Matrix a representation for $L(x)$ w.r.t. the homogeneous coordinate system

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I'm trying to find a matrix representation for $L(x) = x + a, a = $$\begin{bmatrix}2\\5\end{bmatrix}$ w.r.t. the homogeneous coordinate system.

Here is my attempt below. I am not mathematically-inclined, so please use simple English when dealing with my poor brain.


$L(x) = A(x)$... so $A(x) = x + $$\begin{bmatrix}2\\5\end{bmatrix}$. I'm pretty sure I need to find $A$.

I'll let x = $\begin{bmatrix}x_1\\x_2\end{bmatrix}$. So $x + a = \begin{bmatrix}x_1 + 2\\x_2 + 5\end{bmatrix}$, which says that $$L(x) = \begin{bmatrix}x_1 + 2\\x_2 + 5\end{bmatrix} = A(x)$$.

Okay, so, to find $A$, I'll plug in the standard basis vectors for $\Bbb{R^2}$, $ e_1$ and $e_2$.

$A(e_1) = A(1, 0)^T = (3, 5)^T$, $A(e_2) = A(0, 1)^T = (2, 6)^T$

So $A = \begin{bmatrix} 3 & 2\\5 & 6 \end{bmatrix}$.

Okay, but now, with respect to the homogeneous co-ordinate system, where you need an extra column and row for extra calculations or something. I'm guessing it would be in a form like, maybe, $\begin{bmatrix} 3 & 2 & 0\\5 & 6 & 0\\ 0&0&1 \end{bmatrix}$??

EDIT, possibly correct:

Interesting: I read that, to convert a vector in $\Bbb{R^2}$ to the H.C.S in $\Bbb{R^3}$, you take your vector $a = \begin{bmatrix}2\\5\end{bmatrix}$ and place it into the first 2 rows of the third column of the identity matrix in $\Bbb{R^3}$. This then becomes: $a = \begin{bmatrix}1&0&2\\0&1&5\\0&0&1\end{bmatrix}$.

I'm guessing this is correct.


Please help point me in the right direction. I feel like I'm close, but not there.

Cheers, -Jon

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It seems that you know homogeneous coordinates. Using these coordinates, a point $X$ in an affine space that corresponds to the vector $x=[x_1,x_2]^T$ is represented by the triple $X=[x_1,x_2,1]^T$ and the matrix that represents a translation of vector $a=[2,5]^T$ is : $$ T_a=\begin{bmatrix} 1 & 0 & 2\\0 & 1 & 5\\ 0&0&1 \end{bmatrix} $$ that, applied to the point $X$ ( with the usual row-column product rule) gives: $$ T_a(X)=\begin{bmatrix} 1 & 0 & 2\\0 & 1 & 5\\ 0&0&1 \end{bmatrix}\begin{bmatrix} x_1\\x_2\\1 \end{bmatrix}= \begin{bmatrix} x_1+2\\x_2+5\\1 \end{bmatrix}$$

That are the homogeneous coordinates of the translated point.

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You can not do that since $L$ is not linear transformation.

If $L$ is linear, then $L(\vec{0}) = \vec{0}$ but in your case $L(\vec{0}) = a\ne \vec{0}$