From a more fundamental viewpoint, if
[tex]T= \begin{bmatrix}1 & 0 \\ 0 & 1 \\ 0 & 0\end{bmatrix}[/tex]
Then T maps [itex]R^2[/itex] to [itex]R^3[/itex]. (x, y, z) will be in the image of T if and only if there exist (a, b) such that
[tex]\begin{bmatrix}1 & 0 \\ 0 & 1 \\ 0 & 0\end{bmatrix}\begin{bmatrix}a \\ b\end{bmatrix}= \begin{bmatrix}a \\ b \\ 0\end{bmatrix}= \begin{bmatrix}x \\ y \\ z\end{bmatrix}[/tex]
which gives the equations a= x, b= y, 0= z. Since a and b could be any numbers so can x and y- but z must be 0. That is, any vector in the image of T must be of the form (x, y, 0)= (x, 0, 0)+ (0, y, 0)= x(1, 0, 0)+ y(0, 1, 0). Yes, the image is two-dimensional and a basis is the set of vectors {(1, 0, 0), (0, 1, 0)}, the columns of the matrix you show. However, it is important to remember that a basis is a set of vectors, not a matrix. Even though you used a matrix to calculate it, you should show the basis as a set of vectors, not as a matrix.