I am trying to separate symmetric part from the $XY^T$ matrix so that I can have a matrix $\hat{X}{\hat{Y}}^{T}$ without any symmetries. For this, I removed the symmetric part and got a skew-symmetric $A$, I want to now decompose this skew-symmetric matrix $A = CD^T - C^T D$ such that $A = \hat{X}{\hat{Y}}^{T}$. Which type of decomposition can I use? Will the decomposition be unique? I am new to matrix decomposition, any help is really appreciated.

Update: $X, \hat{X}, Y, \hat{Y}$ are real and sparse, and $A, B, C, D$ are real


1 Answer 1


The question is a little open-ended, but let me answer two of your subquestions.

The decomposition will not be unique, unless you heavily restrict the class of matrices you are looking at. If $A=\hat X\hat Y^T$ and $Q$ is an orthogonal matrix (meaning $QQ^T=I$), then also $A=(\hat XQ)(\hat YQ)^T$. Whether $(\hat X,\hat Y)\mapsto(\hat X Q,\hat Y Q)$ is an allowable change of decomposition depends on context.

If you are willing to work with complex matrices (entries are complex numbers instead of real), notice that $iA$ is Hermitean and therefore can be decomposed as $iA=U^{-1}DU$, where $U$ is unitary and $D$ is real and diagonal. This gives you $A=-iU^{-1}DU$.

In the end, what is convenient depends on context. What are you trying to achieve with your decomposition?

  • $\begingroup$ I have $X$ and $Y$ sparse matrices, I want to remove symmetries between $X$ and $Y$ such that we get a new $\hat{X}$ and $\hat{Y}$ which are sparse and have very fewer symmetries between them. $\endgroup$
    – newbie
    Commented Feb 24, 2018 at 20:42
  • $\begingroup$ @DushyantSahoo If you choose $Q$ to be a permutation of the basis, multiplying by it preserves sparsity. Therefore my example of non-uniqueness is valid for you. $\endgroup$ Commented Feb 24, 2018 at 20:49

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