text: # Give a Formula For a Linear Transformation From $\R^2$ to $\R^3$
  
  ## Problem 339
  
  Let $\{\mathbf{v}_1, \mathbf{v}_2\}$ be a basis of the vector space $\R^2$, where
  $\mathbf{v}_1=\begin{bmatrix} 1 \\ 1 \end{bmatrix} \text{ and } \mathbf{v}_2=\begin{bmatrix} 1 \\ -1 \end{bmatrix}.$ The action of a linear transformation $T:\R^2\to \R^3$ on the basis $\{\mathbf{v}_1, \mathbf{v}_2\}$ is given by
  \begin{align*}
  T(\mathbf{v}_1)=\begin{bmatrix}
  2 \\
  4 \\
  6
  \end{bmatrix} \text{ and } T(\mathbf{v}_2)=\begin{bmatrix}
  0 \\
  8 \\
  10
  \end{bmatrix}.
  \end{align*}
  
  Find the formula of $T(\mathbf{x})$, where
  $\mathbf{x}=\begin{bmatrix} x \\ y \end{bmatrix}\in \R^2.$
  
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  ## Solution.
  
  We give two solutions.
  
  ### Solution 1 (linear combination)
  
  Since we know the values of $T$ on the basis vectors $\mathbf{v}_1, \mathbf{v}_2$, if we express the vector $\mathbf{x}$ as a linear combination of $\mathbf{v}_1, \mathbf{v}_2$, we can find $F(\mathbf{x})$ by the linearity of the linear transformation $T$.
  
  So let us find the scalars $c_1, c_2$ such that
  $\mathbf{x}=c_1\mathbf{v}_1+c_2\mathbf{v}_2.$ We write this as
  \begin{align*}
  \begin{bmatrix}
  x \\
  y
  \end{bmatrix}=c_1\begin{bmatrix}
  1 \\
  1
  \end{bmatrix}+c_2\begin{bmatrix}
  1 \\
  -1
  \end{bmatrix}
  =
  \begin{bmatrix}
  1 & 1\\
  1& -1
  \end{bmatrix}\begin{bmatrix}
  c_1 \\
  c_2
  \end{bmatrix}.
  \end{align*}
  
  The matrix $\begin{bmatrix} 1 & 1\\ 1& -1 \end{bmatrix}$ is invertible (as its determinant is $-2$) and its inverse matrix is
  $\begin{bmatrix} 1 & 1\\ 1& -1 \end{bmatrix}^{-1}=\frac{1}{2}\begin{bmatrix} 1 & 1\\ 1& -1 \end{bmatrix}.$
  
  Thus, we have
  \begin{align*}
  \begin{bmatrix}
  c_1 \\
  c_2
  \end{bmatrix}&=\begin{bmatrix}
  1 & 1\\
  1& -1
  \end{bmatrix}^{-1}\begin{bmatrix}
  x \\
  y
  \end{bmatrix}\6pt] &=\frac{1}{2}\begin{bmatrix} 1 & 1\\ 1& -1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}\\[6pt] &=\frac{1}{2}\begin{bmatrix} x+y \\ x-y \end{bmatrix} \end{align*} Therefore, we obtain the linear combination \[\mathbf{x}=\frac{1}{2}(x+y)\mathbf{v}_1+\frac{1}{2}(x-y)\mathbf{v}_2.
  
  Now we compute $T(\mathbf{x})$ as follows.
  We have
  \begin{align*}
  T(\mathbf{x})&=T\left(\frac{1}{2}(x+y)\mathbf{v}_1+\frac{1}{2}(x-y)\mathbf{v}_2 \right)\6pt] &=\frac{1}{2}(x+y)T(\mathbf{v}_1)+\frac{1}{2}(x-y)T(\mathbf{v}_2) && \text{by linearity of T}\\[6pt] &=\frac{1}{2}(x+y)\begin{bmatrix} 2 \\ 4 \\ 6 \end{bmatrix}+\frac{1}{2}(x-y)\begin{bmatrix} 0 \\ 8 \\ 10 \end{bmatrix} \\[6pt] &=\begin{bmatrix} x+y \\ 6x-2y \\ 8x-2y \end{bmatrix}. \end{align*} Hence the formula is \[T(\mathbf{x})=\begin{bmatrix} x+y \\ 6x-2y \\ 8x-2y \end{bmatrix}.
  
