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

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 $n+1$. –  Vectk Nov 6 '12 at 3:46

2 Answers 2

up vote 3 down vote accepted

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.

In your case :

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

I chose the coefficient from the highest degree to the smallest. That is important only to set some ground rules so you will know how to revert your cartesian product into polynomial again, if you want to do so. *You can choose any other order.

Now because these are isomorphics subspaces you'll get the same Kernel and the same Image.

I hope It helps.

Guy

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