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I'm currently learning eigenvectors and Linear Transformations. And this basis change part is particularly confusing for me..

I feel like there are two types of questions,

  1. First type, you have a matrix with basis $B =\{b_1,b_2\}$ and then you change it to basis $C = \{c_1,c_2,c_3\}$.

ex) $T(b_1) = 3c_1-2c_2+5c_3$ and $T(b_2) = 4c_1+7c_2-c_3$.

In this case the coefficients of the $c_1,c_2,c_3$ become the column vectors for the transformation matrix, so $\pmatrix{3&4\cr -2&7\cr 5&-1\cr}$. (NOTE: original was [(3,-2,5),(4,7,-1)] but the intent was clear from context.)

  1. Second type, you are given $T\pmatrix{x_1\cr x_2\cr} = \pmatrix{x_1+3x_2\cr 5x_2-x_1\cr 4x_1+x_2\cr}$. Then the transformation matrix is transformed into a matrix whose first column vector is coefficients of $x_1$ and that of $x_2$ in the second column, so $\pmatrix{1&3\cr -1&5\cr4&1\cr}$. (NOTE: Original was [(1,-1,4),(3,5,1)].)

What's the difference between these two, and how come there are differences in getting the transfomration matrix between the two? Aren't they both just basis change?

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  • $\begingroup$ In case (2), this is correct only if your bases are the standard bases. If your bases are not both the standard basis, you have a matrix that you need to multiply in front and/or in back. $\endgroup$ – Christopher Carl Heckman Apr 26 '16 at 6:59
  • $\begingroup$ you mean when T(x1,x2) is not x1 and x2? it could be T(4*x1,6*x2)? $\endgroup$ – Hello Apr 26 '16 at 7:02
  • $\begingroup$ No ... I mean your basis for $\mathbb R^2$ might be $\displaystyle\left\{ {3\choose 2}, {-1\choose 3}\right\}$ or something like that. $\endgroup$ – Christopher Carl Heckman Apr 26 '16 at 7:03
  • $\begingroup$ In that case, how would I solve it ? $\endgroup$ – Hello Apr 26 '16 at 7:05
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    $\begingroup$ The explanation is too long. Check out this PDF: math.unm.edu/~loring/links/linear_s06/matrixRep.pdf $\endgroup$ – Christopher Carl Heckman Apr 26 '16 at 7:15

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