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Working through "Linear Algebra and Its Applications" by Gilbert Strang. This comes from chapter 5.6 on similarity transformations.

THe similar matrix $B = M^{-1}AM$ is closely connected to $A$, if we go back to linear transformations. Remember the key idea: Every linear transformation is represented by a mtrix The matrix depends on the choice of basis!. If we change the basis by $M$ we change the matrix A to a similar matrix $B$.

Similar matrices represent the same transformation $T$ with respect to different bases.

He then gives this small example graph.

similar matrices

which I interpreted as this...

$A$ is a linear transformation projecting any vector in $\mathbb{R}^2$ onto the column space of $A$ using (1, 0) and (0, 1) as basis vectors that define our point of view.

If we change the basis vectors according to (1, 1) and (1, -1) by $B = M^{-1}AM$ with $M$ being a matrix of the new basis (eigenvectors of $A$ in this example. Then the new matrix $B$ will transform any point in $\mathbb{R}^2$ with the new basis vectors.

But here is where I am confused... It says $V_2$ is projected to zero and $V_1$ projects to $V_1$, but if I use $\Lambda$ (which is equal to $B$ here) as the transformation, I can see $V_1, V_2$ both project to (1, 0) which doesn't make sense

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    $\begingroup$ $V_1$ is $(1,-1)$ in the original basis, but it's $(1,0)$ in the new basis (which is $V_1,V_2$). Similarly, $V_2$, $(1,1)$, and $(0,1)$. $\endgroup$ – Gerry Myerson Aug 5 '19 at 2:54
  • $\begingroup$ so with the new bases the two new projections (1, 0) are equal to the old projections in the old bases on the left picture? $\endgroup$ – deltaskelta Aug 5 '19 at 2:55
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    $\begingroup$ $V_1$ projects to $V_1$, which is $(1,0)$ in the new basis. $V_2$ projects to the zero vector, which is $(0,0)$ in any basis. Maybe I don't understand your question. The new basis vectors are not the same as the old, so their images under the transformation are not the same as the images of the old basis vectors. $\endgroup$ – Gerry Myerson Aug 5 '19 at 2:59
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    $\begingroup$ You're not listening. $\Lambda$ is the matrix for the transformation with respect to the new basis. With respect to the new basis, $V_2$ is $(0,1)$, and $\Lambda V_2$ is the zero vector. $\endgroup$ – Gerry Myerson Aug 5 '19 at 3:29
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    $\begingroup$ Yes, it looks like you missed that. If you understand the situation now, you can write up how you see it and post that as an answer. $\endgroup$ – Gerry Myerson Aug 5 '19 at 3:37
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Thanks to the comments I was able to see that after the basis change occurs, $V_1$ is the new (0, 1) so when that is transformed by $\Lambda$, it projects to 0 and $V_1$ is the new (1, 0) which stays right in place when we multiply $\Lambda (1, 0)^T$

I think the important parts that I was missing was that the projection stays the same (we are still projecting to the same line) but out point of view of where the gridlines are has changed (which is what the basis vectors tell us).

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