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I have an $m\times n$ binary matrix, $A$ i.e. $a_{ij} \in \begin{Bmatrix} 0,1 \end{Bmatrix}$. The matrix $S$ is defined as the following: $S=AA^{\top}$ and is hence a symmetric positive semidefinite matrix. If rank{$A$}=rank{$S$}=$r$, then I can find the following decomposition for $S$: $$ S = Q \Lambda Q^{\top} $$ where $Q$ is an $m\times r$ matrix and $\Lambda$ an $r \times r$ diagonal matrix containing eigenvalues of $S$. Columns of $Q$ are independent normalized eigenvectors of $S$. Hence $Q^{\top}Q=I$ is the identity matrix. My confusion started from reading this in wikipedia, that for the complex normal matrix $S$, there exists the $unitary$ matrix $Q$ such that $A=Q\Lambda Q^{\top}$. Doesn't unitary mean that $QQ^{\top}=I$ too?(wikipedia.com) For the above $m$ by $r$ eigenvectors matrix $Q$, $Q^{\top}Q=I$ but $QQ^{\top} \neq I$. So, how can I construct such a unitary matrix $Q$

Moreover based on the same Wiki page, every $m\times m$ real symmetric matrix has m linearly independent real eigenvectors. Does that also hold for singular matrices? How can one find eigenvectors corresponding to zero eigenvalues ? I guess it would be an arbitrary set of $m-r$ independent vectors, that together with the $r$ eigenvectors corresponding to nonzero eigenvalues, will make a basis for $\mathbb{C}^m$. Would $QQ^{\top}=I$ for any such $Q$, given that $Q^{\top}Q=I$? Can anyone clarify this for me please.

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Sort of, for a unitary matrix the formula is QQ*=I, where Q* denotes the conjugate transpose/adjoint of Q. Refer here as the technique of decomposition of A is known as Singular Value decomposition –  Zarks Sep 20 '11 at 4:05
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1 Answer

up vote 2 down vote accepted

In the definition of the eigendecomposition of $S$, you say that $Q$ comprises only those vectors for the nonsingular part of S. Typically, one writes a complete eigendecomposition where $Q$ is a square matrix of size $m\times m$. As you hint in your final sentences, the eigenvectors corresponding to the zero eigenvalues should span the nullspace of $S$. They can't be arbitrary vectors such that $Q$ spans $C^m$; they must also be orthogonal to the eigenvectors corresponding to the nonzero eigenvalues. When you perform the full eigendecomposition, the $Q$ is square, and in this case orthogonal, so then $QQ^T = Q^TQ = I$.

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Since the eigenvectors are orthonormal, $Q^\{top}Q=I$ is trivial. How do we prove the other part ($QQ^\{top}=I$) ? –  Austin Sep 20 '11 at 10:42
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