Example IAV Injective, Archetype V

Archetype V is defined by \begin{equation*} \ltdefn{T}{P_3}{M_{22}},\quad\lt{T}{a+bx+cx^2+dx^3}= \begin{bmatrix} a+b & a-2c\\ d & b-d \end{bmatrix} \end{equation*}

To establish that the linear transformation is injective, begin by supposing that two polynomial inputs yield the same output matrix, \begin{equation*} \lt{T}{a_1+b_1x+c_1x^2+d_1x^3}=\lt{T}{a_2+b_2x+c_2x^2+d_2x^3} \end{equation*}

Then \begin{align*} \zeromatrix &=\begin{bmatrix} 0&0\\0&0 \end{bmatrix}\\ &=\lt{T}{a_1+b_1x+c_1x^2+d_1x^3}-\lt{T}{a_2+b_2x+c_2x^2+d_2x^3}&&\text{Hypothesis}\\ &=\lt{T}{(a_1+b_1x+c_1x^2+d_1x^3)-(a_2+b_2x+c_2x^2+d_2x^3)}&&\knowl{./knowls/definition.LT.knowl}{\text{Definition LT}}\\ &=\lt{T}{(a_1-a_2)+(b_1-b_2)x+(c_1-c_2)x^2+(d_1-d_2)x^3}&&\text{Operations in $P_3$}\\ &= \begin{bmatrix} (a_1-a_2)+(b_1-b_2) & (a_1-a_2)-2(c_1-c_2)\\ (d_1-d_2) & (b_1-b_2)-(d_1-d_2) \end{bmatrix}&&\text{Definition of $T$} \end{align*}

This single matrix equality translates to the homogeneous system of equations in the variables $a_i-b_i$, \begin{align*} (a_1-a_2)+(b_1-b_2)&=0\\ (a_1-a_2)-2(c_1-c_2)&=0\\ (d_1-d_2)&=0\\ (b_1-b_2)-(d_1-d_2)&=0 \end{align*}

This system of equations can be rewritten as the matrix equation \begin{equation*} \begin{bmatrix} 1&1&0&0\\1&0&-2&0\\0&0&0&1\\0&1&0&-1 \end{bmatrix} \colvector{(a_1-a_2)\\(b_1-b_2)\\(c_1-c_2)\\(d_1-d_2)}=\colvector{0\\0\\0\\0} \end{equation*}

Since the coefficient matrix is nonsingular (check this) the only solution is trivial, i.e. \begin{align*} a_1-a_2&=0&b_1-b_2&=0&c_1-c_2&=0&d_1-d_2&=0 \end{align*} so that \begin{align*} a_1&=a_2&b_1&=b_2&c_1&=c_2&d_1&=d_2 \end{align*} so the two inputs must be equal polynomials. By Definition ILT, $T$ is injective.