I know how to approach finding a matrix of a linear transformation with respect to bases, but I am stumped as to how to approach this in the case of 'determining' and 'proving' whether or not 2 bases $\mathscr C_2$ and $\mathscr C_3$ exist for $\Bbb R^2$ and $\Bbb R^3$, respectively

The linear transformation T : $\Bbb R^3 \rightarrow \Bbb R^2$ in question is defined by T$\begin{pmatrix} x \\ y \\ z \end{pmatrix}$ = $\begin{pmatrix} 2x + y \\ y + 2z\end{pmatrix}$

Where the matrix of T with respect to $\mathscr C_3$ and $\mathscr C_2$ is given as

$Matrix_{\mathscr C_3,\mathscr C_2}$(T) = $\begin{pmatrix} 1 && 0 && 0 \\ 0 && 1 && 0\end{pmatrix}$

  • 1
    $\begingroup$ So you want the matrix representation of the linear transformation $T$, assuming the standard basis in each space? Or are you looking for something else? $\endgroup$
    – Arthur
    Oct 25 '18 at 16:37
  • 2
    $\begingroup$ @Arthur, I think what they want is to find bases of $\mathbb R^2$ and $\mathbb R^3$ such that the matrix of the given transformation equals $\begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\end{bmatrix}$ with respect to those bases. $\endgroup$
    – memerson
    Oct 25 '18 at 16:51
  • $\begingroup$ @memerson That seems more reasonable than my suggestion, now that you say it. $\endgroup$
    – Arthur
    Oct 25 '18 at 17:23
  • $\begingroup$ Yes, I already have the matrix representation. The bases $\mathscr C_3$ and $\mathscr C_2$ are to be found (if they exist), where the matrix representation of the transformation T with respect to those bases is given $\endgroup$
    – nrg1998
    Oct 25 '18 at 17:25
  • $\begingroup$ You can always find bases for which the matrix has some number of $1$s along the main diagonal and the rest of the matrix is zero. That’s a consequence of the rank-nullity theorem. The key to this solving this problem is determining how many $1$s there should be for $T$. $\endgroup$
    – amd
    Oct 25 '18 at 18:54

It turns out that for any linear map $T:V\to W$ with $V$ and $W$ finite-dimensional, you can always find a pair of bases such that the $\dim W\times\dim V$ matrix of $T$ has the form $$\left[\begin{array}{c|c}I_r & 0 \\ \hline 0&0\end{array}\right],$$ where $r=\operatorname{rank}T$. This is a consequence of the rank-nullity theorem. The proof gives you a way to construct these bases: With $n=\dim V$ and $m=\dim W$, let $\{v_r,\dots,v_n\}$ be a basis for $\ker T$. Extend this to a complete basis for $V$. The vectors $w_1=Tv_1,\dots,w_{r-1}=Tv_{r-1}$ are distinct and linearly independent (why?), so they form a basis for the image of $T$. Extend this to a complete basis of $W$. The matrix relative to these bases then has the desired form.

So, if the only thing that this problem asks you to do is to verify that such bases exist, all you need to do is compare the rank of $T$ to the rank of the given matrix. The latter is obviously 2, so is $\operatorname{rank}T$ also equal to 2? If you also have to produce a pair of bases, the above procedure tells you how to find such a pair. If you’re dealing with real vector spaces, it should be clear from the construction that there’s an infinite number of basis pairs that work.

  • $\begingroup$ I can see why that would be the case, since the number of lin. indep. columns in the identity is equal to the rank of T. Are the vectors $w_1$ = T $v_1$,...,$w_{r-1} = *T* $v_{r-1}$ linearly independent because the transformation matrix is defined as the identity? i.e. the linear combo of the w basis vectors only has 1 non-zero co-efficient? Also, the questions specifically asks to determine if the bases $\mathscr C_3$ and $\mathscr C_2$ could exist as bases for $\Bbb R^3$ and $\Bbb R^2$, respectively. How would one prove this? $\endgroup$
    – nrg1998
    Oct 27 '18 at 16:53
  • $\begingroup$ @nrg1998 The above is a constructive proof that those bases exist. As for why the $Tv_i$ are linearly independent, proving that is usually given as an exercise in linear algebra courses. You should try to prove it for yourself (hint: try proof by contradiction). It’s not to do with the form of the matrix because you don’t have the matrix yet—that argument would be circular. $\endgroup$
    – amd
    Oct 27 '18 at 19:13

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