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Is this matrix injective or surjective? $M=\begin{pmatrix} 1 & 2 & -3\\ 2 & 3 & -5 \end{pmatrix}$

I need to calculate rank.

$M^{T}= \begin{pmatrix} 1 & 2\\ 2 & 3\\ -3 & -5 \end{pmatrix}$. If we form it with Gauss, we get (it's formed correctly): $\begin{pmatrix} 6 & 12\\ 0 & 5\\ 0 & 0 \end{pmatrix}$

$\Rightarrow rank(M)= 2 \Rightarrow$ not surjective because $M \in \mathbb{R}^{3}$ but $rank(M) \neq 3$

Is it injective? No because $dim(Ker(M)) = 3-2= 1 \neq 0$


Is it good? If not please explain not too complicated. I think there can be trouble at beginning when I transposed?

Edit: I'm not sure if $M$ is really in $\mathbb{R}^{3}$, I said that because we have $2$ lines but $3$ columns. That's fine?

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    $\begingroup$ What do you means for $M \in \mathbb{R}^3$? It is $M: \mathbb{R}^3 \to \mathbb{R}^2$. $\endgroup$ – Emilio Novati Mar 23 '17 at 13:49
  • $\begingroup$ @EmilioNovati Does that mean $M \in \mathbb{R}^{3}$ because it has 3 columns and $M^{T}$ is $\mathbb{R}^{2}$ because it has 2 columns? $\endgroup$ – tenepolis Mar 23 '17 at 13:52
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    $\begingroup$ NO. $\mathbb{R}^3$ and $\mathbb{R}^2$ are the vector spaces between which the matrix operates, not the vector space in which the matrix lies. $\endgroup$ – Emilio Novati Mar 23 '17 at 13:57
  • $\begingroup$ @EmilioNovati Ohh but what is the dimension then? $\endgroup$ – tenepolis Mar 23 '17 at 13:59
  • $\begingroup$ see my answer at: math.stackexchange.com/questions/2170009/… $\endgroup$ – Emilio Novati Mar 23 '17 at 15:07
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This matrix expresses a transformation $M: \mathbb{R}^3 \to \mathbb{R}^2$, so it can't be injective. To prove that it's surjective, though, you just need to find two vectors in $\mathbb{R}^3$ whose images are not scalar multiples of each other (this means that the images are linearly independent and therefore span $\mathbb{R}^2$). $u = (1, 0, 0)$ and $v = (0, 1, 0)$ work for this: $Mu = (1, 2)$ and $Mv = (2, 3)$.

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  • $\begingroup$ So how can I know if it's surjective? I don't want look for vectors that work. There is really no easier way? $\endgroup$ – tenepolis Mar 23 '17 at 14:24
  • $\begingroup$ Computing the rank of the matrix is fine, as well: the rank of the matrix is the dimension of its image, and we know that the matrix can be surjective only if its image has dimension 2. $\endgroup$ – Connor Harris Mar 23 '17 at 14:37
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Your application $M$ goes from $\mathbb{R}^3$ to $\mathbb{R}^2$ so it can't be injective for a dimentional reason. Your calculation for the rank is right because the rank of a matrix $A$ is equal to that of its transpose. Feel free to ask if I wasn't clear .

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  • $\begingroup$ I got the rank but how can I know if it's surjective? $\endgroup$ – tenepolis Mar 23 '17 at 14:22
  • $\begingroup$ The rank of a matrix is defined to be the dimension of the image and you correctly found that it is $2$ that it is also the dimension of $\mathbb{R}^2$ sothe image is the entire space and your matrix is surjective $\endgroup$ – Tommaso Scognamiglio Mar 23 '17 at 14:23

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