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Lets say we have a point that is a part of 3d space, but fixed to the $z=1$ plane. That is, we have, in column matrix form, assuming the canonical $\mathbb R^3$ basis:

$$ v_0 = \begin{bmatrix} x\\ y\\ 1\\ \end{bmatrix} \ $$

To translate this point, we can use an affine transformation. For instance, to translate this point by $(T_x, T_y) = (1, 1)$, we could left multiply $v_0$ by a matrix generated using this translation vector. Let's call it $T$.

$$ v_1 = T v_0 = \begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{bmatrix} \begin{bmatrix} x\\ y\\ 1\\ \end{bmatrix} = \begin{bmatrix} x + 1\\ y + 1\\ 1\\ \end{bmatrix} \ $$

My issue is that the matrix $T$ can be interpreted as a change of basis matrix, and this operation is clashing with my knowledge of what that means.

Our original vector $v_0$ was defined based on the canonical basis, as I mentioned. However, by left multiplying $T$ by $v_0$, what we are conceptually doing, in my understanding, is calculating what the matching coefficients of the linear combination on our original canonical base are of a linear combination of a different basis, the components of which are the set of vectors that form the columns of $T$.

Performing this operation assumes that the original coordinates no longer describe a linear combination of the original canonical basis, but they are now describing one of the vectors present on the columns of $T$.

Could you explain what is the reasoning for this sudden change of basis of the coordinates, and why does it end up correctly describing a translation in our original basis? Thanks.

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Simple answer

You don't have to bother with the base of change that the matrix of translation on affine plane represents.
Just take it as another kind of application of matrix multiplication, and always work with the standard basis.
Note also that - though always exists - the inverse of $T$ is not involved here, whereas the change of base formula contains it.

Translating by a plane vector $v=(a,b)$ on the affine plane $z=1$ can be expressed by $$\pmatrix{1&0&a\\0&1&b\\0&0&1}\pmatrix{x\\y\\1}=\pmatrix{x+a\\y+b\\1}$$ That's it, and really not much more.


the change of basis

If we want to regard this matrix as a change of base matrix, the new basis vectors, as you wrote, are just its columns.
Calling the standard basis $e_1,e_2,e_3$, the new basis is $e_1,e_2,\,v+e_3$.
These are exactly the images of the basis vectors under the linear transformation on the whole $\Bbb R ^3$ that $T$ determines, i.e. $w\mapsto T\cdot w$.
Geometrically, this is a shearing, which, restricted to the affine plane $z=1$ gives back the translation by $v$, and on level $z$, it simply translates by $zv$, in particular keeping the $x,y$ plane fixed.

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