Below is a shear transformation matrix A. If you multiply a vector in $\mathbb{R}^{2}$ by the matrix A you will get back a sheared vector in $\mathbb{R}^{2}$.

A = $\begin{bmatrix} 1&2 \\0&1 \end{bmatrix}$

However, the above matrix is also the inverse matrix (A is the inverse of B) of the following:

B = $\begin{bmatrix} 1&-2 \\0& 1 \end{bmatrix}$

If you multiply a vector (standard basis) by the matrix A you will get back the coordinates of that vector relative to the coordinate system defined by matrix B. Correct?

Therefore, a matrix's "meaning" is dependent on the context we use it in. Is that right?

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    $\begingroup$ You cannot multiply $A$ with a vector in $\mathbb{R}^3$. Do you mean $\mathbb{R}^2$ instead? $\endgroup$ – Jalaj Jan 28 '12 at 2:35
  • $\begingroup$ Corrected. New to this. $\endgroup$ – koin Jan 28 '12 at 3:20
  • $\begingroup$ The commas were neither corrected nor addressed. $\;$ $\endgroup$ – user57159 Jan 28 '12 at 3:42
  • $\begingroup$ A matrix always represents a linear transformation. What is really changing is our interpretation of the elements of $\mathbb{R}^2$, as either locations on the plane, or as coordinate vectors that describe linear combinations of certain vectors. $\endgroup$ – Arturo Magidin Jan 28 '12 at 4:24

I think you are saying that we need to be careful to distinguish between "active" and "passive" ways to think about transformations.

You might be interested to read this wikipedia page helpful http://en.wikipedia.org/wiki/Active_and_passive_transformation

Terry Tao has a detailed discussion at https://profiles.google.com/114134834346472219368/buzz/AWqcUGXVjcs


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