  ### Solution 2 (Matrix representation)
  
  In the second solution, we use the matrix representation for the linear transformation $T$.
  Let $A$ be the matrix of $T$ with respect to the standard basis $\{\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}\}$ of $\R^2$.
  Thus, we have $T(\mathbf{x})=A\mathbf{x}$ by definition.
  
  To find the matrix $A$, we compute
  \begin{align*}
  A\begin{bmatrix}
  1 & 1\\
  1& -1
  \end{bmatrix}&=A\begin{bmatrix}
  \mathbf{v}_1 & \mathbf{v}_2 \\
  \end{bmatrix}\\[6pt] &=\begin{bmatrix}
  A\mathbf{v}_1 & A\mathbf{v}_2
  \end{bmatrix}\\[6pt] &=\begin{bmatrix}
  2 & 0 \\
  4 & 8 \\
  6 &10
  \end{bmatrix}
  \end{align*}
  It follows that we have
  \begin{align*}
  A&=\begin{bmatrix}
  2 & 0 \\
  4 & 8 \\
  6 &10
  \end{bmatrix}\begin{bmatrix}
  1 & 1\\
  1& -1
  \end{bmatrix}^{-1}\\[6pt] &=\begin{bmatrix}
  2 & 0 \\
  4 & 8 \\
  6 &10
  \end{bmatrix}\frac{1}{2}\begin{bmatrix}
  1 & 1\\
  1& -1
  \end{bmatrix}\\[6pt] &=\begin{bmatrix}
  1 & 1 \\
  6 & -2 \\
  8 &-2
  \end{bmatrix}.
  \end{align*}
  
  We now be able to find $T(\mathbf{x})$ as follows.
  We have
  \begin{align*}
  T(\mathbf{x})&=A\mathbf{x}\\[6pt] &=\begin{bmatrix}
  1 & 1 \\
  6 & -2 \\
  8 &-2
  \end{bmatrix}\begin{bmatrix}
  x \\
  y
  \end{bmatrix}\\[6pt] &=\begin{bmatrix}
  x+y \\
  6x-2y \\
  8x-2y
  \end{bmatrix},
  \end{align*}
  which is, of course, the same formula that we obtained in solution 1.
  
  ## Related Question.
  
  A similar problem for a linear transformation from $\R^3$ to $\R^3$ is given in the post “Determine linear transformation using matrix representation“.
  
  Instead of finding the inverse matrix in solution 1, we could have used the Gauss-Jordan elimination to find the coefficients.
  See the post “Give a formula for a linear transformation if the values on basis vectors are known” for a similar problem and its solution using this alternative method.
  
  Sponsored Links
  
  ### 3 Responses
  
  1. 03/16/2017
  
  […] the post “Give a formula for a linear transformation from $R^2$ to $R^3$” for a similar problem. Two solutions are given. One uses the inverse matrix and the other […]
  
  2. 03/16/2017
  
  […] A similar problem with a linear transformation from $R^2$ to $R^3$ is given in the post “Give a formula for a linear transformation from $R^2$ to $R^3$“. […]
  
  3. 03/26/2017
  
  […] Give a formula for a linear transformation from […]
  
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explanation: The provided text is a detailed solution to a linear algebra problem. It describes a problem where a linear transformation \( T: \mathbb{R}^2 \to \mathbb{R}^3 \) is defined by its action on a basis of \(\mathbb{R}^2\). The problem asks to find the formula for \( T(\mathbf{x}) \) for any vector \(\mathbf{x} \in \mathbb{R}^2\). The text provides two solutions to this problem, one using linear combinations and the other using matrix representation.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Give a Formula For a Linear Transformation From $\R^2$ to $\R^3$
  
