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I have a linear transformation $$T(x_1,x_2,x_3) = (x_1-x_2+2x_3,2x_1+x_2,-x_1-2x_2+2x_3) $$

Assuming an ordered basis $\{u_1,u_2,u_3\}$ I get the linear transformation of the basis vectors as below: $$T(u_1) = u_1+2u_2-u_3,\\ T(u_2) = -u_1+u_2-2u_3, \\T(u_3) = 2u_1+u_3\\$$ There are two matrix representations of the above transformation as below: $$[x_1,x_2,x_3]\left[ \begin{matrix} 1 & 2 & -1 \\ -1 & 1 & -2 \\ 2 & 0 & 2 \\ \end{matrix}\right] = [y_1,y_2,y_3] $$ and $$\left[ \begin{matrix} 1 & -1 & 2 \\ 2 & 1 & 0 \\ -1 & -2 & 2 \\ \end{matrix}\right]\left[\begin{matrix} x_1 \\ x_2 \\ x_3 \\ \end{matrix}\right] = \left[\begin{matrix} y_1 \\ y_2 \\ y_3 \\ \end{matrix}\right] $$

Here $y_1,y_2,y_3 $ element of Range(T). I believed both are same, however I am not able to determine certain relations from each. Example, row reducing 1 I get $$\left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ \end{matrix}\right]$$ which gives me the basis for R(T) and linear combination of these basis vectors yield $(x_1,x_2, x_1-x_2)$ i.e. I get the relationship between the coordinates and the augmented matrix gives the constraint on $T(u_1),T(u_2),T(u_3)$ as $-2T(u_1)+4T(u_2)+3T(u_3) = 0$, this relationship however I do not get from the second representation. $$\\$$ The second representation gives us relationship between coordinates and also gives us the Null space as $$\{(-a/2,a,(3/4)a):a e R\}$$ and I am not able to get the null space from the first representation.

Are there any differences between the two representations? And few relations that can be derived from one and not from another?

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    $\begingroup$ You seem to be confusing elements of the basis, $u_i$, which are vectors, with coordinates $x_i$ wrt to the standard basis, which are scalars. $\endgroup$ – ancientmathematician May 14 at 14:47
  • $\begingroup$ $T(u_3)=2u_1+2u_3$. The matrices are correct, however. $\endgroup$ – amd May 14 at 19:48
  • $\begingroup$ @ancientmathematician I assume representation 1 corresponds to elements of the basis $u_i$ and 2 corresponds to coordinates $x_i$. However once we multiply the matrices the result is the same, hence I am confused. Can you please elaborate a bit further? $\endgroup$ – Rohan Frederick May 15 at 8:13
  • $\begingroup$ Do you know what the matrix of a linear transformation is? If you write down the exact definition and carry it through you will see that one of the matrices you havew above is the matrix of $T$; the other is not the matrix of $T$. [Which is right and which is wrong depends on your exact definition] $\endgroup$ – ancientmathematician May 15 at 15:23
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As per my Analysis both of them are same and can be used interchangeably. Putting a bit more effort I got the same results from both the representations. Working shown with non-standard representation.

Row Reducing $$\left[ \begin{array}{ccc|ccc} 1 & 2 & -1 & 1 & 0 & 0\\ -1 & 1 & -2 & 0 & 1 & 0 \\ 2 & 0 & 2 & 0 & 0 & 1 \\ \end{array}\right]$$ we get $$\left[ \begin{array}{ccc|ccc} 1 & 0 & 1 & 1/3 & -2/3 & 0\\ 0 & 1 & -1 & 1/3 & 1/3 & 0 \\ 0 & 0 & 0 & -1/6 & 1/3 & 1/4 \\ \end{array}\right]$$

Above gives us that (1,0,1) and (0,1,-1) are the basis for Range Space and linear combination of these vectors give us the transformation as $T(x_1,x_2,x_3) = (x_1,x_2,x_1-x_2)$ also with $$[y_1,y_2,y_3]\left[ \begin{matrix} 1/3 & -2/3 & 0\\ 1/3 & 1/3 & 0 \\ -1/6 & 1/3 & 1/4 \\ \end{matrix}\right]$$ We get $$x_1 = \frac{y_1+y_2}{3} -y_3/6\\ x_2 = \frac{y_2-2y_1}{3} +y_3/3\\ x_3 = y_3/4$$

The Null space is also given by the third row as: $$\frac{-T(u_1)}{6}+\frac{T(u_2)}{3}+\frac{T(u_3)}{4} = 0\\ \implies T(\frac{-(u_1)}{6}+\frac{(u_2)}{3}+\frac{(u_3)}{4}) = 0\\ $$ Null space is spanned by vector $\frac{-(u_1)}{6}+\frac{(u_2)}{3}+\frac{(u_3)}{4}$ or linear combination of $\frac{-(u_1)}{2}+(u_2) +\frac{(3u_3)}{4}$ $$\therefore Ker(T) = \{(\frac{-\alpha}{2},\alpha,\frac{3\alpha}{4}):\alpha\ e\ R\}$$

Please correct me in case of any issue.

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  • $\begingroup$ @ancientmathematician, please correct me in case of any issue $\endgroup$ – Rohan Frederick May 21 at 13:11

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