I was trying to do this exercise and I don't know if what I did is okay or not. Let $T:P_2[\mathbb R] \to \mathbb R^4$ be a linear transformation and let $B=\{1 + x^2, x, 2x + 1\}$ and $$B'= \{(0,0,1,3), (0,1,1,0), (1,1,2,1), (1,0,0,2)\}$$ be bases of $P_2[\mathbb R]$ and $\mathbb R^4$, respectively. If the matrix representing $T$ is $M =\left[\matrix{1&3&2\cr 0&1&1\cr 1&2&1\cr 2&4&2\cr}\right]$,

a)Find the bases $\ker (T)$ and ${\rm Im} (T)$. Check dimension theorem.

a)First of all I did this - rename the basis $B=\{u_0,u_1,u_2\}$ and $B'=\{v_1,v_2,v_3,v_4\}$. So: $$\eqalign{T(u_0) &= v_1 + v_3 + 2v_4\cr T(u_1) &= 3v_1 + v_2 + 2v_3 + 4v_4\cr T(u_2) &= 2v_1 + v_2 + v_3 + 2v_4\cr}$$

Then, row reduce the given matrix to get the kernel, so: $$\left[\matrix{1&0&-1\cr 0&1&1\cr 0&0&0\cr 0&0&0\cr}\right] \left[\matrix{a\cr b\cr c\cr}\right] = \left[\matrix{0\cr 0\cr 0\cr0\cr}\right]$$ $a - c =0$ and $b + c=0$, so $a=c, b=-c$, so a basis is $\left\{\pmatrix{1\cr -1\cr 1\cr}\right\}$.

Then, for the image I did this: $$\eqalign{ a + 3b + 2c &= w\cr b + c &= x\cr a + 2b + c &= y\cr 2a + 4b + 2c &= z\cr}$$ After gauss, finally reached to $w=x +y$ -- I don't know what to do now.

  • $\begingroup$ Your kernel basis is correct. To find a basis for the image, you need the column space of $M$. You should have learned how to do this, right before this exercise. $\endgroup$ – Christopher Carl Heckman May 14 '16 at 23:52
  • $\begingroup$ Well, i never did an exercise like this. Can explain me what to do now to get the image?Is the kernel okay or i need to express it as a polynomial form? $\endgroup$ – Tincho May 14 '16 at 23:55
  • 1
    $\begingroup$ To find the column space of a matrix $A$, you find the RREF, and then pick the columns from $A$ that have a pivot. There are pivots in columns 1 and 2 of your RREF, so $\left\{ \pmatrix{1\cr 0\cr 1\cr 2\cr}, \pmatrix{3\cr 1\cr 2\cr 4\cr}\right\}$ will be a basis for the column space of $A$ (and the image of $T$). $\endgroup$ – Christopher Carl Heckman May 15 '16 at 0:22
  • $\begingroup$ That´s right. Suppose that i want to do this using a + 3b + 2c = w b + c = x a + 2b + c = y 2a + 4b + 2c = z Is it okay?Supposedly yes, but i don´t know what to do with the result after doing GAUSS :S ---> w = x + y $\endgroup$ – Tincho May 15 '16 at 0:27
  • $\begingroup$ You'll get conditions that the vectors have to satisfy. You should also get that $z=2y$. $\endgroup$ – Christopher Carl Heckman May 16 '16 at 6:04

You can do this with just Gaussian elimination:

$$\left[\begin{array}{ccc|c} 1&3&2&w\\ 0&1&1&x\\ 1&2&1&y\\ 2&4&2&z\end{array}\right] \to \left[\begin{array} {ccc|c} 1&3&2&w\\ 0&1&1&x\\ 0&-1&-1&y-w\\ 0&-2&-2&z-2w\end{array}\right] \to \left[\begin{array} {ccc|c} 1&3&2&w\\ 0&1&1&x\\ 0&0&0&y-w+x\\ 0&0&0&z-2w+2x\end{array}\right]$$ You have two rows which are almost entirely zero. In order to guarantee that there is a solution, you must have $y-w+x=0$ and $z-2w+2x=0$. Any vector that meets those conditions will be in the column space.


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