  ## Problem 339
  
  Let $\{\mathbf{v}_1, \mathbf{v}_2\}$ be a basis of the vector space $\R^2$, where
  $\mathbf{v}_1=\begin{bmatrix} 1 \\ 1 \end{bmatrix} \text{ and } \mathbf{v}_2=\begin{bmatrix} 1 \\ -1 \end{bmatrix}.$ The action of a linear transformation $T:\R^2\to \R^3$ on the basis $\{\mathbf{v}_1, \mathbf{v}_2\}$ is given by
  \begin{align*}
  T(\mathbf{v}_1)=\begin{bmatrix}
  2 \\
  4 \\
  6
  \end{bmatrix} \text{ and } T(\mathbf{v}_2)=\begin{bmatrix}
  0 \\
  8 \\
  10
  \end{bmatrix}.
  \end{align*}
  
  Find the formula of $T(\mathbf{x})$, where
  $\mathbf{x}=\begin{bmatrix} x \\ y \end{bmatrix}\in \R^2.$
  
  Sponsored Links
  
  ## Solution.
  
  We give two solutions.
  
  ### Solution 1 (linear combination)
  
  Since we know the values of $T$ on the basis vectors $\mathbf{v}_1, \mathbf{v}_2$, if we express the vector $\mathbf{x}$ as a linear combination of $\mathbf{v}_1, \mathbf{v}_2$, we can find $F(\mathbf{x})$ by the linearity of the linear transformation $T$.
  
  So let us find the scalars $c_1, c_2$ such that
  $\mathbf{x}=c_1\mathbf{v}_1+c_2\mathbf{v}_2.$ We write this as
  \begin{align*}
  \begin{bmatrix}
  x \\
  y
  \end{bmatrix}=c_1\begin{bmatrix}
  1 \\
  1
  \end{bmatrix}+c_2\begin{bmatrix}
  1 \\
  -1
  \end{bmatrix}
  =
  \begin{bmatrix}
  1 & 1\\
  1& -1
  \end{bmatrix}\begin{bmatrix}
  c_1 \\
  c_2
  \end{bmatrix}.
  \end{align*}
  
  The matrix $\begin{bmatrix} 1 & 1\\ 1& -1 \end{bmatrix}$ is invertible (as its determinant is $-2$) and its inverse matrix is
  $\begin{bmatrix} 1 & 1\\ 1& -1 \end{bmatrix}^{-1}=\frac{1}{2}\begin{bmatrix} 1 & 1\\ 1& -1 \end{bmatrix}.$
  
  Thus, we have
  \begin{align*}
  \begin{bmatrix}
  c_1 \\
  c_2
  \end{bmatrix}&=\begin{bmatrix}
  1 & 1\\
  1& -1
  \end{bmatrix}^{-1}\begin{bmatrix}
  x \\
  y
  \end{bmatrix}\6pt] &=\frac{1}{2}\begin{bmatrix} 1 & 1\\ 1& -1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}\\[6pt] &=\frac{1}{2}\begin{bmatrix} x+y \\ x-y \end{bmatrix} \end{align*} Therefore, we obtain the linear combination \[\mathbf{x}=\frac{1}{2}(x+y)\mathbf{v}_1+\frac{1}{2}(x-y)\mathbf{v}_2.
  
  Now we compute $T(\mathbf{x})$ as follows.
  We have
  \begin{align*}
  T(\mathbf{x})&=T\left(\frac{1}{2}(x+y)\mathbf{v}_1+\frac{1}{2}(x-y)\mathbf{v}_2 \right)\6pt] &=\frac{1}{2}(x+y)T(\mathbf{v}_1)+\frac{1}{2}(x-y)T(\mathbf{v}_2) && \text{by linearity of T}\\[6pt] &=\frac{1}{2}(x+y)\begin{bmatrix} 2 \\ 4 \\ 6 \end{bmatrix}+\frac{1}{2}(x-y)\begin{bmatrix} 0 \\ 8 \\ 10 \end{bmatrix} \\[6pt] &=\begin{bmatrix} x+y \\ 6x-2y \\ 8x-2y \end{bmatrix}. \end{align*} Hence the formula is \[T(\mathbf{x})=\begin{bmatrix} x+y \\ 6x-2y \\ 8x-2y \end{bmatrix}.
  
