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"square matrices have as many eigenvectors as they have linearly independent dimensions; i.e. a 2 x 2 matrix would have two eigenvectors, a 3 x 3 matrix three,
by alephone 11y ago
"square matrices have as many eigenvectors as they have linearly independent dimensions; i.e. a 2 x 2 matrix would have two eigenvectors, a 3 x 3 matrix three, and an n x n matrix would have n eigenvectors, each one representing its line of action in one dimension."
This is not quite right. First of all, if v is an eigenvector of A, then a*v for any non-zero a is also an eigenvector of A, so it makes no sense to say "there are n eigenvectors". Perhaps you meant to say n linearly independent eigenvectors?
But this is also not correct. An n x n matrix does not necessarily have to have n linearly independent eigenvectors; consider for instance A = [1, 1; 0, 1], which has dim N(A - I) = 1 (hence 1 linearly independent eigenvector). Basically, not every matrix is diagonalizable. This example also shows that rank(A) is also not necessarily equal to the number of linearly independent eigenvectors (is that what you meant by linearly independent dimensions?).
A more correct statement would be to say every symmetric matrix has n linearly independent eigenvectors (and its eigenvectors can be chosen to form an orthonormal basis for R^n).
- vonnik 11y agoThanks for making that point. The statement "square matrices have as many eigenvectors as they have linearly independent dimensions" is meant to inform the statements that follow.
- jordigh 11y agoThis statement makes no sense. Independence is not a property of dimensions. A set of vectors is said to be independent if the only linear combination of them that is zero is the trivial linear combination. Otherwise, they are dependent. Or do you have a definition of independence for dimensions? That would first require a different definition of dimension. Ordinarily, dimension is just the cardinality of an independent, spanning set (we call this a basis). It is a single number. A linear space does not have a plurality of dimensions, nor do matrices have an associated plurality of dimensions. Please don't mix technical words without defining them. In other areas of life, humans get along fine by using very imprecise, poetic, ambiguous, and metaphoric language. This is not how mathematics works. Everything you say must have a precise definition. :-(
- vonnik 11y agoHi Jordi - I'm sure you're right. You have an amazing resume. Please help us make ND4J/ND4S better. https://github.com/deeplearning4j/nd4j https://github.com/deeplearning4j/nd4j Nasa JPL is using the framework for climatic modeling, so maybe we can help save the planet together. :) One response I would have to both your comment and @alephone's rewording, which is perfectly correct, is that they are incomprehensible to most people. So the question is: How do you bridge the gap to make non-mathematicians understand? The other would be: It's hard to talk about math. There's always slippage when you try to speak in an everyday way about technical things. I'm not sure how to get around that. Too much precision, in this case, undoes the goal of reaching another audience.
- deleted 11y ago[deleted]
- tanderson92 11y ago> One response I would have to both your comment and @alephone's rewording, which is perfectly correct, is that they are incomprehensible to most people. They are incomprehensible to those who do not understand the mathematics which are being discussed. > So the question is: How do you bridge the gap to make non-mathematicians understand? By teaching them mathematics. > Too much precision, in this case, undoes the goal of reaching another audience. No. In this case what failed was the author's understanding of the criterion for diagonalizability: the operator being normal or self-adjoint (depending on the underlying field). It isn't hard to add a caveat emptor saying "in some cases operators may not have a full set of eigenvectors, so it is only appropriate to say a matrix can have at most as many linearly independent eigenvectors as its order", provided the author realizes that this is indeed the case. Precision is important because there are phenomenally important cases where operators have no eigenvectors at all! Consider an arbitrary rotation in R^2 (fun fact: in odd dimensions all real operators have an eigenvalue, why?); clearly, this fixes no vector.
- vonnik 11y agoI've added a caveat along the lines of your suggestion. > By teaching them mathematics. The terse complacence of this statement, and your easy use of technical terms, make me doubt whether you know what it means to teach mathematics, to empathize with those who lack knowledge, and ultimately to help math exit the ghetto of jargon.
- thearn4 11y agoGood points. Honestly, I feel like a discussion on PCA using the SVD as a background is much more intuitive, and would fit with some of the author has claimed about eigenvectors.