I am self-studying a chapter on Matrix/Linear transformations, and have encountered a question that I am unsure how to answer.

So far, to prove that a transformation is one-to-one, I have shown that the column vectors in the matrix are linearly independent (equivalently, show that the matrix is invertible).

The form of this question confuses me as I do not know how to represent the transformation as a matrix:

$T(x_1,x_2,...,x_n) = (0,x_1,x_2,...,x_n)$

It was given that $n \geq 2 $. Any tips or clarifications at all would be appreciated!


1 Answer 1


Each column of the matrix representation of $T$ is the vector with the coordinates of the basis vectors. For instance, if we take the default basis on $\mathbb{R}^n$, then the first basis vector is $(1,0,0,\dotsc)$, which gets mapped to $(0,1,0,0,\dotsc)$, so the first column will be $(0,1,0,0,\dots)^\top$. Because the original basis vectors were linearly independent (by definition of a basis), we immediately see the columns of this matrix are as well (do you see why?).

  • $\begingroup$ Okay, so for any transformation T, we can determine the matrix by determining how the transformation would affect the basis vectors? And since the original input into T was a basis for Rn, then the output is a basis for R(n-1), so it will be linearly independent? $\endgroup$ May 29, 2017 at 17:21
  • $\begingroup$ @JessC That is how you do it in general; for each basis vector, you determine the image of that basis vector, and express those images in the basis of the 'outcome space'. In this case, however, the map $T$ is one-to-one but not onto as it's called. It is a map to $\mathbb{R}^{n+1}$, and you can't make all vectors from $\mathbb{R}^{n+1}$ with the column of the matrix of $T$: the first coordinate is missing. Linear independence is not sufficient to be a basis. $\endgroup$
    – SvanN
    May 29, 2017 at 17:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .