# Show that operator $S$ is invertible

Problem: $S$ is an operator on $\mathbb{R}_2[x]$ and is given by $S(a+bx+cx^2)=(a+b+2c)+(2a+3b+2c)x+(a+3b+2c)x^2$. Show that operator S is invertible.

Attempt: If we can show that there is an $S^{-1}$, then we are done. Let A be the left inverse of S, then $A \circ S = I$ where $I$ is the identity polynomial. Then let $A(a+b+2c)+(2a+3b+2c)x+(a+3b+2c)x^2 = 1$. We can solve this to get $a=0$, $b=2$, and $c=-3$. I am stuck here. How do I show that $A$ is also a right inverse and thus $S^{-1}$?

• You want an operator $T$ such that $T(S(p)) = S(T(p)) = p$, for all $p$ of the form $p(x) = ax^2 + bx + c$. Commented Oct 9, 2017 at 5:26

You are taking a complicated road. Let's try somethig easy. $\mathbb{R}_2[x]$ has finite dimension and $S$ maps this space into it self. So, if we check $$\dim \ker S =0$$ Automatically is a isomorphism. Ok, let $f = a + bx + cx^2 \in \ker S$. Then $S(f) =0$, i.e., $$(a+b+c) + (2a + 3b + 2c)x + (a + 3b + 2c)x^2 = 0$$ Then $$a+b+c =0 \\ 2a + 3b + 2c = 0 \\ a + 3b + 2c = 0$$ A 3x3 linear system. But note: $(1, 1, 1), (2, 3, 2), (1, 3, 2)$ are LI (can you check it?), so this system has only the trivial solution. So $f = 0$ and our work was done.

Note: what I did i basically the same of: take a basis (in this case, we already have a basis: {1, x, x^2}) and check if it's invertible (you can take de determinant to see that, it's really simple in this low dimensional problem). But I think it's important to remember Rank–Nullity Theorem and I like to use it.

If you prefer the matrix stuff, here we go: $$\beta = \{1, x, x^2\}$$ is a basis to $\mathbb{R}_2[x]$ and we have $$S(1) = 1 + 2x + x^2\\ S(x) = 1 + 3x + 3x^2\\ S(x^2) = 1 + 2x + 2x^2$$ So, in $\beta$ we can write $S$ as the matrix $$[\mathrm{S}]_{\beta} = \begin{bmatrix}1&1&1\\2&3&2 \\1&3&2\end{bmatrix}$$ (compare with the linear systen we did) As observed by Michael Lee we have $\det [\mathrm{S}] = 1 \neq 0$ so, it's invertible. Nice. But how to find it's inverse? Well, we can invert this matrix. Reverse an matrix is really a terrible pain... you can do Gaussian Elimination or cofactor/adjoint stuff... all options sound terrible to a human. I calculated the adjoint matrix because I have nothing to do:

$$Cof[\mathrm{S}]_{\beta} = \begin{bmatrix}0&-2&3\\1&1&-2 \\-1&0&1\end{bmatrix}$$

$$[\mathrm{S}^{-1}]_{\beta} = \frac{1}{\det [\mathrm{S}]} (Cof[\mathrm{S}]_{\beta})^{t} = \begin{bmatrix}0&1&-1\\-2&1&0 \\3&-2&1\end{bmatrix}$$

I checked here and all computations are ok. Take a vector $$f = a +bx + cx^2 \Rightarrow [f]_{\beta} = \begin{bmatrix}a\\b\\c\end{bmatrix}$$ Then $$[S^{-1}f]_{\beta} = [S^{-1}]_{\beta} [f]_{\beta} = \begin{bmatrix}0&1&-1\\-2&1&0 \\3&-2&1\end{bmatrix} \cdot \begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}b-c\\-2a+b\\3a+2b+c\end{bmatrix} \\ \therefore S^{-1}(a + bx + cx^2) = (b-c) + (-2a+b)x + (3a + 2b + c)x^2$$ We found the inverse (ufa).

• Alternatively, $$\det\begin{pmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ 1 & 3 & 2 \end{pmatrix} = 1\neq 0$$ Commented Oct 9, 2017 at 5:21
• Checking linear independence of a set of $n$ vectors of length $n$ has the same computational complexity as calculating the determinant of an $n\times n$ matrix. Commented Oct 9, 2017 at 5:24
• @MichaelLee I do not have the patience to writing matrices in latex (sorry), but I edited my answer to explain it (I hope so). Commented Oct 9, 2017 at 5:26
• A small remark : "Reverse a matrix is really a terrible pain..." if ypu do it by hand. But is it the future ? I don't think so ; we have always a computing device close to us that can do the work in an instant manner... Commented Oct 9, 2017 at 6:52