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I'm having a bit of difficulty obtaining an answer to this problem. Specifically, finding a basis for the kernel of a transformation, $\ker(T)$ .

Let \begin{align*} p_1(t)&=-1+t \\ p_2(t)&=-t+t^2 \\ p_3(t)&=-1+t^3, \quad\text{and}\\ p_4(t)&=2t-t^2-t^3. \end{align*} Let $W=\operatorname{span}\{p_1,p_2,p_3,p_4\}$, a subspace of $\mathbf P_3$.

Let $T: W \to \mathbf P_2$ be a linear transformation so that $T(p_i) = q_i$ where \begin{align*} q_1(t)&=-1+t+t^2 \\ q_2(t)&=2+4t-t^2 \\ q_3(t)&=1+11t+t^2 \quad\text{and}\\ q_4(t)&= -4-14t+t^2. \end{align*}

  1. Find a basis for $\operatorname{im}(T)$. The elements of this basis must be polynomials in $\mathbf P_2$.
  2. Find a basis for $\ker(T)$. The elements of this basis must be polynomials in $W$.

I've been able to find (1) by taking the image of each element in $W$ (given by $q_1...q_4$) putting them into a matrix and row reducing to find the column space. Here I know that $\{q_1,q_2\}$ form a basis for $\operatorname{im}(T)$.

However I'm having trouble with finding $\ker(T)$. I believe the next step is to find the null space of the matrix with vector columns of $q_1 \ldots q_4$, but I am not sure of this. The answer given is $g=-3p_1-2p_2+p_3$. This seems to come about if I find the null space of a matrix with columns $q_1 \ldots q_3$, but what about $q_4$? I am likely misunderstanding something here.

Thank you!

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I do not know if this is the best way to do this, but I tried to determine $T$ from the given data points $T(p_i) = q_i$.

I got the matrix (see appendix) $$ T = \begin{pmatrix} a-1 & a-2 & a & a \\ b-11 & b-10 & b-6 & b \\ c-1 & c & c-1 & c \\ d & d & d & d \end{pmatrix} $$ where $a,b,c,d$ are free variables. So we have many candidates for $T$.

Solving for $T x = 0$: $$ \begin{pmatrix} a-1 & a-2 & a & a \\ b-11 & b-10 & b-6 & b \\ c-1 & c & c-1 & c \\ d & d & d & d \end{pmatrix} \to \begin{pmatrix} a-1 & a-2 & a & a \\ b-11 & b-10 & b-6 & b \\ c-1 & c & c-1 & c \\ 1 & 1 & 1 & 1 \end{pmatrix} \to \\ \begin{pmatrix} a & a-1 & a+1 & a+1 \\ b & b+1 & b+5 & b+11 \\ c & c+1 & c & c+1 \\ 1 & 1 & 1 & 1 \end{pmatrix} \to \begin{pmatrix} 0 & -1 & 1 & 1 \\ 0 & 1 & 5 & 11 \\ 0 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix} \to \\ \begin{pmatrix} 0 & 0 & 1 & 2 \\ 0 & 0 & 5 & 10 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \end{pmatrix} \to \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & -2 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 0 & -2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$ This gives the solutions $x = (2s, -s, -2s, s)^\top = s (2,-1,-2,1)^\top = s (2 -t -2 t^2 + t^3)$ for $s \in \mathbb{R}$, thus $\ker T$ is created by one basis vector, which agrees with your solution.

Determining $T$

We know $Q = T P$ with

P =

  -1   0  -1   0
   1  -1   0   2
   0   1   0  -1
   0   0   1  -1

Q =

   -1    2    1   -4
    1    4   11  -14
    1   -1    1    1
    0    0    0    0

and then linearize $T$ into $x = (t_{11}, t_{12}, t_{13}, t_{14}, t_{21}, \dotsc, t_{44})^\top$ to get a system $A x = b$ with

A =

  -1   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0  -1   1   0   0   0   0   0   0   0   0   0   0   0   0   0
  -1   0   0   1   0   0   0   0   0   0   0   0   0   0   0   0
   0   2  -1  -1   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0  -1   1   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0  -1   1   0   0   0   0   0   0   0   0   0
   0   0   0   0  -1   0   0   1   0   0   0   0   0   0   0   0
   0   0   0   0   0   2  -1  -1   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0  -1   1   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0  -1   1   0   0   0   0   0
   0   0   0   0   0   0   0   0  -1   0   0   1   0   0   0   0
   0   0   0   0   0   0   0   0   0   2  -1  -1   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0  -1   1   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0  -1   1   0
   0   0   0   0   0   0   0   0   0   0   0   0  -1   0   0   1
   0   0   0   0   0   0   0   0   0   0   0   0   0   2  -1  -1

and

b =

   -1
    2
    1
   -4
    1
    4
   11
  -14
    1
   -1
    1
    1
    0
    0
    0
    0

Then Gauss-Jordan elimination gives the row-echelon form:

