I'm trying to solve the following problem:

Prove that mapping $\varphi$: $\mathbb{R}_{2}[x]\rightarrow \mathbb{R}_{2}[x]$, which maps $[ax^2+bx+c]$ $\rightarrow$ $[a(x+1)^2+a(x+1)+b+c]$ is a linear map. Also find it's matrix in basis {$3;2x;x^2$} and find the basis of the kernel of $\varphi$.

My solution is the following:

In order to prove that the mapping is a linear map, I created the following equation: $a(x+1)^2+a(x+1)+b+c+f(x+1)^2+f(x+1)+g+h=(a+f)(x+1)^2+(a+f)(x+1)+b+g+c+h$

which holds. Next I created the following equation: $\alpha a(x+1)^2+\alpha a(x+1)+\alpha b+\alpha c = \alpha(a(x+1)^2+a(x+1)+b+c)$, which is also correct.

Thus I prove that the mapping is a linear map. In order to find the matrix in the basis, I calculated the following:

  • $\varphi (3)=3$
  • $\varphi (2x)=2$
  • $\varphi(x^2)=(x+1)^2+(x+1) = x^2+3x+2$

Therefore, the matrix in the basis is $\bigl(\begin{smallmatrix} 0 &0 &1 \\ 0 &0 &3 \\ 3& 2 & 2 \end{smallmatrix}\bigr)$. Is my solution correct? Also, how would I calculate the kernel? If I had a matrix of the map, I would just solve it for $0$, but I don't have one. Can I just use the matrix in the basis I created earlier or do I need to create a new matrix of the map? Thanks

  • $\begingroup$ Have another look at your matrix. Namely, $\phi(x^2)=2+3x+x^2=\frac{2}{3}.3+\frac{3}{2}.2x+1.x^2$. $\endgroup$ Jul 1, 2019 at 19:13
  • $\begingroup$ @Alexandros what should it look like? $\endgroup$
    – james F.
    Jul 1, 2019 at 19:18
  • $\begingroup$ Just take the coefficients and put them vertically in the last column of your matrix. $\endgroup$ Jul 1, 2019 at 19:20

1 Answer 1


The matrix is incorrect. Your computations of $\varphi$ are correct, but you have to write them as linear combinations of the given basis vectors. Denote the basis by $\beta = \{3,2x,x^2\}$. \begin{align} \begin{cases} \varphi(3) &= 3 &= 1 \cdot (3) + 0 \cdot (2x) + 0 \cdot (x^2) \\\\ \varphi(2x) &= 2 &= \dfrac{2}{3} \cdot (3) + 0 \cdot (2x) + 0 \cdot (x^2) \\\\ \varphi(x^2) &= x^2 + 3x + 2 &= \dfrac{2}{3} \cdot (3) + \dfrac{3}{2}\cdot (2x) + 1 \cdot (x^2) \end{cases} \end{align} Now we insert the coefficients into a matrix. So, the matrix of $\varphi$ relative to the basis $\beta$ in the domain and target space is \begin{align} [\varphi]_{\beta}^{\beta} = \begin{pmatrix} 1 & \frac{2}{3} & \frac{2}{3} \\ 0 & 0 & \frac{3}{2} \\ 0 & 0 & 1 \end{pmatrix} \end{align}

To find the kernel, you really just apply the definition. The kernel is defined as the set of polynomials $f(x)$ which get mapped to the zero polynomial $0$, under $\varphi$. Symbolically, \begin{align} \ker(\varphi) = \{f(x) \in \Bbb{R}_2[x]: \varphi(f(x)) = 0\} \end{align} A general element $f(x)$ can be written as $ax^2 + bx + c$. So this means $ax^2 + bx + c \in \ker(\varphi)$ if and only if \begin{align} \varphi(ax^2 + bx + c) &= 0 \end{align} equivalently, \begin{align} a x^2 + (3a)x + (a+b+c) = 0x^2 + 0x + 0 \end{align} This happens if and only if $a=0$ and $b=-c$.

The method I showed above can be used in any circumstance; you just close your eyes, and apply the definition directly, and crank out the solution of the resulting system of equations. However, in your particular example, if you pause for a moment, you might be able to arrive at the solution quicker.

Notice that in $[\varphi]_{\beta}^{\beta}$, there are only two independent columns (the first and third). This implies that $\text{rank}(\varphi) = 2$. Hence, by the rank-nullity theorem, you can conclude that \begin{equation} \dim \ker(\varphi) = \dim \Bbb{R}_2[x] - \text{rank}(\varphi) = 3-2 = 1 \end{equation} So, the kernel is one-dimmnsional, which means if you find a non-zero $f(x)$ such that $\varphi(f(x)) = 0$, then $\ker(\varphi) = \text{span}\{f(x)\}$. By a quick observation, you can notice that \begin{equation} \varphi(x-1) = 0 \end{equation} Hence, it immediately follows that \begin{align} \ker(\varphi) = \text{span} \{x-1\} = \{ax^2 + bx + c \in \Bbb{R}_2[x] : a=0, \quad b=-c\} \end{align} This is in agreement with what we found above.

Additional Remarks:

In this particular case, it is arguable whether or not the second way is really much faster. But if you had a larger matrix, then there are some situations where using rank-nullity to determine the dimension of the kernel allows you to get the answer much quicker, because sometimes just based on observation, some elements are clearly in the kernel (such as $x-1$ in your example).

So, my point is that before you dive head-first into computations involving row-reduction or whatever, take a moment to understand your operator, and decide which method to use for solving.

Also, I answered another question regarding how to write matrices of linear operators with respect to given bases in this answer. Perhaps it might be of help to you, to really solidify what's going on.

  • $\begingroup$ Thanks. How did you get this: $x^2+(3a)x+(a+b+c)=0x^2+0x+0$ ? $\endgroup$
    – james F.
    Jul 1, 2019 at 19:44
  • $\begingroup$ $\varphi(ax^2 + bx + c) = a(x+1)^2 + a(x+1) + b + c = ax^2 + 3ax + (a+b+c)$, and we are setting that to be the zero polynomial, hence all the coefficients have to be $0$. $\endgroup$
    – peek-a-boo
    Jul 1, 2019 at 19:45

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