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Question Find the formula of a linear transformation φ:R2→R3\varphi: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} given by the matrix M(φ)AB=[122331]M(\varphi)_{\mathcal{A}}^{\mathcal{B}}=\begin{bmatrix} 1 & 2 \\ 2 & 3 \\ 3 & 1 \end{bmatrix}, where A={(1,0),(1,−1)}\mathcal{A}=\{(1,0),(1,-1)\} and B={(1,0,0),(0,1,1),(−1,0,1)}\mathcal{B}=\{(1,0,0),(0,1,1),(-1,0,1)\}. Also, find the matrix M(φ∘ψ)ABM(\varphi \circ \psi)_{\mathcal{A}}^{\mathcal{B}} for the linear transformation ψ:R2→R2\psi: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2} given by ψ((x1,x2))=(x2,x1+x2)\psi((x_1, x_2))=(x_2, x_1+x_2).

Studdy Solution

STEP 1

Assumptions
1. A\mathcal{A} is a basis for R2\mathbb{R}^{2} and consists of vectors (1,0)(1,0) and (1,−1)(1,-1).
2. B\mathcal{B} is a basis for R3\mathbb{R}^{3} and consists of vectors (1,0,0)(1,0,0), (0,1,1)(0,1,1), and (−1,0,1)(-1,0,1).
3. φ:R2→R3\varphi: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} is a linear transformation represented by the matrix M(φ)ABM(\varphi)_{\mathcal{A}}^{\mathcal{B}} in the bases A\mathcal{A} and B\mathcal{B}.
4. ψ:R2→R2\psi: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2} is a linear transformation given by the formula ψ((x1,x2))=(x2,x1+x2)\psi\left(\left(x_{1}, x_{2}\right)\right)=\left(x_{2}, x_{1}+x_{2}\right).
5. We need to find the formula for φ\varphi and the matrix representation of the composition φ∘ψ\varphi \circ \psi in the given bases.

STEP 2

To find the formula for φ\varphi, we will express the action of φ\varphi on the basis vectors of R2\mathbb{R}^{2} in terms of the basis vectors of R3\mathbb{R}^{3}.

STEP 3

The matrix M(φ)ABM(\varphi)_{\mathcal{A}}^{\mathcal{B}} tells us how φ\varphi maps the basis vectors of A\mathcal{A} to linear combinations of the basis vectors of B\mathcal{B}.

STEP 4

Write down the matrix M(φ)ABM(\varphi)_{\mathcal{A}}^{\mathcal{B}}.
M(φ)AB=[122331] M(\varphi)_{\mathcal{A}}^{\mathcal{B}}=\left[\begin{array}{cc} 1 & 2 \\ 2 & 3 \\ 3 & 1 \end{array}\right]

STEP 5

The first column of M(φ)ABM(\varphi)_{\mathcal{A}}^{\mathcal{B}} corresponds to the image of the first basis vector of A\mathcal{A} under φ\varphi, and the second column corresponds to the image of the second basis vector of A\mathcal{A} under φ\varphi.

STEP 6

Using the matrix, we can write the images of the basis vectors of A\mathcal{A} as follows:
φ((1,0))=1⋅(1,0,0)+2⋅(0,1,1)+3⋅(−1,0,1) \varphi((1,0)) = 1 \cdot (1,0,0) + 2 \cdot (0,1,1) + 3 \cdot (-1,0,1)
φ((1,−1))=2⋅(1,0,0)+3⋅(0,1,1)+1⋅(−1,0,1) \varphi((1,-1)) = 2 \cdot (1,0,0) + 3 \cdot (0,1,1) + 1 \cdot (-1,0,1)

STEP 7

Calculate the images of the basis vectors of A\mathcal{A}:
φ((1,0))=(1,2,2+3)=(1,2,5) \varphi((1,0)) = (1,2,2+3) = (1,2,5)
φ((1,−1))=(2,3,3+1)=(2,3,4) \varphi((1,-1)) = (2,3,3+1) = (2,3,4)

STEP 8

Now we have the formula for φ\varphi on the basis vectors of A\mathcal{A}. To find the formula for any vector (x1,x2)(x_1, x_2) in R2\mathbb{R}^{2}, we express it as a linear combination of the basis vectors of A\mathcal{A}.

STEP 9

Let (x1,x2)(x_1, x_2) be an arbitrary vector in R2\mathbb{R}^{2}. We can write it as:
(x1,x2)=x1⋅(1,0)+(x2−x1)⋅(1,−1) (x_1, x_2) = x_1 \cdot (1,0) + (x_2 - x_1) \cdot (1,-1)

STEP 10

Apply φ\varphi to the vector (x1,x2)(x_1, x_2) using linearity of φ\varphi:
φ((x1,x2))=x1⋅φ((1,0))+(x2−x1)⋅φ((1,−1)) \varphi((x_1, x_2)) = x_1 \cdot \varphi((1,0)) + (x_2 - x_1) \cdot \varphi((1,-1))

STEP 11

Substitute the images of the basis vectors obtained earlier:
φ((x1,x2))=x1⋅(1,2,5)+(x2−x1)⋅(2,3,4) \varphi((x_1, x_2)) = x_1 \cdot (1,2,5) + (x_2 - x_1) \cdot (2,3,4)

STEP 12

Distribute and combine like terms to find the formula for φ((x1,x2))\varphi((x_1, x_2)):
φ((x1,x2))=(x1+2(x2−x1),2x1+3(x2−x1),5x1+4(x2−x1)) \varphi((x_1, x_2)) = (x_1 + 2(x_2 - x_1), 2x_1 + 3(x_2 - x_1), 5x_1 + 4(x_2 - x_1))

STEP 13

Simplify the expression:
φ((x1,x2))=(2x2−x1,3x2−x1,4x2−x1) \varphi((x_1, x_2)) = (2x_2 - x_1, 3x_2 - x_1, 4x_2 - x_1)

STEP 14

We have now found the formula for φ\varphi. Next, we will find the matrix representation of the composition φ∘ψ\varphi \circ \psi in the given bases.

