I have very basic question. In the Linear Algebra and Its Applications by Lay gives a definition of the mapping. The author also says that $[0,0; 0,1]$ is not a map between $\mathbb{R}^n$ and $\mathbb{R}^m$. Can you please elaborate why? Suppose I had a vector $[1,1]$. After the mapping it became $[1,0]$. Why this does not satisfy the mapping rule? Is this a direct consequence of the fact that the transformation matrix is singular? So, will this always be the case for singular matrix?

Update: Here is verbatim what was said in the book (Table 4 is a figure below): enter image description here

Update 1: Here is the example that satisfies ONTO mapping from $\mathbb{R}^2$ and $\mathbb{R}^2$ enter image description here

about the "onto" term:




enter image description here

  • 1
    $\begingroup$ It appears he claims it is not a mapping ONTO $\mathbb{R}^m$, which is a big deal. Saying "into/between" is very, very different from "onto." $\endgroup$
    – Randall
    Aug 30, 2017 at 15:51
  • $\begingroup$ The author probably says (or wanted to say) that this is not an onto mapping. The vector [1,0] can not be obtained as the image of any vector in the domain. It will be the case for every singular matrix from an $n$ dimensional space to $n$ dimensional space because by definition a matrix which is singular sends a direction to $0$ which means that it has to reduce the dimension of the space. Food for thought: infinite dimensional linear maps can send a direction to $0$ but still be onto. $\endgroup$
    – Sina
    Aug 30, 2017 at 15:52
  • $\begingroup$ From your update it is quite clear that the author is defining the word "onto". That is why it is in bold. The projection is a mapping into $\Bbb R^2$, but it is not onto $\Bbb R^2$. "Onto" is an additional condition that mappings can meet. This condition - that every point in the codomain is the image of something under the map - is also called being "surjective". Occasionally the prefix "epi" is also used for such maps - most commonly in "epimorphism". $\endgroup$ Aug 30, 2017 at 16:59
  • $\begingroup$ Thank you a lot for the answers. The "onto" term was new to me (I now updated my question with some links). So, if got it correctly, the my original mapping was not onto because in $\mathbb{R}^m$ there are vectors (like, [1,1]) that can never be obtained after transformation. Correct? So, whether the transformation matrix is singular or not is related to whether the mapping is one-to-one, but is unrelated to whether the transformation is onto. Does it make sense? $\endgroup$
    – student
    Aug 30, 2017 at 18:04

1 Answer 1


Note that the linear map $$ \begin{bmatrix} 0&0\\0&1 \end{bmatrix} $$ which takes vectors from $\mathbb{R}^2\to \mathbb{R}^2$. What vector in $\mathbb{R}^2$ gets mapped to $\begin{bmatrix} 1\\0\end{bmatrix}$?

Taking $b=\begin{bmatrix} 1\\0\end{bmatrix}$ does the mapping satisfy the definition of being onto? Indeed, it should be clear from the geometric interpretation that any vector of the form $$ \begin{bmatrix} a\\0\end{bmatrix} $$ won't be mapped to by the projection.

  • $\begingroup$ I see. Thanks. I also commented above. So, I cannot get any vector in the destination space with X that is different from 0. So, not all space is reachable. Correct? $\endgroup$
    – student
    Aug 30, 2017 at 18:08
  • $\begingroup$ yup! that's exactly right. Since you need some vector with nonzero $x$ value to span $\mathbb{R}^2$, your mapping is not onto (also called "surjective" in case you see this word come up) $\endgroup$ Aug 30, 2017 at 18:09

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