# Find a basis of $\ker T$ and $\dim (\mathrm{im}(T))$ of a linear map from polynomials to $\mathbb{R}^2$

$T: P_{2} \rightarrow \mathbb{R}^2: T(a + bx + cx^2) = (a-b,b-c)$

Find basis for $\ker T$ and $\dim(\mathrm{im}(T))$.

This is a problem in my textbook, it looks strange with me, because it goes from polynomial to $\mathbb{R}^2$. Before that, i just know from $\mathbb{R}^m \rightarrow \mathbb{R}^n$.

Thanks :)

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Would you be able to do it if you pretend that the polynomial is the vector $\begin{pmatrix}a & b & c\end{pmatrix}^\rm{T}$? That's essentially all it is. – EuYu Nov 6 '12 at 3:46
The set of all polynomials of degree at most $n$ with real coefficients is a vector space of dimension $n+1$. – Ink Nov 6 '12 at 3:46

The kernel of the transformation consists of all elements of $P_2$ that are sent by $T$ to the zero-vector of the target space $\mathbb{R}^2$. In your case, we want $(a-b,b-c)=(0,0)$. So we want $a=b$ and $b=c$. Thus the polynomials that are sent to $(0,0)$ are all polynomials of the shape $k+kx+kx^2$, that is, all multiples of the polynomial $1+x+x^2$.

For the dimension of the image, the image is a subset of $\mathbb{R}^2$, so has dimension $\le 2$. If you can find two linearly independent vectors in the image (which I am sure you can), you will have shown that the image has dimension $2$.

Or maybe you can refer to a general theorem. If $T$ is a linear mapping from $U$ to $V$, you may have in your list of theorems a result about the relationship between the dimension of $U$, the dimension of the kernel of $T$, and the dimension of the image of $T$.

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You can treat any polynomial $P_n$ space as an $R^n$ space.

That is, in your case the polynomial is $P_2$ and it can be converted to $R^3$.

The logic is simple, each coefficient of the term in the polynomial is converted to a number in $R^3$.

In the end of this conversion you'll get an isomorphics spaces/subspaces.

The polynomial : $a + bx + cx^2$ can be converted to the cartesian product: $(c, b, a)$