>> rref([A,b])
ans =

 Columns 1 through 15:

    1    0    0   -1   -0   -0   -0   -0   -0   -0   -0   -0   -0   -0   -0
    0    1    0   -1    0    0    0    0    0    0    0    0    0    0    0
    0    0    1   -1   -0   -0   -0   -0   -0   -0   -0   -0   -0   -0   -0
    0    0    0    0    1    0    0   -1   -0   -0   -0   -0   -0   -0   -0
    0    0    0    0    0    1    0   -1    0    0    0    0    0    0    0
    0    0    0    0    0    0    1   -1    0    0    0    0    0    0    0
    0    0    0    0    0    0    0    0    1    0    0   -1   -0   -0   -0
    0    0    0    0    0    0    0    0    0    1    0   -1    0    0    0
    0    0    0    0    0    0    0    0    0    0    1   -1    0    0    0
    0    0    0    0    0    0    0    0    0    0    0    0    1    0    0
    0    0    0    0    0    0    0    0    0    0    0    0    0    1    0
    0    0    0    0    0    0    0    0    0    0    0    0    0    0    1
    0    0    0    0    0    0    0    0    0    0    0    0    0    0    0
    0    0    0    0    0    0    0    0    0    0    0    0    0    0    0
    0    0    0    0    0    0    0    0    0    0    0    0    0    0    0
    0    0    0    0    0    0    0    0    0    0    0    0    0    0    0

and

 Columns 16 and 17:

   -0   -1
    0   -2
   -0   -0
   -0  -11
    0  -10
    0   -6
   -0   -1
    0    0
    0   -1
   -1   -0
   -1    0
   -1    0
    0    0
    0    0
    0    0
    0    0
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Note that the columns of $$ A= \left[\begin{array}{rrrr} -1 & 0 & -1 & 0 \\ 1 & -1 & 0 & 2 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \end{array}\right] $$ are identified with the polynomials $p_1$, $p_2$, $p_3$, and $p_4$. Row reducing gives $$ \DeclareMathOperator{rref}{rref}\rref A= \left[\begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ This tells us that $W$ is three-dimensional with basis $\{p_1,p_2,p_3\}$.

Now, note that the columns of $$ B= \left[\begin{array}{rrrr} -1 & 2 & 1 & -4 \\ 1 & 4 & 11 & -14 \\ 1 & -1 & 1 & 1 \end{array}\right] $$ are identified with the polynomials $q_1$, $q_2$, $q_3$, and $q_4$. Row reducing gives $$ \rref B= \left[\begin{array}{rrrr} 1 & 0 & 3 & -2 \\ 0 & 1 & 2 & -3 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ This tells us that the image of $T$ is two-dimensional with basis $\{q_1,q_2\}$.

The rank-nullity theorem then implies that $\ker T$ is one-dimensional. To find a basis for $\ker T$, note that $\rref B$ tells us that the vector $$ \vec v=\left(1,\,4,\,1,\,2\right) $$ is in the null space of $B$. This tells us that $$ q_1+4\,q_2+q_3+2\,q_4=0 $$ It follows that $$ T(p_1)+4\,T(p_2)+T(p_3)+2\,T(p_4)=0 $$ so that $$ T(p_1+4\,p_2+p_3+2\,p_4)=0 $$ That is, the polynomial $$ p_1+4\,p_2+p_3+2\,p_4=-t^3+2\,t^2+t-2 $$ forms a basis for $\ker T$.

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  • $\begingroup$ You appear to have an error in rref $A$—the upper-right element should be $-1$. $\endgroup$ – amd May 16 '16 at 3:27
  • $\begingroup$ @amd Nope. wolframalpha.com/input/… $\endgroup$ – Brian Fitzpatrick May 16 '16 at 3:47
  • $\begingroup$ @BrianFitzpatrick - this explanation is very thorough but I get lost when you find the basis for ker$T$. I'd find a basis for ker$T$ by finding a basis for the null space of rref$B$ which gives me a basis $\{(-3,-2,1,0),(2,3,0,1)\}$. I understand why ker$T$ should really be one dimensional, but the basis I'm finding (incorrectly) has two vectors. How did you find $\vec{v}=(1,4,1,2)$? $\endgroup$ – econom May 16 '16 at 3:56
  • $\begingroup$ nm. I had transposed two elements of the original matrix. $\endgroup$ – amd May 16 '16 at 7:03
  • $\begingroup$ @econom I’m not sure where $(1,4,1,2)$ came from, but you can work with the two kernel vectors that you’ve found. They describe linear combinations of the $p$’s and you’ll find that when you expand and simplify them (equivalently, multiply those two vectors by $A$), you end up with the same polynomial. You could also have worked with just $q_1$ through $q_3$ to find the image and kernel since rref $A$ shows that $p_4$ is redundant. (Retaining the latter in the computations checks that $T$ is indeed linear, so it’s not completely wasted.) $\endgroup$ – amd May 16 '16 at 7:21

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