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Graph

Linear Algebra Changing Coordinates Problem

How come when solving this problem they decide that the B coordinate vector of v is

\begin{bmatrix} x' \\ y' \end{bmatrix} and not \begin{bmatrix} x \\ y \end{bmatrix} ? Deciding which one is which completely changes the answer of the problem.

Also on the right side of the picture I thought of another way to do it, since \begin{bmatrix} x' \\ y' \end{bmatrix} is just rotated 45 degrees counter clockwise from the original x and y so I just put a 45 degree rotation matrix multiplied by the original coordinates. Then I would solve this for x' and y' and plug them back in the original equation and then I would get the equation of the graph in the form of the standard coordinate system, but this does not work and when I plot it, the ellipse is rotated 45 degrees clockwise and not counterclockwise.

So I am not quite sure why this method does not work, nor do I understand how was the books logic in deciding which vector to be the B coordinate vector of V.

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    $\begingroup$ It is reasonable that you find something confusing here: "choosing a basis" and "coordinates with respect to a basis" are (in a sense that is not so elementary) "dual", which leads to the sort of seeming paradox you have found: is the matrix $A$, or is it $A$-transpose? $\endgroup$ Nov 24, 2023 at 22:47

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To answer the question about the B coordinates, $x’$ and $y’$ are the definitions of the new coordinates. The rest of the problem is then to transform $x$ and $y$ to $x’$ and $y’$ because you are transferring the coordinate system from $v$ to $[v]_B$. The reason you need a $-45$ degree rotation instead of a $45$ degree rotation is because you are rotating the coordinate system to an ellipse that is already rotated 45 degrees. This is similar to rotating the ellipse $-45$ degrees in the original coordinate system.

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