C50 Consider the linear transformation \(S:\quad { M }_{ 12 }\quad \longrightarrow \quad { P }_{ 1 }\\\) from the set of 1 times 2 matrices to the set of polynomials of degree at most 1, defined by \(S\left( \left[ a,b \right] \right) \quad =\quad \left(3a\quad +\quad b\right)\quad +\quad \left(5a\quad +2b\right)x\) Prove that S is invertible. Then show that the linear transformation \(R:\quad { P }_{ 1 }\quad \longrightarrow \quad { M }_{ 12 },\quad R\left( r\quad +sx \right) \quad =\quad \left[ \left( 2r-s \right) ,\left( -5r+3s \right) \right]\) is the inverse of S, that is \(\quad{ S }^{-1} \quad= \quad R\).
S is injective if and only if for \({u}_{1}\quad and\quad {u}_{2}\quad \epsilon\quad {M}_{12},\quad S\left({u}_{1}\right) = S\left({u}_{}\right)\quad \longrightarrow\quad {u}_{1}\quad =\quad {u}_{2}\)
\(\\ { u }_{ 1 }\quad =\quad \left[ { a }_{ 1 },\quad { b }_{ 1 } \right] \quad \quad { u }_{ 2 }\quad =\quad \left[ { a }_{ 2 },\quad { b }_{ 2 } \right] \\ S\left( { u }_{ 1 } \right) \quad =\quad \left( 3{ a }_{ 1 }\quad +\quad { b }_{ 1 } \right) \quad +\quad \left( 5{ a }_{ 1 }\quad +\quad 2{ b }_{ 1 } \right) x\quad \quad S\left( { u }_{ 2 } \right) \quad =\quad \left( 3{ a }_{ 2 }\quad +\quad { b }_{ 2 } \right) \quad +\quad \left( 5{ a }_{ 2 }\quad +\quad 2{ b }_{ 2 } \right) x\\ Taking\quad S\left( { u }_{ 1 } \right) \quad =\quad S\left( { u }_{ 2 } \right) \quad \longrightarrow \quad S\left( { u }_{ 1 } \right) \quad -\quad S\left( { u }_{ 2 } \right) \quad =\quad 0\)
\(S\left( { u }_{ 1 } \right) \quad -\quad S\left( { u }_{ 2 } \right) \quad =\quad 0\)
Using their expressions, we have: \(\left( 3{ a }_{ 1 }\quad +\quad { b }_{ 1 } \right) \quad +\quad \left( 5{ a }_{ 1 }\quad +\quad 2{ b }_{ 1 } \right) x\quad -\quad \left( 3{ a }_{ 2 }\quad +\quad { b }_{ 2 } \right) \quad +\quad \left( 5{ a }_{ 2 }\quad +\quad 2{ b }_{ 2 } \right) x \quad = \quad0\\\)
Reordering the equation \(\left[\left( 3{ a }_{ 1 }\quad +\quad { b }_{ 1 } \right)\quad - \quad \left( 3{ a }_{ 2 }\quad +\quad { b }_{ 2 } \right) \right]\quad +\quad \left[\left( 5{ a }_{ 1 }\quad +\quad 2{ b }_{ 1 } \right)\quad - \quad \left( 5{ a }_{ 2 }\quad +\quad 2{ b }_{ 2 } \right)\right]x\)
\(\left[\left( 3{ a }_{ 1 }\quad - \quad3{ a }_{ 2 }\right)\quad\quad +\quad \left( { b }_{ 1 } \quad - \quad { b }_{ 2 } \right) \right]\quad +\quad \left[\left( 5{ a }_{ 1 }\quad -\quad 5{ a }_{ 2 }\right)\quad + \quad \left( 2{ b }_{ 1 }\quad-\quad2{ b }_{ 2 }\right)\right]x\)
