Finding the pre-image of a vector

Question

I was asked to find ${\alpha}$ such that, a non injective linear transformation is well defined. The problem is solved, but I'm not sure if my procedure was ok, because I'm supporting my idea very weakly... and when solving this things we're asked to support our procedure on theorems and so.

$T: P_2 → P_2 $

$T(t+\alpha t^2)=1+\alpha t + t^2 \\ T(1+t+t^2)=1+2t + t^2 \\ T(1+\alpha t+t^2)=\alpha +2t\\$

I said, if it's a well defined linear transformation, then, the three vectors we're using must be linear independent, so it's a basis of $P_2$. On the other side, if I want it to be a non inyective linear transformation, the dimension of the image should be less than the dimension of the domain, so one of it's vectors must be a linear combination of the others. I did the math, and found that $\alpha = 2$

$a)\ T(t+2t^2)=1+2t + t^2 \\ b)\ T(1+t+t^2)=1+2t + t^2 \\ c)\ T(1+2t+t^2)=2 +2t\\$

Now, I was asked to find all vectors such that $T(P)=t+t^2$

Before I jump into the pool, just by looking, I can create $t+t^2$ with the images of the basis vectors I've been given in two ways, one being $(a-c/2)$ and $(b-c/2)$ . But I wanted to do it by calculating the matrix associated to this linear transformation, and I ended up with this matrix.

$$A= \begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 1 & -1 & 1\\ \end{pmatrix} $$ Now, if I want to find the the subspace that creates $t+t^2$ I must solve the following system (which is calculating the preimage of that vector): $$A= \begin{pmatrix} 0 & 1 & 0 & | & 0 \\ 1 & 0 & 1 & | & 1 \\ 1 & -1 & 1 & | & 1\\ \end{pmatrix} $$

Now... I end up with 2 quations and 3 variables, therefore, I only need one parameter.

$$A= \begin{pmatrix} 0 & 1 & 0 & | & 0 \\ 1 & 0 & 1 & | & 1 \\ \end{pmatrix} \\P_2 = a_1 + a_2 t + a_3 t^2 $$ By looking at the matrix, $ a_1 = a_1, a_2 = 0; a_3 = 1-a_1$

$\\P_2 = a_1 + 0 t + (1-a_1) t^2 \\P_2 = \alpha (1-t^2)+ t^2 $

Answer 1

Your solution lacks a lot of explanations and some parts are very confusing. Here is how I would approach it:

First of all I recommend you working with coordinates from start as it will make things a lot clearer. Consider the basis $B = \{1,t,t^2\}$ then you have:

$$t+\alpha t^2= (0, 1, \alpha)_B\\ 1+t+t^2=(1,1,1)_B \\ 1+\alpha t+t^2=(1, \alpha, 1)_B\\ 1+\alpha t + t^2 = (1, \alpha, 1)_B\\ 1+2t + t^2 = (1,2,1)_B\\ \alpha +2t = (1, 2, 0)_B\\$$

so then:

$$T(0, 1, \alpha)=(1, \alpha, 1)\\ T(1,1,1)= (1,2,1) \\ T(1,\alpha,1)=(\alpha,2,0)\\$$

You want to find $\alpha$ so that $T$ is not injective. Picking $\alpha=2$ you have that $T(0,1,2)=T(1,1,1)=(1,2,1)$. Now for this alpha we have to check that $T$ is well defined.

$\{(0,1,2), (1,1,1), 1,2,1)\}$ is a basis, so we can use the following theorem:

Let $V$ be a vector space of dimension $n$ and let $\{v_1,...,v_n\}$ be some basis of it. Let be any other vector space over the same field as and let $\{w_1,...,w_n\}$ be any $n$ vectors in $W$. Then there exists a unique linear transformation $T:V \rightarrow W$ such that $T(v_i)=w_i$

Using this, the function $T$ defined by the images of the basis $\{(0,1,2), (1,1,1), 1,2,1)\}$ is unique and therefore well defined.

Now for the second part, you found that the associated matrix of $T$ in the canonical basis $B$ is (I haven't checked it):

$$A=\begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 1 & -1 & 1\\ \end{pmatrix}$$

Now you're looking for vectors $(x,y,z)$ such that

$$\begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 1 & -1 & 1\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ z\\ \end{pmatrix} = \begin{pmatrix} 0\\ 1\\ 1\\ \end{pmatrix} $$

So you end up with the linear system

$$y=0\\ x+z=1\\ x-y+z=1\\$$

So $(x,y,z)$ will be the vectors satisfying $y=0, x+z=1$ or equivalently, the vectors of the form $(x,y,z)=(\lambda, 0, 1-\lambda), \lambda\in \mathbb{R}$, or in its original form, $\lambda + (1-\lambda)t^2, \lambda\in \mathbb{R}$