations in vector spaces.
Answer
$$ u_1 = \begin{bmatrix} -11 \\ 14 \\ 13 \end{bmatrix}, \quad u_2 = \begin{bmatrix} 46 \\ -31 \\ -19 \end{bmatrix}, \quad u_3 = \begin{bmatrix} -63 \\ 40 \\ 30 \end{bmatrix} $$
Answer for screen readers
The basis ${u_1, u_2, u_3}$ for $\mathbb{R}^3$ is: $$ u_1 = \begin{bmatrix} -11 \ 14 \ 13 \end{bmatrix}, \quad u_2 = \begin{bmatrix} 46 \ -31 \ -19 \end{bmatrix}, \quad u_3 = \begin{bmatrix} -63 \ 40 \ 30 \end{bmatrix} $$
Steps to Solve
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Understand the Problem We need to find a basis ${u_1, u_2, u_3}$ such that the columns of the change-of-coordinates matrix $P$ represent the coordinates of the basis vectors $u_i$ in the basis ${v_1, v_2, v_3}$.
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Identify the Components The columns of the matrix $P$ give the coordinates of the basis vectors $u_1, u_2, u_3$ relative to the basis ${v_1, v_2, v_3}$: [ u_1 = P[\cdot][1], \quad u_2 = P[\cdot][2], \quad u_3 = P[\cdot][3] ] Specifically, this means: $u_1 = 1 \cdot v_1 + 2 \cdot v_2 - 1 \cdot v_3$,
$u_2 = -3 \cdot v_1 - 5 \cdot v_2 + 0 \cdot v_3$,
$u_3 = 4 \cdot v_1 + 6 \cdot v_2 + 1 \cdot v_3$. -
Calculate Each Basis Vector Substitute the given $v_1$, $v_2$, and $v_3$ values into the equations.
For $u_1$: [ u_1 = 1 \cdot \begin{bmatrix} -2 \ 2 \ 3 \end{bmatrix} + 2 \cdot \begin{bmatrix} -8 \ 5 \ 2 \end{bmatrix} - 1 \cdot \begin{bmatrix} -7 \ 2 \ 6 \end{bmatrix} ] For $u_2$: [ u_2 = -3 \cdot \begin{bmatrix} -2 \ 2 \ 3 \end{bmatrix} - 5 \cdot \begin{bmatrix} -8 \ 5 \ 2 \end{bmatrix} + 0 \cdot \begin{bmatrix} -7 \ 2 \ 6 \end{bmatrix} ] For $u_3$: [ u_3 = 4 \cdot \begin{bmatrix} -2 \ 2 \ 3 \end{bmatrix} + 6 \cdot \begin{bmatrix} -8 \ 5 \ 2 \end{bmatrix} + 1 \cdot \begin{bmatrix} -7 \ 2 \ 6 \end{bmatrix} ]
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Perform the Calculations Completing the computations for each vector:
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For $u_1$: [ u_1 = \begin{bmatrix} -2 \ 2 \ 3 \end{bmatrix} + \begin{bmatrix} -16 \ 10 \ 4 \end{bmatrix} + \begin{bmatrix} 7 \ 2 \ 6 \end{bmatrix} = \begin{bmatrix} -11 \ 14 \ 13 \end{bmatrix} ]
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For $u_2$: [ u_2 = \begin{bmatrix} 6 \ -6 \ -9 \end{bmatrix} + \begin{bmatrix} 40 \ -25 \ -10 \end{bmatrix} = \begin{bmatrix} 46 \ -31 \ -19 \end{bmatrix} ]
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For $u_3$: [ u_3 = \begin{bmatrix} -8 \ 8 \ 12 \end{bmatrix} + \begin{bmatrix} -48 \ 30 \ 12 \end{bmatrix} + \begin{bmatrix} -7 \ 2 \ 6 \end{bmatrix} = \begin{bmatrix} -63 \ 40 \ 30 \end{bmatrix} ]
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The basis ${u_1, u_2, u_3}$ for $\mathbb{R}^3$ is: $$ u_1 = \begin{bmatrix} -11 \ 14 \ 13 \end{bmatrix}, \quad u_2 = \begin{bmatrix} 46 \ -31 \ -19 \end{bmatrix}, \quad u_3 = \begin{bmatrix} -63 \ 40 \ 30 \end{bmatrix} $$
More Information
The change-of-coordinates matrix $P$ helps translate between different basis representations in vector spaces. Understanding this connection is crucial in linear algebra.
Tips
- Forgetting to properly apply the coefficients to each vector when calculating $u_i$.
- Misaligning the components during addition or multiplication.
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