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I am a beginner at matrices and I am trying to find out whether or not the linear transformation defined by the matrix $A$ is onto, and also whether it is 1-1.

Here is the matrix $A$:

$$\begin{bmatrix} 1 & 3 & 4 & -1 & 2\\ 2 & 6 & 6 & 0 & -3\\ 3 & 9 & 3 & 6 & -3\\ 3 & 9 & 0 & 9 & 0 \end{bmatrix}$$

I reduced it to echeleon row form but I am not sure what to do from there, thank you for any help.

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up vote 1 down vote accepted

If the reduced matrix have a zero column, then it is not 1-1 because for example $A(00001)^{T}$ will be zero, this will be the case, beacuse the matrix $A$ that a vector in $\mathbb{R}^5$ to $\mathbb{R}^4$, so it can´t be 1-1. To see if it is into, you have to see that the columns generate $\mathbb{R}^4$, so if the reduced matrix have $4$ non-zero rows, that will be the case.

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i see... thanks for the answer, one more question, is it possible for a matrix to be NOT be one to one AND NOT be onto?? – Ogen Apr 2 '13 at 4:44
You mean, a matrix that is not onto and not 1-1? yes, for example the matrix $0$ (is a trivial example, but i think you can find more easily, if not, tell me. – Dimitri Apr 3 '13 at 2:15

Seen as an operator, your matrix is a map from $\,5-$dimensional space to a $\,4-$dimensional one, so $\;1-1\;$ it certainly cannot be.

Now, if its rank is $\,4\,$ then it is onto (why?), so you really have to reduce it and check this...

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