ok, so linear algebra..

College Life
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<p>besides recasting systems of equations in a nice form (or anything that can be represented as a system of equations), what does it actually enable one to do? </p>

<p>how does one approximate solutions is what I would also be curious about too. eigenvalues/vectors are, of course, quite interesting (way more interesting than I can imagine). But I do know some applications. But I wonder what is the difference between finding all eigenvalues and finding if a particular solution exists (not necessarily an eigenvalue), and which is computationally easier. </p>

<p>Also about analyzing a certain finite set of linear equations for solutions, and how to go about that.</p>

<p>I don’t expect any answers, dw :slight_smile: (and for reasons the reader might not expect).</p>

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<p>Your question is so poorly worded that I am not even sure if I can answer it properly.</p>

<p>The goal of even finding eigenvalues are to find your basis for your eigenspace which is a “general solution”. Particular solutions are well, particular solutions.</p>

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<p>so eigenvectors necessarily form a basis for the vector space?</p>

<p>Also, is part of what makes eigenvectors special the fact that they form a basis by themselves <em>and</em> when operated on by the matrix?</p>

<p>to find a particular solution you just invert the matrix. hm. I’m looking to invert the same vector by a set of matrices. Anyways.</p>

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<p>For the Eigenspace, be specific.</p>

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<p>What do you mean by themselves? </p>

<p>Ax = λx </p>

<p>(A-λI)x = 0</p>

<p>You are basically solving the nullspace and finding the vectors that form this space corresponding to your eigenvalues λ. Those new vectors will form your basis</p>

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<p>A particular solution of what? You trying to test whether an eigenvector is in the eigenspace?</p>

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<p>When I took Linear Algebra my professor told us a brief story…</p>

<p>He was at a conference giving by a well known mathematician. When the topic of linear algebra came up, someone from the audience asked “What is the essence of eigenvalues and eigenvectors? (aka, what are they exactly used for?)”.</p>

<p>The great mathematician stood at the podium and gave this inquiring mind an honest answer. “We don’t really know, but we’re hoping to find out”.</p>

<p>So yeah, eigenvalues are still a bit of an unknown (as far as applications go). </p>

<p>I think the best use of Linear algebra is for multiple regression modeling, and a related field known as Design of Experiment.</p>

<p>I know it’s used in signal processing, although I’m not exactly sure how.</p>

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<p>Yeah… took a semester of this stuff, memorised the stuff, aced the stuff, and still don’t know/care about it’s practical uses. To me and many others, linear algebra is just the busywork prereq for other classes.</p>

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<p>Principal Component Analysis in statistics, which essentially provides a way to remove the least relevant factors to a set of data, is derived using singular value decomposition, which makes use of eigenvectors. Once you’ve derived it, though, you can get away without singular value decomposition - you just need to find eigenvalues and eigenvectors, which represent “components” of the data.</p>

<p>Linear’s also useful for solving systems of linear differential equations, since the technique for solving them involves exponentiating a matrix (essentially by use of the taylor series for e^x, but x is a matrix). In these cases, we want orthogonal eigenvectors, since this lets us decompose a matrix into A = SDS^-1, where S is the eigenvectors as columns and D is the eigenvalues along the diagonals. This has nice properties like A^n = S(D^n)S^-1, making multiplication (and inversion) easy. When you get this form, the Taylor representation of e^A simplifies nicely, and that’s where you get solutions involving e to the power of eigenvalues of a matrix. In the case of repeated eigenvalues, another decomposition trick is needed, though.</p>

<p>Approximating solutions can be done with least squares, since even if Ax=b has no solutions, A<em>Ax = A</em>b, where A* both takes the transpose of A and takes the complex conjugate of all entries, does necessarily have a solution (though not necessarily a unique one, since A*A can be non-invertible). Making use of projections onto solution spaces, it can be shown that this provides the best solution to Ax=b, even if no actual solutions exist.</p>

<p>Not sure what you mean by determining if solutions exist, beyond the standard Ax=b has solutions iff A is invertible (and has non-zero determinant, and Ax=0 has infinitely many solutions, etc.). A necessarily has a number of (not-necessarily distinct) eigenvalues equal to the rank of A, but finding them becomes tough as A gets big, since there’s no formula to solve 5th degree polynomials exactly. I’d imagine one would have to estimate roots and use the A<em>Ax=A</em>b trick to estimate eigenvectors, then orthogonalize them to make use of them, but I don’t know when those would be useful.</p>

<p>However, I must argue that it’s not necessarily true that eigenvectors are orthogonal. In the case where all eigenvalues have multiplicity 1, that’s true. However, if they don’t, multiple eigenvectors from the same eigenvalue need not be orthogonal. That’s when you use Gram-Schmidt (or other techniques) to orthogonalize them.</p>

<p>(I like linear algebra and can deal with abstract algebra…but I don’t claim to understand or know most of the applications).</p>

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<p>I wasn’t expecting such thoughtful replies! I appreciate them all :).</p>