STEP 15

First, we need to find the matrix representation of ψ\psi in the basis A\mathcal{A}, M(ψ)AM(\psi)_{\mathcal{A}}.

STEP 16

The formula for ψ\psi is given by ψ((x1,x2))=(x2,x1+x2)\psi\left(\left(x_{1}, x_{2}\right)\right)=\left(x_{2}, x_{1}+x_{2}\right). We will apply ψ\psi to the basis vectors of A\mathcal{A}.

STEP 17

Calculate ψ\psi for the first basis vector of A\mathcal{A}:
ψ((1,0))=(0,1+0)=(0,1) \psi((1,0)) = (0, 1+0) = (0,1)

STEP 18

Calculate ψ\psi for the second basis vector of A\mathcal{A}:
ψ((1,−1))=(−1,1+(−1))=(−1,0) \psi((1,-1)) = (-1, 1+(-1)) = (-1,0)

STEP 19

Express the images of the basis vectors under ψ\psi as linear combinations of the basis vectors of A\mathcal{A}:
ψ((1,0))=0⋅(1,0)+1⋅(1,−1) \psi((1,0)) = 0 \cdot (1,0) + 1 \cdot (1,-1)
ψ((1,−1))=−1⋅(1,0)+1⋅(1,−1) \psi((1,-1)) = -1 \cdot (1,0) + 1 \cdot (1,-1)

STEP 20

Write down the matrix M(ψ)AM(\psi)_{\mathcal{A}} using the coefficients from the linear combinations:
M(ψ)A=[0−111] M(\psi)_{\mathcal{A}} = \left[\begin{array}{cc} 0 & -1 \\ 1 & 1 \end{array}\right]

STEP 21

Now, to find the matrix representation of the composition φ∘ψ\varphi \circ \psi, we need to multiply the matrices M(φ)ABM(\varphi)_{\mathcal{A}}^{\mathcal{B}} and M(ψ)AM(\psi)_{\mathcal{A}}.

STEP 22

Perform the matrix multiplication:
M(φ∘ψ)AB=M(φ)AB⋅M(ψ)A M(\varphi \circ \psi)_{\mathcal{A}}^{\mathcal{B}} = M(\varphi)_{\mathcal{A}}^{\mathcal{B}} \cdot M(\psi)_{\mathcal{A}}
M(φ∘ψ)AB=[122331]⋅[0−111] M(\varphi \circ \psi)_{\mathcal{A}}^{\mathcal{B}} = \left[\begin{array}{cc} 1 & 2 \\ 2 & 3 \\ 3 & 1 \end{array}\right] \cdot \left[\begin{array}{cc} 0 & -1 \\ 1 & 1 \end{array}\right]

STEP 23

Calculate the product of the matrices:
M(φ∘ψ)AB=[1⋅0+2⋅11⋅(−1)+2⋅12⋅0+3⋅12⋅(−1)+3⋅13⋅0+1⋅13⋅(−1)+1⋅1] M(\varphi \circ \psi)_{\mathcal{A}}^{\mathcal{B}} = \left[\begin{array}{cc} 1 \cdot 0 + 2 \cdot 1 & 1 \cdot (-1) + 2 \cdot 1 \\ 2 \cdot 0 + 3 \cdot 1 & 2 \cdot (-1) + 3 \cdot 1 \\ 3 \cdot 0 + 1 \cdot 1 & 3 \cdot (-1) + 1 \cdot 1 \end{array}\right]

STEP 24

Simplify the entries of the resulting matrix:
M(φ∘ψ)AB=[21311−2] M(\varphi \circ \psi)_{\mathcal{A}}^{\mathcal{B}} = \left[\begin{array}{cc} 2 & 1 \\ 3 & 1 \\ 1 & -2 \end{array}\right]

STEP 25

We have now found the matrix representation of the composition φ∘ψ\varphi \circ \psi in the given bases A\mathcal{A} and B\mathcal{B}.
a) The formula for φ\varphi is φ((x1,x2))=(2x2−x1,3x2−x1,4x2−x1)\varphi((x_1, x_2)) = (2x_2 - x_1, 3x_2 - x_1, 4x_2 - x_1).
b) The matrix representation of the composition φ∘ψ\varphi \circ \psi in the given bases is:
M(φ∘ψ)AB=[21311−2] M(\varphi \circ \psi)_{\mathcal{A}}^{\mathcal{B}} = \left[\begin{array}{cc} 2 & 1 \\ 3 & 1 \\ 1 & -2 \end{array}\right]

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