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The function T:R2→R3 is defined by T(x1x2)=⎛⎝⎜3x1−3x1+4x2−2x2⎞⎠⎟ for ...
Jul 15, 2024
The function T:R2→R3 is defined by T(x1x2)=⎛⎝⎜3x1−3x1+4x2−2x2⎞⎠⎟ for all (x1x2)∈R2 . Show that T is linear. To make sure you are on the right track you should answer the following questions. Is the domain of T a known vector space? Is the codomain of T a known vector space? Does T preserve vector addition? Does T preserve scalar multiplication? Write in the correct mathematical writing mode
Solution by Steps
step 1
The domain of TT is R2R^2, which is a known vector space
step 2
The codomain of TT is R3R^3, which is also a known vector space
step 3
To check if TT preserves vector addition, let u=(u1,u2)\mathbf{u} = (u_1, u_2) and v=(v1,v2)\mathbf{v} = (v_1, v_2) be vectors in R2R^2. Then, T(u+v)=T((u1+v1,u2+v2))T(\mathbf{u} + \mathbf{v}) = T((u_1 + v_1, u_2 + v_2))
step 4
Calculate T(u+v)T(\mathbf{u} + \mathbf{v}): T(u1+v1,u2+v2)=(3(u1+v1)−3(u1+v1)+4(u2+v2)−2(u2+v2))=(3u1+3v1−3u1−3v1+4u2+4v2−2u2−2v2) T(u_1 + v_1, u_2 + v_2) = \begin{pmatrix} 3(u_1 + v_1) \\ -3(u_1 + v_1) + 4(u_2 + v_2) \\ -2(u_2 + v_2) \end{pmatrix} = \begin{pmatrix} 3u_1 + 3v_1 \\ -3u_1 - 3v_1 + 4u_2 + 4v_2 \\ -2u_2 - 2v_2 \end{pmatrix}
step 5
Calculate T(u)+T(v)T(\mathbf{u}) + T(\mathbf{v}): T(u1,u2)+T(v1,v2)=(3u1−3u1+4u2−2u2)+(3v1−3v1+4v2−2v2)=(3u1+3v1−3u1−3v1+4u2+4v2−2u2−2v2) T(u_1, u_2) + T(v_1, v_2) = \begin{pmatrix} 3u_1 \\ -3u_1 + 4u_2 \\ -2u_2 \end{pmatrix} + \begin{pmatrix} 3v_1 \\ -3v_1 + 4v_2 \\ -2v_2 \end{pmatrix} = \begin{pmatrix} 3u_1 + 3v_1 \\ -3u_1 - 3v_1 + 4u_2 + 4v_2 \\ -2u_2 - 2v_2 \end{pmatrix}
step 6
Since T(u+v)=T(u)+T(v)T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}), TT preserves vector addition
step 7
To check if TT preserves scalar multiplication, let cc be a scalar and u=(u1,u2)\mathbf{u} = (u_1, u_2) be a vector in R2R^2. Then, T(cu)=T((cu1,cu2))T(c\mathbf{u}) = T((cu_1, cu_2))
step 8
Calculate T(cu)T(c\mathbf{u}): T(cu1,cu2)=(3(cu1)−3(cu1)+4(cu2)−2(cu2))=(c(3u1)c(−3u1+4u2)c(−2u2)) T(cu_1, cu_2) = \begin{pmatrix} 3(cu_1) \\ -3(cu_1) + 4(cu_2) \\ -2(cu_2) \end{pmatrix} = \begin{pmatrix} c(3u_1) \\ c(-3u_1 + 4u_2) \\ c(-2u_2) \end{pmatrix}
step 9
Calculate cT(u)cT(\mathbf{u}): cT(u1,u2)=c(3u1−3u1+4u2−2u2)=(c(3u1)c(−3u1+4u2)c(−2u2)) cT(u_1, u_2) = c \begin{pmatrix} 3u_1 \\ -3u_1 + 4u_2 \\ -2u_2 \end{pmatrix} = \begin{pmatrix} c(3u_1) \\ c(-3u_1 + 4u_2) \\ c(-2u_2) \end{pmatrix}
step 10
Since T(cu)=cT(u)T(c\mathbf{u}) = cT(\mathbf{u}), TT preserves scalar multiplication
Answer
TT is linear.
Key Concept
Linearity of a transformation
Explanation
A transformation is linear if it preserves vector addition and scalar multiplication.
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