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A reasonable analogy for understanding similar matrices?

  1. Dec 12, 2015 #1
    I don't really feel that I understand what it means for two matrices to be similar.

    Of course, I understand the need to understand ideas on their own terms, and that in math analogies are very much frowned upon. In asking if you know of any "reasonable" analogies for what it means for two matrices to be similar, I'm also asking for a broader kind of informal help with the ideas of similar matrices, Jordan Forms and diagonalization.

    My cousin (who has not studied LA beyond High School vectors and matrices, solving systems of equations, etc) recently asked me if similar matrices have the same angle between the vectors in the matrix, drawing an obvious analogy to geometry with similar triangles. I replied that similar matrices have the same set of eigenvalues (necessary but not sufficient) and must all share the same Jordan Form...of course, I don't really know what a Jordan Form is - I've seen the method for constructing them and I know that they are as near to a diagonalization as is possible for a given matrix that cannot be diagonalized; and that the diagonal form of a diagonalizable matrix is its Jordan Form.

    Which brings me to what I do and don't understand (and suspect) about diagonalization.

    A diagonalization is a change of basis to a full eigenvector basis for a matrix/transformation. So if you convert all your vectors of interest into that same eigenbasis, then you can just multiple that vector by the diagonal lambda matrix, which is much more computationally efficient and analytically insightful than doing the transformation in some other basis -- as the former merely involves a simple scaling of the basis vectors to achieve the same transformation.

    So I went into MATLAB's eigshow(), and went to compare a matrix to it's diagonal form

    [tex]\frac{1}{4}\begin{bmatrix}1 & 3\\ 4 & 2\end{bmatrix} \ \text{versus} \ \begin{bmatrix}-0.5 & 0\\ 0 & 1.25\end{bmatrix}[/tex]

    And it definitely seems that both those similar matrices are performing the same transformation, just in a different basis.

    RXBDmJl.jpg

    So, a diagonal form is doing the same transformation, but in a different (more manageable) basis...?

    Is the Jordan Form of a matrix doing the same transformation just in a different basis?

    Is there a connection between Jordan Form and SVD?
     
  2. jcsd
  3. Dec 12, 2015 #2

    mathwonk

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    "Is the Jordan Form of a matrix doing the same transformation just in a different basis? "

    yes, thats what similar means in general. i am assuming by "A is similar to C" you mean that C has form B^-1AB for some invertible B. Angles however are usually not preserved, since the basis does not have to be mutually orthogonal. For certain matrices, e.g. symmetric ones I believe, you can actually find an orthonormal basis that diagonalizes it.

    So for linear maps or matrices that can be orthogonally diagonalized, it just means you grab hold of the axis system in space and rotate it around (without changing angles between axes) into a new position and in that position your map just stretches along the axes.

    Or said backwards, you just look for those lines along which your map merely stretches things, those are guaranteed to be orthogonal (in good cases), and you choose them as your new axes.

    So in general finding all matrices similar to yours just means looking at the matrices of your map in all possible bases. The theorem on jordan form says that if the characteristic polynomial splits into linear factors in your chosen field, which can always be achieved in some finite extension field, then the matrix is similar to exactly one jordan matrix, except for permuting the order of the blocks, or deciding whether your jordan matrices should be "upper or lower diagonal".

    I have written several free sets of course notes on this, in math 845, 4050, 8000, on my webpage, the most detailed being math 845:

    http://alpha.math.uga.edu/~roy/

    The 845 notes are a bit abstract, using the notion of modules, but the fact you deduced is stated on page 47 of the 845-1 notes. The 4050 notes were aimed more at undergrads and tried to avoid the concept of modules, (a generalization of vector spaces where the scalars come from just a ring, which I take to be commutative, but not necessarily a field).

    there are also excellent free notes out there by sergei treil, at brown ("linear algebra done wrong"), and others.
     
    Last edited: Dec 12, 2015
  4. Dec 12, 2015 #3

    HallsofIvy

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    An "m by n matrix" is a representation of a linear transformation from an n dimensional vector space to an m dimensional space in specific bases for the vector spaces. Two such matrices are "similar" if and only if they represent the same linear transformation, just in different bases for one or both of the vector spaces.
     
  5. Dec 12, 2015 #4
    Isn't similarity an equivalence relation between square matrices ?
     
  6. Dec 13, 2015 #5

    mathwonk

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    well of course everyone makes his own definitions, but probably yes, similarity is mostly used for square matrices, and the same basis is used both times. What Halls is referring to is the same idea for maps between two different spaces, with two different bases of course, and maybe most people call that simply equivalence?
     
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