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Is it possible to draw any parallels between the SVD and QR decomposition of a matrix?

Moreover, for a given matrix $\mathbf{A}\in\mathbb{R}^{n\times m}$, under what conditions, the $\mathbf{U}$ matrix coming from singular value decomposition of $\mathbf{A}$ is equal to $\mathbf{Q}$ matrix obtained via QR-decomposition of $\mathbf{A}$

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  • $\begingroup$ Why does your title differ from the question ? $\endgroup$
    – user65203
    Jul 6, 2017 at 19:29
  • $\begingroup$ I realized that difference after posting. Anyways let me come up with a better title $\endgroup$
    – NAASI
    Jul 6, 2017 at 19:39
  • $\begingroup$ check out this post on QR for SVD from Cleve Moler's blog $\endgroup$
    – user11260
    Dec 16, 2019 at 8:02

1 Answer 1

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I don't know if there really is a link between them... Notice that those decompositions are not unique, so the question is a bit unclear. But here is some thoughts on your second question. Maybe it can help you ...

Consider $A \in \Bbb C^{m \times n}$, $m \geq n$. (In the other case, just take the transpose...)

We first fix some notations :

1) A $QR$ decomposition of $A$ is any decomposition $A = QR = Q_1R_1$, where $Q = \begin{pmatrix} Q_1 & Q_2 \end{pmatrix} \in \Bbb C^{m \times m}$ is a unitary matrix, $Q_1 \in \Bbb C^{m \times n}$, and $R = \begin{pmatrix} R_1 \\ 0\end{pmatrix} \in \Bbb C^{m \times n}$, $R_1 \in \Bbb C^{n \times n}$ is upper triangular. We say such a $R$ is upper triangular as well (even if it's not a square matrix).

2) A $SVD$ of $A$ is any decomposition $A = U \Sigma V^*$, where $\Sigma = \begin{pmatrix} \Sigma_1 \\ 0\end{pmatrix} \in \Bbb C^{m \times n}$, $\Sigma_1 \in \Bbb C^{n \times n}$ is diagonal with non negative entries (we say such a $\Sigma$ is diagonal with non negative diagonal entries) and $U \in \Bbb C^{m \times m}$, $V \in \Bbb C^{n \times n}$ are unitary.

Suppose that $A$ has full column rank and let $A = QR$ be a $QR$ decomposition for $A$. There exists a $SVD$ of $A$ such that $Q = U$ if and only if $A^*A$ is diagonal.

Suppose $A^*A$ is diagonal. Then, $A^*A = R^*Q^*QR = R^*R = R_1^*R_1$ so $R_1$ must be diagonal (using the fact that $A$ has full column rank and $R_1$ is upper triangular). Then $A = QR = QRI$ and we can multiply some columns of $R$ and the corresponding rows of $I$ by $-1$ so that we get $A = QR'J = U \Sigma V$, where $R'$ is diagonal with non negative diagonal entries, $Q$ and $J$ are unitary. This is a $SVD$ decomposition for $A$ with $Q = U$.

Now, suppose $Q = U$ for some $SVD$ of $A$. Then we have $\Sigma_1$ is diagonal with strictly positive entries (since $A$ has full column rank) and $\Sigma V^* = R \Rightarrow \Sigma_1 V^* = R_1$ so $V^*$ is upper triangular. Since $V^*$ is unitary and upper triangular, it must be diagonal. But the columns of $V$ form a basis of eigenvectors of $A^*A$. This means $e_i \in \Bbb R^n$ is an eigenvector of $A^*A$ for all $i = 1, ..., n$. Therefore, $A^*A$ is diagonal.

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  • $\begingroup$ This is a very nice result, @Desura. Does it exist in a paper or book anywhere that you know of? Or is this your own work? $\endgroup$
    – kdbanman
    Dec 15, 2019 at 6:58
  • $\begingroup$ @kdbanman I'm sorry, I deduced it by myself so I cannot think of any book or paper. $\endgroup$
    – Desura
    Dec 16, 2019 at 7:58

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