Let $U = P_3(\Bbb R)$, the vector space of polynomials of degree at most $3$ in a formal variable $t$.

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

(i) Determine the matrix of f with respect to the bases {$1$,$t$,$t^2,t^3$} and {1}.

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

My attempt

(i) $$ \left[ \begin{matrix}{} 0&1&2&3\\ \end{matrix} \right] $$

(ii) As it is already in reduced row echelon form we can proceed,

Would I be correct in saying that this matrix has rank 1 (because it has 1 leading 1) and therefore by the rank and nullity theorem it has nullity 3?

Therefore we have the equation $x_2+2x_3+3x_4= 0$, therefore we can set the value of any three variables and work out the last one. So the basis I got was:

Let $x_1=1,x_2=1,x_3=1$ then we have $x_4= -1$

Let $x_1=1,x_2=0,x_3=0$ then we have $x_4= 0$

Let $x_1=0,x_2=0,x_3=1$ then we have $x_4= -\frac{2}{3}$

Therefore a basis for the kernel is {$(1+t+t^2-t^3),(1),(t^2-\frac{2}{3}t^3)$}

Is this correct?

  • $\begingroup$ Is this correct? $\endgroup$ Dec 7 '15 at 21:14
  • $\begingroup$ $f(t^2-\frac23)\ne0$ $\endgroup$
    – amd
    Dec 7 '15 at 21:44
  • $\begingroup$ Sorry I'll correct that mistake $\endgroup$ Dec 7 '15 at 21:51
  • $\begingroup$ Is it correct now? $\endgroup$ Dec 7 '15 at 21:52

Looks like you’ve found a basis for the kernel, but there’s a more systematic way to go about this once you have the rref matrix. Find the columns that don’t have pivots (leading entries). Those will be the free coordinates in the kernel basis. Set each one to one and the rest to zero in turn, and solve for the remaining coordinates. You’ll find very quickly that “solving” for the missing values amounts to reading them from the column in the rref matrix that corresponds to the coordinate that’s been set to $1$, then negating them.

In this case, the vectors constructed in this way will be $(1,x_2,0,0)^T$, $(0,x_2,1,0)^T$ and $(0,x_2,0,1)^T$. The missing coordinates are, respectively, $0$, $-2$ and $-3$, i.e., the negations of the entries in the first, third and last columns.

  • $\begingroup$ Thank you so much for this, super helpful! $\endgroup$ Dec 7 '15 at 22:03

Here is another approach which uses the polynomial structure. First, let s=t-1. Then every polynomial can be written as $u=a_0+a_1 s + a_2 s^2 + a_3 s^3$ and now the map is $f(u) = a_1$. Thus, the kernel is spanned by $1,s^2,s^3$ since these polynomials and their combinations have $a_1=0$. Thus, $1,(t-1)^2,(t-1)^3$ is a basis!


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