I have a linear transformation:

$$f:\mathbb{R}^3 \rightarrow \mathbb{R}[x]_2$$

$$f((1,1,1))=2x^2-3x,\ f((1,2,3))=-3x,\ f((1,2,4))=2x^2-4x$$

I need to find kernel, image, dimension of image and dimension of kernel.

I tried to find formula describing this transformation:


$$\begin{cases} x_1+y_1+z_1=2 \\ x_1+2y_1+3z_1=0 \\x_1+2y_1+4z_1=2 \end{cases}$$

$$\begin{cases}x_2+y_2+z_2=-3 \\x_2+2y_2+3z_2=-3 \\x_2+2y_2+4z_2=-4 \end{cases}$$

$$\begin{cases}x_3+y_3+z_3=0 \\x_3+2y_3+3z_3=0 \\x_3+2y_3+4z_3=0\end{cases}$$

$$\begin{cases}x_1=6 \\ y_1=-6 \\ z_1=2 \\ x_2=-4 \\ y_2=2 \\ z_2=-1 \\ x_3=0 \\ y_3=0 \\ z_3=0 \end{cases}$$



$$\begin{cases}6a-6b+2c=0 \\ -4a+2b-c=0\end{cases}$$

$$\begin{cases}b=-a \\ c=-6a\end{cases}$$

$$\ker(f)=\operatorname{Lin}((1,-1,6)) \\\dim\ker(f)=1$$

Unfortunately, I don't know how it is possible to find image (and its dimension) of $f$.

I think that $\dim\operatorname{Im}(f)=3-1=2$, but I am not sure about it.

  • 2
    $\begingroup$ If your calculations are correct, then $$Im f=\{(6a-6b+2c)x^2+(-4a+2b-c)x|a,b,c\in\mathbb{R}\}$$ $\endgroup$ – blue Dec 11 '17 at 20:28

Your expression for $f$ is correct. Here is a slightly easier way to get it:

$$f(0,0,1) = f(1,2,4) - f(1,2,3) = 2x^2 - x$$ $$f(0,1,0) = f(1,2,3) - f(1,1,1)-2\cdot f(0,0,1) = -6x^2 + 2x$$ $$f(0,0,1) = f(1,1,1) - f(0,1,0) - f(0,0,1) = 6x^2 - 4x$$


$$f(a,b,c) = a\cdot f(1,0,0) + b\cdot f(0,1,0) + c\cdot f(0,0,1) = (6a-6b+2c)x^2+(-4a+2b-c)x$$

Your calculations for the kernel are correct. $\{(1,-1,6)\}$ is the basis for $\operatorname{Ker} f$, and hence $\dim\operatorname{Ker} f = 1$.

Now, for the image, the set

$$\{f(1,0,0), f(0,1,0), f(0,0,1)\} = \{2x^2 - x, -6x^2 + 2x, 6x^2 - 4x\}$$

certainly spans $\operatorname{Im} f$. It only remains to be seen whether it is linearly independent. We obtain:

$$6x^2 - 4x = 6\cdot(2x^2-2) + (-6x^2 + 2x)$$


$$\operatorname{Im} f = \operatorname{span} \{2x^2 - x, -6x^2 + 2x, 6x^2 - 4x\} = \operatorname{span} \{2x^2 - x, -6x^2 + 2x\}$$

The set $\{2x^2 - x, -6x^2 + 2x\}$ is linearly independent so it is a basis for $\operatorname{Im}f$. We conclude that $\dim\operatorname{Im}f = 2$.


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