[Math] Computing the inverse of linear transformations using matrices

Question

For each of the following linear transformations T, determine whether T is invertible, and compute T^-1 if it exists.

(a) T: P₂(R) $\to$ P₂(R) defined by T(f(x)) = f ''(x) + 2 f '(x) – f(x).

My answer:

Using the standard basis of P₂(R), $\beta$ = {1, x, x²} I got that the matrix representation of T is
\begin{matrix}
-1 & 2 & 2 \\
0 & -1 & 4 \\
0 & 0 & -1 \\
\end{matrix}

Taking its inverse, I got
\begin{matrix}
-1 & -2 & -10 \\
0 & -1 & -4 \\
0 & 0 & -1 \\
\end{matrix}

However, from here, I don't know how to use the inverse matrix to define T^-1.

Similarly,

(b) T: R³ $\to$ P₂(R) defined by T(a, b, c) = (a+b+c)+(a–b+c)x+ a x²

Matrix representation using basis of R³, {(1, 0, 0), (0,1,0), (0,0,1)} and basis of P₂(R), {1, x, x²}
\begin{matrix}
1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 0 & 0 \\
\end{matrix}

Taking the inverse I get
\begin{matrix}
0 & 0 & 1 \\
1/2 & -1/2 & 0 \\
1/2 & 1/2 & -1 \\
\end{matrix}

And again, I don't know how to use this inverse to define T^-1: __________.

Similarly,

(c) T: P₂(R) $\to$ R³ defined by T(f(x)) = (f(-1), f(0), f(1))

Using the same bases as in (b), I got the matrix representation
\begin{matrix}
1 & -1 & 1 \\
1 & 0 & 0 \\
1 & 1 & 1 \\
\end{matrix}

with inverse
\begin{matrix}
0 & 1 & 0 \\
-1/2 & 0 & 1/2 \\
1/2 & -1 & 1/2 \\
\end{matrix}

Again, I don't know how to use this information to define T. Please help me, especially in the last two problems where the vector spaces are different. Thanks!

Answer 1

This is simply a question of what "matrix of a linear transformation" means. For your third question you have $$C^{-1}\pmatrix{a\cr b\cr c\cr} =\pmatrix{b\cr-\frac12a+\frac12c\cr \frac12a-b+\frac12c\cr}\ ,$$ assuming your calculated inverse is correct (I haven't checked). Using the bases you have given in (b), this means $$\eqalign{T^{-1}(a\color{red}{(1,0,0)}+&b\color{red}{(0,1,0)}+c\color{red}{(0,0,1)})\cr &=b\color{blue}{1}+\bigl(-\frac12a+\frac12c\bigr)\color{blue}{x} +\bigl(\frac12a-b+\frac12c\bigr)\color{blue}{x^2}\ ,\cr}\tag{$*$}$$ that is, $$T^{-1}(a,b,c)=b+\bigl(-\frac12a+\frac12c\bigr)x +\bigl(\frac12a-b+\frac12c\bigr)x^2\ .$$

You can easily check that $T(T^{-1}({\bf v}))=\bf v$ and $T^{-1}(T(f))=f$.