\(\left[3\left( { a }_{ 1 }\quad - \quad{ a }_{ 2 }\right)\quad\quad +\quad \left( { b }_{ 1 } \quad - \quad { b }_{ 2 } \right) \right]\quad +\quad \left[5\left( { a }_{ 1 }\quad -\quad { a }_{ 2 } \right)\quad + \quad 2\left({ b }_{ 1 }\quad-\quad{ b }_{ 2 }\right)\right]x\\\)
For all x in R we obtain this system of equation: \(\begin{cases}3\left( { a }_{ 1 }\quad - \quad{ a }_{ 2 }\right)\quad\quad +\quad \left( { b }_{ 1 } \quad - \quad { b }_{ 2 } \right)\quad= \quad0 \\5\left( { a }_{ 1 }\quad -\quad { a }_{ 2 } \right)\quad + \quad 2\left({ b }_{ 1 }\quad-\quad{ b }_{ 2 }\right)\quad =\quad0 \end{cases}\\\)
We can transform to the product of matrices. \(\begin{pmatrix}3 & 1\\ 5 & 2 \end{pmatrix}\begin{pmatrix} { a }_{ 1 }\quad - \quad{ a }_{ 2 } \\ { b }_{ 1 }\quad-\quad{ b }_{ 2 } \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \end{pmatrix}\)
The determinant of the matrix is 6 - 5 = 1
The only solution of that system of equation is the triaval one. Then \({ a }_{ 1 }\quad - \quad{ a }_{ 2 }\quad =\quad 0\\{ b }_{ 1 }\quad-\quad{ b }_{ 2 }\quad=\quad0\)
\({ a }_{ 1 }\quad = \quad{ a }_{ 2 }\\{ b }_{ 1 }\quad=\quad{ b }_{ 2 }\)
This means \({u}_{1}\quad =\quad {u}_{2}\) S is injective
The linear transformation \(S:\quad { M }_{ 12 }\quad \longrightarrow \quad { P }_{ 1 }\\\) is surjective if for every v in {P}_{1} there is a u in S so that \(S\left(u\right)\quad =\quad v\)
Let v = c + dx, where c and d are real numbers, be an arbitrary function of {P}_{1} We must find u so that \(S\left(u\right)\quad =\quad v\)
\(c\quad +\quad dx\quad =\quad \left(3a\quad +\quad b\right)\quad +\quad \left(5a\quad +2b\right)x\)
\(\begin{cases} 3a\quad +\quad b\quad= \quad c \\ 5a\quad +\quad2b\quad = \quad d \end{cases}\)
We are looking a and b in term of c and d. The derterminant of the matrix is 1 The ony solution is the couple (2c - d, -5c + 3d) For any input u(2c - d, -5c + d) we will get v(c,d) S is surjective
Since S is injective and surjective, then S is invertible.
\(SoR\left(v\right)\quad =\quad S\left[R\left(r+sx\right)\right]\quad = \quad S\left[2r-s,\quad -5r+3s\right]\\ =\quad \left[3\left(2r-s\right)\quad + \quad \left(-5r+3s\right) \right]\quad + \quad \left[5\left(2r-s\right)\quad + \quad 2\left(-5r+3s\right)\right]x\\ =\quad \left(6r-3s-5r+3s\right)+\left(10r-5s-10r+6s\right)x\\=\quad r\quad +\quad sx\\=\quad v\\SoR\quad =\quad{I}(P_{1})\)
\(RoS\left(u\right)\quad = \quad R\left[\left(3a+b\right)\quad + \quad \left(5a+2b\right)x\right]\\ = \quad \left[2\left(3a+b\right)-\left(5a+2b\right),\quad -5\left(3a+b\right)+3\left(5a+2b\right)\right]\\=\quad \left[6a+2b-5a-2b,\quad-15a-5b+15a+6b\right]\\=\quad \left[a,\quad b\right]\\ =\quad u\\ RoS\quad = \quad I({M}_{12})\)
We shown that \(R\quad =\quad {S}^{-1}\)