Link to Problem: HERE
Determine if the lienar transformation \(S:P_3 \rightarrow M_{22}\) is (a) injective, (b) surjective, (c) invertible.
\[S(a+bx+cx^2+dx^3)=\left[\begin{array} {rrr} -a+4b+c+2d & 4a-b+6c-d\\ a+5b-2c+2d & a+2c+5d \\ \end{array}\right]\]
Injectivity can be determined by the kernal of S as per theorem KILT. The kernal can be easily found by augmenting the transformation matrix and row reducing the homogenious system.
\[-a+4b+c+2d=0\\ 4a-b+6c-d=0\\ a+5b-2c+2d=0\\ a+2c+5d=0\]
A <- matrix(c(-1, 4, 1, 1, 4, -1, 5, 0, 1, 6, -2, 2, 2, -1, 2, 5), 4, 4)
pracma::rref(A)## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1\[\left[\begin{array} {rrr} -1 & 4 & 1 & 2\\ 4 & -1 & 6 & -1 \\ 1 & 5 & -2 & 2 \\ 1 & 0 & 2 & 5 \end{array}\right] \xrightarrow{RREF} \left[\begin{array} {rrr} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \] The identify matrix is nonsingular and nonsingular matrices of homogenious systems have a trivial null space. Therefor by Theorem KILT, the transformation is injective.
Surjectivity is now easily calculated by using the kernal of S (calculated above) and theorem RPNDD which states:
\[dim(U) - n(S) = r(S)\]
The dimension of U is 4 and the kernal is 0, leaving \(r(S)=4\). Therefore by theorem RSLT S is surjective.
Invertibility is always simple to addressed once injectivity and surjectivity (or lack thereof) have been established. By theorem ILTIS, a transformation is invertible if it is both injective and surjective. As S meets both these requirements, S is invertible.
Thus, S has been established as injective, surjective and invertible.