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?

  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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 $Mv = (2, 3)$.

  • $\begingroup$ 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

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 .

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.