Consider the linear transformation \(S : M_{12} → P_1\) from the set of 1 × 2 matrices to the set of polynomials of degree at most 1, defined by \(S\left(\begin{bmatrix} a & b \end{bmatrix}\right) = (3a+b)+(5a+2b)x\)
Prove that S is invertible. Then show that the linear transformation \(R:P_1 →M_{12}\), \(R(r+sx) = \begin{bmatrix} (2r-s) & (-5r+3s) \end{bmatrix}\)
To do that, we need to prove that S is both injective and surjective:
\[\underline{Injective}\]
Suppose \(c, d \in M_{12}\) and assume that \(S\left(\begin{bmatrix} a & b \end{bmatrix}\right) = S\left(\begin{bmatrix} c & d \end{bmatrix}\right)\)
\(\Rightarrow S\left(\begin{bmatrix} a & b \end{bmatrix}\right) = (3a+b)+(5a+2b)x\) And \(S\left(\begin{bmatrix} c & d \end{bmatrix}\right) = (3c+d)+(5c+2d)x\)
So:
\[ \begin{align} \begin{bmatrix} a & b \end{bmatrix} &= I_s (\begin{bmatrix} a & b \end{bmatrix}) \\ &= (S^{-1} \circ S) (\begin{bmatrix} a & b \end{bmatrix})\\ &=S^{-1} (S(\begin{bmatrix} a & b \end{bmatrix}))\\ &=S^{-1} (S(\begin{bmatrix} c & d \end{bmatrix}))\\ &=(S^{-1} \circ S) (\begin{bmatrix} c & d \end{bmatrix})\\ &=I_s (\begin{bmatrix} c & d \end{bmatrix})\\ &=\begin{bmatrix} c & d \end{bmatrix} \end{align} \] Hence, S is injective.
\[\underline{Surjective}\]
Suppose \(p \in P_1\) , then \(S^{-1}(p)\) is a vector in \(M_{12}\) ,
Compute:
$$ \[\begin{align} S(S^{-1}(p)) &= (S \circ S^{-1}) (p) \\ & = I_P (p) \\ & =p \end{align}\] $$
So, there is an element from \(M_{12}\), when used as an input to S, \(S^{-1}(p)\), that produces the desired output, p, and hence S is surjective.
Since S is both injective and surjective, then, S is invertible.
To show that, we need to show the followings:
\[ \begin{array}{cc} S : M_{12} \rightarrow P_1 \\ R : P_1 \rightarrow M_{12} \\ \end{array} \Rightarrow \begin{align} S \circ R &= I_{P_1}\\ R \circ S &= I_{M_{12}} \end{align} \]
First let’s show the first direction:
\[ S : M_{12} \rightarrow P_1 \Rightarrow S \circ R = I_{P_1} \]
$$ \[\begin{align} (S \circ R) (r+sx) & = S(R(r+sx)) \\ & = S\left( \begin{bmatrix} (2r-s) & (-5r+3s) \end{bmatrix}\right)\\ &= \left[3(2r-s) +(-5r+3s)\right] + \left[5(2r-s) + 2(-5r+3s)\right]x\\ &= (3a+b) + (5a+2b)x\\ &=I_{p_1}(r+sx) \end{align}\] $$
Now let’s show the second direction:
\[ R : P_1 \rightarrow M_{12} \Rightarrow R \circ S = I_{M_{12}} \]
So;
$$ \[\begin{align} (R \circ S)(\begin{bmatrix} a & b\end{bmatrix}) &= R(S(\begin{bmatrix} a & b \end{bmatrix}))\\ &= R[ (3a+b) +(5a+2b)x]\\ &= \begin{bmatrix} (2(3a+b) - (5a+2b)) & (-5(3a+b) + 3(5a+2b))\end{bmatrix}\\ & = \begin{bmatrix} (2r-s) & (-5r + 3s) \end{bmatrix}\\ & = I_{M_{12}} (\begin{bmatrix} a & b\end{bmatrix}) \end{align}\] $$
By showing
\[ \begin{array}{cc} S : M_{12} \rightarrow P_1 \\ R : P_1 \rightarrow M_{12} \\ \end{array} \Rightarrow \begin{align} S \circ R &= I_{P_1}\\ R \circ S &= I_{M_{12}} \end{align} \]
We show that \(S^{-1} = R\)