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Symmetry of Orthogonally diagonalizable matrix

  1. Apr 25, 2015 #1
    Can someone confirm or refute my thinking regarding the diagonalizability of an orthogonal matrix and whether it's symmetrical?

    A = [b1, b2, ..., bn] | H = Span {b1, b2, ..., bn}. Based on the definition of the span, we can conclude that all of vectors within A are linearly independent. Furthermore, we can then conclude that Rank(A) = n.

    If Rank(A) = n then none of the vectors in A can be made as a linear combination of the other n-1 vectors. Since A can be row-reduced to an identity matrix and the transpose of the identity matrix is itself. Can it be concluded that the original matrix A is also symmetrical (AT = A)?
  2. jcsd
  3. Apr 25, 2015 #2


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    This is incorrect. However, if A is orthogonal, then you can use that an orthogonal matrix is invertible (the transpose is the inverse), and that the columns of an invertible matrix are linearly independent.
  4. Apr 25, 2015 #3
    When you say that this is incorrect, are you referring to my rational regarding symmetrical matrices or vector linear independence?
  5. Apr 25, 2015 #4


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    I was referring to the part of your post that I quoted. There's nothing about the definition of "span" that allows you to conclude that ##\{b_1,\dots,b_n\}## is linearly independent.
  6. Apr 25, 2015 #5
    Unless of course you know somehow that the dimension of ##\text{span}\{b_1,...,b_n\}## is ##n##.
  7. Apr 26, 2015 #6


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    Do you intend here that H is itself n dimensional? If so, you need to say that. If not, your following statement is not true.

  8. Apr 27, 2015 #7
    Orthogonal matrices are a particular case of unitary matrices, which are in turn a particular case of normal matrices. Any normal matrix ##A## is orthogonally diagonalizable, i.e. represented as ##A=UDU^*##, where ##U## is a unitary and ##D## is diagonal matrix (generally with complex coefficients).

    Orthogonal matrices do not need to be symmetric, a rotation matrix $$R_\alpha = \left(\begin{array}{cc} \cos \alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{array}\right)$$ with ##\alpha\ne \pi n## can serve as an example.

    On the other hand, there are symmetric orthogonal matrices, for example the identity matrix ##I## or the matrices $$\left(\begin{array}{cc} 1 & 0 \\ 0 &-1 \end{array}\right), \qquad \left(\begin{array}{cc} \cos \alpha & \sin\alpha \\ \sin\alpha &- \cos\alpha \end{array}\right) $$ (the last two matrices are unitarily equivalent).
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