# Is there any connection between QR and SVD of a matrix?

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 $\mathbf{A}$

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

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.

• 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? Dec 15, 2019 at 6:58
• @kdbanman I'm sorry, I deduced it by myself so I cannot think of any book or paper. Dec 16, 2019 at 7:58