In the vector space R2 consider the two bases given by [u1, u2] with u1 = (1, 2)T, u2 = (2, 5)T and [v1, v2] with v1 = (3, 2)T, v2 = (4, 3)T (both with respect to the standard basis). Follow these steps to find transition matrices from [v1, v2] to [u1, u2] and back again: d) Find the transition matrix from [u1, u2] to [v1, v2]. e) Use your result from part c) to find the coordinates of w = v1 − v2 with respect to the basis [u1, u2]. (Note that [w][v1,v2] = (1, −1)T.)

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
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Chapter2: Systems Of Linear Equations
Section2.3: Spanning Sets And Linear Independence
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In the vector space R2 consider the two bases given by [u1, u2] with u1 = (1, 2)Tu2 = (2, 5)T and [v1v2] with v1 = (3, 2)Tv2 = (4, 3)T (both with respect to the standard basis).

Follow these steps to find transition matrices from [v1v2] to [u1u2] and back again:


d) Find the transition matrix from [u1u2] to [v1v2].
e) Use your result from part c) to find the coordinates of w = v1 − v2 with respect to the basis [u1u2]. (Note that [w][v1,v2] = (1, −1)T.)

In the vector space \( \mathbb{R}^2 \), consider the two bases given by \([u_1, u_2]\) with \( u_1 = (1, 2)^T, u_2 = (2, 5)^T \) and \([v_1, v_2]\) with \( v_1 = (3, 2)^T, v_2 = (4, 3)^T \) (both with respect to the standard basis).

**For (a),**

We need to find the transition matrix \( V \) from \([v_1, v_2]\) to the standard basis \([e_1, e_2]\).

Notice that \( v_1 = (3, 2)^T \)

\[
= \begin{bmatrix} 3 \\ 2 \end{bmatrix}
\]

\[
= 3 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + 2 \begin{bmatrix} 0 \\ 1 \end{bmatrix}
\]

\[ 
= 3e_1 + 2e_2 
\]

And similarly, \( v_2 = (4, 3)^T = 4e_1 + 3e_2 \).

Thus, the transition matrix \( V \) from \([v_1, v_2]\) to the standard basis \([e_1, e_2]\) is \( M_{ve} = \begin{bmatrix} 3 & 4 \\ 2 & 3 \end{bmatrix} \).

**Step 2**

**For (b),**

We need to find the transition matrix \( U \) from \([u_1, u_2]\) to the standard basis \([e_1, e_2]\).

Notice that \( u_1 = (1, 2)^T \)

\[
= \begin{bmatrix} 1 \\ 2 \end{bmatrix}
\]

\[
= 1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + 2 \begin{bmatrix} 0 \\ 1 \end{bmatrix}
\]

\[
= e_1 + 2e_2 
\]

And similarly, \( u_2
Transcribed Image Text:In the vector space \( \mathbb{R}^2 \), consider the two bases given by \([u_1, u_2]\) with \( u_1 = (1, 2)^T, u_2 = (2, 5)^T \) and \([v_1, v_2]\) with \( v_1 = (3, 2)^T, v_2 = (4, 3)^T \) (both with respect to the standard basis). **For (a),** We need to find the transition matrix \( V \) from \([v_1, v_2]\) to the standard basis \([e_1, e_2]\). Notice that \( v_1 = (3, 2)^T \) \[ = \begin{bmatrix} 3 \\ 2 \end{bmatrix} \] \[ = 3 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + 2 \begin{bmatrix} 0 \\ 1 \end{bmatrix} \] \[ = 3e_1 + 2e_2 \] And similarly, \( v_2 = (4, 3)^T = 4e_1 + 3e_2 \). Thus, the transition matrix \( V \) from \([v_1, v_2]\) to the standard basis \([e_1, e_2]\) is \( M_{ve} = \begin{bmatrix} 3 & 4 \\ 2 & 3 \end{bmatrix} \). **Step 2** **For (b),** We need to find the transition matrix \( U \) from \([u_1, u_2]\) to the standard basis \([e_1, e_2]\). Notice that \( u_1 = (1, 2)^T \) \[ = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \] \[ = 1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + 2 \begin{bmatrix} 0 \\ 1 \end{bmatrix} \] \[ = e_1 + 2e_2 \] And similarly, \( u_2
For (c),

We need to find the transition matrix from \([v_1, v_2]\) to \([u_1, u_2]\).

So,

\[ M_{VU} = M_{eU} M_{Ve} \]

\[ = M_{Ue}^{-1} M_{Ve} \]

\[ = \begin{bmatrix} 1 & 2 \\ 2 & 5 \end{bmatrix}^{-1} \begin{bmatrix} 3 & 4 \\ 2 & 3 \end{bmatrix} \]

\[ = \frac{1}{\text{det} \begin{pmatrix} 1 & 2 \\ 2 & 5 \end{pmatrix}} \begin{pmatrix} 5 & -2 \\ -2 & 1 \end{pmatrix} \begin{bmatrix} 3 & 4 \\ 2 & 3 \end{bmatrix} \]

\[ = \begin{bmatrix} 5 & -2 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 3 & 4 \\ 2 & 3 \end{bmatrix} \]

\[ = \begin{bmatrix} 11 & 14 \\ -4 & -5 \end{bmatrix} \]
Transcribed Image Text:For (c), We need to find the transition matrix from \([v_1, v_2]\) to \([u_1, u_2]\). So, \[ M_{VU} = M_{eU} M_{Ve} \] \[ = M_{Ue}^{-1} M_{Ve} \] \[ = \begin{bmatrix} 1 & 2 \\ 2 & 5 \end{bmatrix}^{-1} \begin{bmatrix} 3 & 4 \\ 2 & 3 \end{bmatrix} \] \[ = \frac{1}{\text{det} \begin{pmatrix} 1 & 2 \\ 2 & 5 \end{pmatrix}} \begin{pmatrix} 5 & -2 \\ -2 & 1 \end{pmatrix} \begin{bmatrix} 3 & 4 \\ 2 & 3 \end{bmatrix} \] \[ = \begin{bmatrix} 5 & -2 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 3 & 4 \\ 2 & 3 \end{bmatrix} \] \[ = \begin{bmatrix} 11 & 14 \\ -4 & -5 \end{bmatrix} \]
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