My professor does not give examples in his lecture notes, i'm trying to develop the method to do this exam question:

Let $U = P3(\mathbb{R})$, (the vector space of polynomials of degree at most 3 in a formal variable $t$).

Then $f : U → R$ is the linear map defined by $f(u) = u′(1)$, where $u′(1)$ is the derivative of $u$ (with respect to $t$) evaluated at $t = 1$.

1) Determine the matrix of $f$ with respect to the bases $\{1,t,t2,t3\}$ and $\{1\}$.

2) Determine the $rank$ and the $nullity$ of $f$, and find a $basis$ of the $kernel$ of $f$.


The method is simple $P_3(\Bbb{R})$ has $\{1,x,x^2,x^3\}$ as a basis and $\Bbb{R}$ is one dimensional with $\{1\}$ as a basis. The derivatives of the basis vectors of $U$ are ${0,1,2x,3x^2}$ and so the matrix of $f$ in the given bases is

$$F=\begin{bmatrix} 0&1&2&3\end{bmatrix}$$

Obviously $f$ is of rank $1$ because the codomain is one dimensional and $f$ is not identically $0$.

The kernel is the set of polynomials

$$\{a_0+a_1x+a_2x^2+a_3x^3,\, a_1+2a_2+3a_3=0\}$$

It is a hyperplane of $U$



for a polynomial $a+bx+cx^2+dx^3$ your linear map act as: $$ f(a+bx+cx^2+dx^3)=b+2c+3d $$

In vector notation with the standard basis $\{1,x,x^2,x^3\}$ and $\{1\}$ this is the transformation: $$ \begin{bmatrix} a\\b\\c\\d \end{bmatrix} \to [b+2c+3d] $$ that is represented by the matrix: $$ f=[0,1,2,3] $$ $$ [0,1,2,3]\begin{bmatrix} a\\b\\c\\d \end{bmatrix} = b+2c+3d $$

Now you can easily see what is the kernel of this transformation and find the answer to the question 2).


The general idea is described here: https://www.youtube.com/watch?v=0nTKoXOC8qI&index=59&list=PLlXfTHzgMRUIqYrutsFXCOmiqKUgOgGJ5 For an example with polynomials, see a few videos down the list. Those videos are for transformations onto the space itself. When going from one space to another, the only difference is that you have to choose different a basis in each space.

In your example, the bases are given. The result will be a $1\times 4$ matrix, each column corresponding to each of the elements of the first basis, representing the decomposition of the result of the transformation with respect to the other basis. This may be a little bit confusing since the target space is one-dimensional and decomposition is trivial.

$1\rightarrow 1'= 0 \text{, and 0 evaluated at 1 is 0}$

$t\rightarrow t'= 1 \text{, and 1 evaluated at 1 is 1}$

$t^2\rightarrow (t^2)'= 2t \text{, and 2t evaluated at 1 is 2}$

$t^3\rightarrow (t^3)'= 3t^2 \text{, and $3t^2$ evaluated at 1 is 3}$

Therefore the matrix is $A=\left[\begin{matrix} 0 & 1 & 2 & 3 \end{matrix}\right]$

A basis (chosen arbitrarily) for the null space of this matrix is

$$\left \{ \begin{bmatrix}\ 1\\0\\0\\0 \end{bmatrix},\begin{bmatrix}\ 0\\2\\-1\\0 \end{bmatrix},\begin{bmatrix}\ 0\\3\\0\\-1 \end{bmatrix}\right\}$$ which corresponds to the following basis for the kernel $$ \left\{ 1, 2x - x^2, 3x-x^3 \right\} $$

All of this is described in utmost detail with exercises in the Linear Algebra section on the Lemma system at http://lem.ma.

  • $\begingroup$ How do you choose the basis for the null space? $\endgroup$
    – HELP
    Apr 17 '16 at 19:01
  • $\begingroup$ I'm not sure how to answer this question without starting at the very beginning. I will soon record a number of videos on null spaces of matrices. For now, may I point you to a video where the null space is calculated for a matrix like this in the context of a linear system: lem.ma/6o. $\endgroup$
    – Lemma
    Apr 17 '16 at 20:04
  • $\begingroup$ Here's a link to practice the null space on its own: lem.ma/V1 $\endgroup$
    – Lemma
    Apr 23 '16 at 16:07

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