I would like to understand how the derivative affects a basis of a vector space of polynomials.

For instance suppose we have the collection of quadratic polynomials with real coefficients which has ${1, t, t^2}$ as a basis. Every quadratic polynomial can be written as $a*1+b*t+c*t^2$, that is, as a linear combination of the basis functions $1$, $t$, and $t^2$.

Now, if we differentiate each term, then we will get new values which could be used as a basis for the lower dimension?

I mean, is the derivative of basis functions always also another basis? For example, in the previous example, we had basis for the second order polynomial,but after derivative , we have basis for the linear form, so can we represent derivative operator of order $n$ let say as a transformation of basis from $n$ degree to some other $s$ degree space or whatever it is called?

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    $\begingroup$ Yes, it is a linear transformation on the set of polynomials. It has a one-dimensional kernel. If the original space is the three-dimensional set of quadratics (and lower), then it is just like a linear transformation in $\mathbb{R}^3$. On the other hand, the set of all polynomials is infinite-dimensional, which is something new. $\endgroup$ – Empy2 Sep 20 '13 at 12:35
  • $\begingroup$ Hi dato: I moved a little of the grammar around in the question, trying to preserve your intent. Please review it and rollback if you disagree with my changes. Thanks! $\endgroup$ – rschwieb Sep 20 '13 at 12:51
  • $\begingroup$ everything is ok thanks $\endgroup$ – dato datuashvili Sep 20 '13 at 12:53

For your example

It would be good to notice here that differentiation is a linear transformation. Let $V=\{\alpha_0+\alpha_1 t+\alpha_2t^2\mid \alpha_i\in \Bbb R\}$, and consider what differentiation does to this space.

We already know that $D(\alpha f+\beta g)=\alpha D(f)+\beta D(g)$ so that it's a linear transformation. Let's find the matrix with respect to the basis $\{1,t,t^2\}$ that you suggested.

A little work shows that the matrix is $A=\begin{bmatrix}0&1 &0\\0&0 &2 \\0&0 &0 \end{bmatrix}$.

Remembering that the column vector $[1,0,0]^\top$ represents $1$, $[0,1,0]^\top$ represents $t$ and $[0,0,1]^\top$ represents $t^2$, you can verify that multiplying these vectors on the left with the matrix outputs the derivative.

Several of the properties of this transformation become clear. One is that it is a nonsingular transformation since it sends $1$ to $0$. You might have already suspected that if you noticed that taking the derivative three times would turn any vector in $V$ into $0$, so obviously $A^3=0$ can't happen for a nonsingular matrix.

Since the kernel of $A$ is clearly $\{[\alpha,0,0]^\top\mid \alpha\in \Bbb R\}$, the kernel is one dimensional, and so the images of any basis will span a $2$ dimensional subspace.

For the particular basis you chose, the image of the basis contains $0$, so the image of the basis is obviously not a basis.

For sure you could look at the image of $A$ as a subspace of $V$. Obviously it's generated by $\{1,2t\}$, but that's the same as the subspace generated by $\{1,t\}.$

For more polynomials

All of this can be done with the other finite dimensional spaces of degree $n$-or-less polynomials.

It can even be done for the entire vector space $\Bbb R[x]$ if you want, but this is an infinite dimensional space with basis $\{1,x,x^2,x^3,\dots\}$. The linear transformation would be a matrix with sides indexed by $\Bbb N$, and it would still have kernel exactly the constant polynomials, so it's a singular transformation. But $\{x,x^2,x^3,\dots\}$ is clearly a basis for the image, so the image has the same dimension as the original space.

For more complex functions

Differentiation is certainly applicable to any space of (differentiable) functions. Furthermore, we know that the only thing in the kernel of the derivative transformation is the constant functions, so it's possible that differentiation is nonsingular on some function spaces.

As an example, consider the real space of functions spanned by $\sin(x)$ and $\cos(x)$. The derivative obviously sends this basis to another basis :) Work out the matrix for this transformation: it will be nonsingular!

  • $\begingroup$ sorry how matrix you got? $\endgroup$ – dato datuashvili Sep 20 '13 at 12:43
  • $\begingroup$ I just asked myself "what is the matrix that sends $[1,0,0]\mapsto [0,0,0]$, $[0,1,0]\mapsto [1,0,0]$ and $[0,0,1]\mapsto [0,2,0]$?" That corresponds to $1\mapsto 0$, $t\mapsto 1$ and $t^2\mapsto 2t$. It's just basic linear algebra of finding a matrix for a transformation and given basis. $\endgroup$ – rschwieb Sep 20 '13 at 12:45
  • $\begingroup$ ui,i forget that result should be derivative of terms,sorry $\endgroup$ – dato datuashvili Sep 20 '13 at 12:47
  • $\begingroup$ does this principle works for any kind of functions?like exponentials for exmaple $\endgroup$ – dato datuashvili Sep 20 '13 at 12:49
  • $\begingroup$ if i multiply this matrix,to previous basis,i should get new basis right?but it does not work,because new basis is $0$,$1$,$2*t$ $\endgroup$ – dato datuashvili Sep 20 '13 at 12:53

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