I understand the definition of a symmetric matrix in terms of how it’s components are related. But functionally, what does it entail about the linear transformation it represents? For example, block tri-diagonal matrices have special relations between entries but they also, functionally, tell us that some non-trivial vector subspace is invariant under the linear transformation with respect to a particular basis. Incidentally, what do skew-symmetric matrices represent, functionally ?
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1$\begingroup$ Are you aware of the spectral theorem for symmetric matrices? $\endgroup$– Ben GrossmannDec 24, 2020 at 14:41
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$\begingroup$ You could also use the characterization that $A$ is symmetric if and only if for all vectors $x,y$, we have $$ \langle Ax,y \rangle = \langle x, Ay \rangle $$ where $\langle x,y \rangle$ denotes a dot-product (and $A$ is a matrix with real entries). $\endgroup$– Ben GrossmannDec 24, 2020 at 14:44
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1$\begingroup$ If you want to extend the definition to something invariant of the choice of inner product, you might ask which matrices are symmetric relative to some possibly oblique basis. The matrices that have this property, as it turns out, are precisely the diagonalizable matrices with real eigenvalues. $\endgroup$– Ben GrossmannDec 24, 2020 at 14:57
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1$\begingroup$ If you ignore the Gram matrix (i.e. ignore the inner product structure of the space), then as I said in my most recent comment, the only thing you can say about a symmetric matrix (with real entries) is that it is diagonalizable with real eigenvalues $\endgroup$– Ben GrossmannDec 24, 2020 at 15:08
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1$\begingroup$ If you want to stick to the real numbers, then skew-symmetric matrices are such that the vector space can be decomposed into a direct sum of invariant subspaces of the matrix such that each subspace has dimension at most $2$ and on each vector space, the matrix either applies a scaled $90^\circ$ rotation or the $0$ transformation. In contrast, diagonalization is about a decomposition into one-dimensional invariant subspaces. $\endgroup$– Ben GrossmannDec 24, 2020 at 15:24
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In the comments (and in the linked discussion) on the question, I make the following claim:
$M$ is symmetric relative to at least one choice of (possibly oblique) basis if and only if $M$ is diagonalizable with real eigenvalues. $M$ is skew-symmetric relative to at least one choice of basis if and only if $M$ is a direct sum of scaled $90^\circ $ rotations and zero transformations.
First, the symmetric case. If $M$ is symmetric, then the spectral theorem states that $M$ is diagonalizable with real eigenvalues. Conversely, if $M$ is diagonalizable with real eigenvalues, then there is a basis relative to which the matrix of $M$ is diagonal with real diagonal entries. Since this diagonal matrix is symmetric, $M$ is symmetric relative to this choice of basis.
For case where $M$ is skew-symmetric, there are two common approaches. For the easy direction: if $M$ is a direct sum of $90^\circ$ rotations and zero transformations, then there is a basis relative to which the matrix of $M$ is the block-diagonal skew-symmetric matrix $$ \pmatrix{0 & \kappa_1 \\-\kappa_1 & 0 \\ && \ddots \\ &&& 0 & \kappa_p\\ &&&-\kappa_p & 0 \\ &&&&&0 \\ &&&&&&\ddots\\ &&&&&&& 0}. $$ There are two approaches for the converse. One is essentially to apply the spectral theorem for Hermitian matrices, noting that if $M$ is skew-symmetric then the complex matrix $iM$ is Hermitian. Alternatively, we can systematically construct a basis relative to which the matrix of $M$ has the above block-diagonal form as is outlined in this post and the proof linked therein.
answered Dec 24, 2020 at 16:21