Basis of the image of a linear transformation Let $E$ be a three dimensional real vector space and $B = \{u_{1}, u_{2}, u_{3}\}$ be a basis of $E$. The linear transformation $f: \mathbb{R^{5}} \rightarrow E$ is defined as: $f(a,b,c,d,e) = (-b-c+d)u_{1} + (2a + b + 3c-3d)u_{2} + (b + c - d)u_{3} $ for any $a,b,c,d,e \in \mathbb{R} $.
I need to find a basis for the image of $f$.
So what I did was I found the matrix of the transformation $f$ which is $ \begin{bmatrix}0 & -1 & -1 & 1 & 0\\ 2 & 1 & 3 & -3 & 0\\ 0 & 1 & 1 & -1 & 0\end{bmatrix} $ and reduced it to row-echelon form. The rows with the pivots should be vectors that form a basis.
The matrix I obtained in row-echelon form was $\begin{bmatrix}1 & 0 & 1 & -1 & 0\\ 0 & 1 & 1 & -1 & 0\\ 0 & 0 & 0 & 0 & 0\end{bmatrix}$ and so the basis should be $\{(1, 0,0),(0,1,0)\}$ but the book says a possible solution is: $ \{u_{1} - u_{3}, u_{2}\}$ wich is completly different from what I have...
Could some one help me understand what I'm doing wrong?
 A: A vector subspace can have more than one basis, usually infinite, so the fact that your answer is different to the one in the book doesn't mean that you're wrong. You could see if your basis and the one from the book generate the same subspace, checking that the four vectors written in matrix form give you rank 2, which would mean the two from your basis are linearly dependent to the other two and viceversa. Even when you have some basis, you could make a linear combination of some of its vectors and substitute the result for one of them in the basis, and you would get a new basis (the vector you substitute from the basis has to take part in the linear combination).
That said, when you obtain the row-echelon form you get information about the linear independence of those vectors, and even about the relation they have if they are dependent, but you don't get new valid vectors, since the sum and scalar product of vectors would happen in a column of the matrix, not in a row. This means that the first two rows of your final row-echelon form aren't some vectors you can take as part of the subspace.
For example, consider the subspace generated by the vector $\begin{pmatrix} 1\\1\\1\end{pmatrix}$. Pretty simple, isn't it? If you consider the row-echelon form of that vector in matrix form you would get the vector $\begin{pmatrix} 1\\0\\0\end{pmatrix}$, which clearly doesn't generate the same subspace as the initial one.
So what do you do? As I said, the row-echelon form gives you information. Another important thing to have in mind is that those numbers in the initial matrix aren't the vectors theirselves; those numbers are the coordinates of the vectors respect the basis $\{u_1,u_2,u_3\}$; we'll use this at the end.
As you can see at the row-echelon form, the first two columns were linearly independent and the three others are linearly dependent respect the first two. So you can take the first two columns (from the initial matrix) as the coordinates of the two vectors for your basis. This means that
$\begin{bmatrix} 0\\2\\0\end{bmatrix}$ is in fact $2u_2$, and
$\begin{bmatrix} -1\\1\\1\end{bmatrix}$ is in fact $-u_1+u_2+u_3$.
So your basis could be $\{-u_1+u_2+u_3,2u_2\}$, but now we can make a linear combination of them: $2(-u_1+u_2+_3)-(2u_2)=-2u_1+2u_3$, so we can have $\{-2u_1+2u_3,2u_2\}$ as a basis. Finally we can consider $\frac{-1}{2}(-2u_1+2u_3)$ and $\frac{1}{2}(2u_2)$, so our basis is the same as the book's one: $\{u_1-u_3,u_2\}$.
