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When we translate a point $p_3 = (x,y,z)$ to coordinates $p_4 =(x + t_x , y + t_y ,z + t_z,1)$ we use $4 \times 4$ Translation matrix using homogenous coordinates, hence we add a $1$ to fourth coordinate of $p_4$.

But I could not find a matrix $M : R^3 \rightarrow R^4$ for which $M p_3 = p_4$. Note that I am talking about a $1$ on the fourt coordinate what ever vector $p_3$ be.

Why can not we find such a generic translation matrix. Omit the case of matrix addition i.e. $M$ may only be multiplication of other matrices.

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You cannot represent a translation in $3$-dimensional space with a matrix smaller than $4\times 4$. All $3\times 3$ matrices represent transformations which leave the origin fixed, because multiplying a matrix for the zero vector always yields the zero vector. Homogeneous coordinates might be used as a trick to solve this, but you need to add a vector entry and so switch to a $4\times 4$ matrix.

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THanks but the explanation I am looking for is about using vector spaces and other abstract math. – camoka Dec 5 '12 at 15:35
I know what you said but what I look for is somewhat a proof related to pure linear algebra – camoka Dec 5 '12 at 15:36
@camoka: In "pure linear algebra" terms, the issue is to be explained in terms of "linear" vs "projective" transformations. A translation is not a linear transformation, because it does not fix the origin. But it is a projective transformation (indeed, it is an affine one). So, if you call $V$ your vector space, you can represent your translation $T$ by imbedding $V$ into a projective space $P(V)$ (which, in coordinates, amounts to introducing homogeneous coordinates) and then using the machinery of representation of projective transformations. – Giuseppe Negro Dec 5 '12 at 20:02
More precisely, to represent $$T\colon V \to V$$ as a projective transformation, you need to regard it as a mapping $$T\colon P(V) \to P(V).$$ Both $T\colon V\to P(V)$ and $T\colon P(V)\to V$ won't work. – Giuseppe Negro Dec 5 '12 at 20:02
This is why you will need to add a "row" dimension and a "column" dimension, and so use a $4\times 4$ matrix. Neither a $4\times 3$ nor a $3\times 4$ matrix will work. – Giuseppe Negro Dec 5 '12 at 20:05

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