Identifying Isomorphisms

Question

I understand that the conditions for isomorphism is that

1. close under multiplication/addition
2. T(-v) = -T(v)
3. preservation of identity

However, I'm confused about number 2. I know that 1 and 3 are both isomorphic.

But in case 2, how do we determine f(-1) or the matrix indices without the function f itself? Is it necessary to determine the isomorphism.

Answer 1

An isomorphism has to be invertible by definition. Let's see than which of the three you mention it's so:

1. $T:P\rightarrow P$ defined as $T(f)=f'$: this cannot be an isomorphism because by integrating $f'$ you get $f+c$ so a family of functions. Differentiating a polynomial get's rid of the constants and you cannot retrieve them in any way by integrating.

2. $T:P_3 \rightarrow \mathbb{R}^{2\times 2}$ defined as $T(f) = \left[\begin{matrix}f(-1)&f(0)\\f(1)&f(4)\end{matrix}\right]$: for this let's choose a base $\mathcal{B} = \{1,x,x^2,x^3\}$ for $P_3$ and than evaluate the the image of the base by $T$: $$T(1) = \left[\begin{matrix}1&1\\1&1\end{matrix}\right]\\ T(x) = \left[\begin{matrix}-1&0\\1&4\end{matrix}\right] \\ T(x^2) = \left[\begin{matrix}1&0\\1&16\end{matrix}\right] \\ T(x^3) = \left[\begin{matrix}-1&0\\1&64\end{matrix}\right] $$ let's choose the standard base for $\mathbb{R}^{2\times 2}$: $\mathcal{C} = \left\{ \left[\begin{matrix}1&0\\0&0\end{matrix}\right],\left[\begin{matrix}0&1\\0&0\end{matrix}\right] ,\left[\begin{matrix}0&0\\1&0\end{matrix}\right] ,\left[\begin{matrix}0&0\\0&1\end{matrix}\right] \right\}$ and write down the matrix of the transformation $$A=\left[\begin{matrix}1&-1&1&-1\\1&0&0&0\\1&1&1&1\\1&4&16&64\end{matrix}\right]$$ the determinant of $A$ is positive, so the map is invertible

3. $T:D_3\rightarrow P_2$ defined as $T\left(\left[\begin{matrix}a&0&0\\0&b&0\\0&0&c\end{matrix}\right]\right) = ax^2+bx+c$: let's use this basis for $D_3$ and $P_2$ $$\mathcal{B} = \{1,x,x^2\}\;\;\;\;\;\mathcal{C} = \left\{\left[\begin{matrix}1&0&0\\0&0&0\\0&0&0\end{matrix}\right], \left[\begin{matrix}0&0&0\\0&1&0\\0&0&0\end{matrix}\right], \left[\begin{matrix}0&0&0\\0&0&0\\0&0&1\end{matrix}\right]\right\}$$ and see where $T$ maps the basis to write down the matrix of this map $$T\left(\left[\begin{matrix}1&0&0\\0&0&0\\0&0&0\end{matrix}\right]\right) = x^2 \\ T\left(\left[\begin{matrix}0&0&0\\0&1&0\\0&0&0\end{matrix}\right]\right)=x\\ T\left(\left[\begin{matrix}0&0&0\\0&0&0\\0&0&1\end{matrix}\right]\right) = 1$$ then, easily enough, the matrix representation of this map is $$A = \left[\begin{matrix}0&0&1\\0&1&0\\1&0&0\end{matrix}\right]$$ which has determinant $-1$ so it's invertible.