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For example, given the transformation matrix (column matrix): $$A=\begin{bmatrix}1&0&1\\0&2&0\\1&0&1\end{bmatrix}$$

Eigenvectors $E_1=(1,0,1), E_2=(0,1,0), E_3=(-1,0,1)$

"The geometrical interpretation of this would be the orthogonal projection on the plane $-x+z=0$, because eigenvector one and two are both perpendicular to the third eigenvector. After its been projected onto the plane, it is stretched by a factor of two."

This seems like an important skill, I have no good idea on where I can find examples like these, so that I can become familiar with common transformation matrices.

Any resource and help is welcome.

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  • $\begingroup$ This is a special case of the theory of eigenvalues and eigenvectors. Do you know that theory? $\endgroup$ – Lee Mosher Mar 17 '15 at 19:36
  • $\begingroup$ @LeeMosher Yes, I'm familiar with eigenvalues/eigenvectors. $\endgroup$ – B. Lee Mar 17 '15 at 19:36
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    $\begingroup$ the orthogonal projection should make the projection vector shorter than the one projected or keep it at the same length. $\endgroup$ – abel Mar 17 '15 at 19:46
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    $\begingroup$ what you can do is study the geometric transformations first. like orthogonal projection, reflection and rotation in $R^2$ well. you can tell what the eigenvalues and eigenvector by studying the transformation itself; not the other way around. $\endgroup$ – abel Mar 17 '15 at 19:50
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    $\begingroup$ The above A is not an orthonormal transformation because the 1st and 3rd columns are identical. $\endgroup$ – ja72 Mar 17 '15 at 19:56
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The matrix itself is just a collection of values. What they can do to a vector, or more generally to a region made up of vectors, is what's important. To geometrically think of the transformation, take the determinant, trace, and eigenvalues. If the absolute value of the determinant is greater than 1 then the transformation stretches out the region, where each point in the region is a vector, it has an area greater than when it started. If its less than 1, then it compresses the region. If its equal to 1, it preserves the area of the region. The trace will tell you how fast the determinant changes area, honestly there is little intuition for the trace so I'll keep it at that. The eigenvalues and eigenvectors, tell you about points, eigenvectors in the region, where the $matrix*vector$ is just the $eigenvalue*vector$. These two are most useful if you apply the matrix transformation more than once, even infinitely.

The eigenvalues for this matrix are 2,0, and 2. This means that the applying this matrix transformation infinitely to the eigenvectors will make two of the values infinite, and one of the values 0. This is because $matrix*eigenvector=eigenvalue*eigenvector$, which can be factored out and reapplied to give $matrix^n*eigenvector=eigenvalue^n*eigenvector$ The determinant is 0, so the area of the region is reduced under the transformation, which isn't surprising, can you figure out why? The trace is 4 so the area transformation changes throughout the region.

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  • $\begingroup$ Wow! Amazing answer. I have only one question to ask. If I diagonalize a matrix, what information does this give me? Is a diagonalized matrix easier to analyze than its counterpart matrix of eigenvectors? Thank you so much! $\endgroup$ – B. Lee Mar 17 '15 at 20:11
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    $\begingroup$ I would think diagonalized matrix would be easier to deal with, but it varies depending on context. Here's why. The determinant is equal to the product of the eigenvalues, and the trace is equal to the sum of the eigenvalues, obviously a sentence mentioning that does not give the identities justice. So in my opinion, a diagonal matrix gives more information on the surface level. $\endgroup$ – Zach466920 Mar 17 '15 at 20:21

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