  ### Solution 2 (Matrix representation)
  
  In the second solution, we use the matrix representation for the linear transformation $T$.
  Let $A$ be the matrix of $T$ with respect to the standard basis $\{\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}\}$ of $\R^2$.
  Thus, we have $T(\mathbf{x})=A\mathbf{x}$ by definition.
  
  To find the matrix $A$, we compute
  \begin{align*}
  A\begin{bmatrix}
  1 & 1\\
  1& -1
  \end{bmatrix}&=A\begin{bmatrix}
  \mathbf{v}_1 & \mathbf{v}_2 \\
  \end{bmatrix}\\[6pt] &=\begin{bmatrix}
  A\mathbf{v}_1 & A\mathbf{v}_2
  \end{bmatrix}\\[6pt] &=\begin{bmatrix}
  2 & 0 \\
  4 & 8 \\
  6 &10
  \end{bmatrix}
  \end{align*}
  It follows that we have
  \begin{align*}
  A&=\begin{bmatrix}
  2 & 0 \\
  4 & 8 \\
  6 &10
  \end{bmatrix}\begin{bmatrix}
  1 & 1\\
  1& -1
  \end{bmatrix}^{-1}\\[6pt] &=\begin{bmatrix}
  2 & 0 \\
  4 & 8 \\
  6 &10
  \end{bmatrix}\frac{1}{2}\begin{bmatrix}
  1 & 1\\
  1& -1
  \end{bmatrix}\\[6pt] &=\begin{bmatrix}
  1 & 1 \\
  6 & -2 \\
  8 &-2
  \end{bmatrix}.
  \end{align*}
  
  We now be able to find $T(\mathbf{x})$ as follows.
  We have
  \begin{align*}
  T(\mathbf{x})&=A\mathbf{x}\\[6pt] &=\begin{bmatrix}
  1 & 1 \\
  6 & -2 \\
  8 &-2
  \end{bmatrix}\begin{bmatrix}
  x \\
  y
  \end{bmatrix}\\[6pt] &=\begin{bmatrix}
  x+y \\
  6x-2y \\
  8x-2y
  \end{bmatrix},
  \end{align*}
  which is, of course, the same formula that we obtained in solution 1.
  
  ## Related Question.
  
  A similar problem for a linear transformation from $\R^3$ to $\R^3$ is given in the post “Determine linear transformation using matrix representation“.
  
  Instead of finding the inverse matrix in solution 1, we could have used the Gauss-Jordan elimination to find the coefficients.
  See the post “Give a formula for a linear transformation if the values on basis vectors are known” for a similar problem and its solution using this alternative method.
  
  Sponsored Links
  
  ### 3 Responses
  
  1. 03/16/2017
  
  […] the post “Give a formula for a linear transformation from $R^2$ to $R^3$” for a similar problem. Two solutions are given. One uses the inverse matrix and the other […]
  
  2. 03/16/2017
  
  […] A similar problem with a linear transformation from $R^2$ to $R^3$ is given in the post “Give a formula for a linear transformation from $R^2$ to $R^3$“. […]
  
  3. 03/26/2017
  
  […] Give a formula for a linear transformation from […]
  
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  ##### 12 Examples of Subsets that Are Not Subspaces of Vector Spaces
  
  Each of the following sets are not a subspace of the specified vector space. For each set, give a reason...
  
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distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text is a detailed solution to a linear algebra problem. It describes a problem where a linear transformation \( T: \mathbb{R}^2 \to \mathbb{R}^3 \) is defined by its action on a basis of \(\mathbb{R}^2\). The problem asks to find the formula for \( T(\mathbf{x}) \) for any vector \(\mathbf{x} \in \mathbb{R}^2\). The text provides two solutions to this problem, one using linear combinations and the other using matrix representation.
  
  Label: question-and-answer