# Estimating singular values from QR decomposition

1. Oct 24, 2015

### vibe3

I have a matrix for which I know its QR decomposition: $A = QR$. I want to estimate the largest and smallest singular values of $A$ ($\sigma_1$ and $\sigma_n$) however in my application it is too expensive to compute the full SVD of $A$.

Is it possible to estimate the largest/smallest singular values from the QR decomposition? The only result I've been able to find so far is
$$\left| \prod_i r_{ii} \right| = \prod_i \sigma_i$$
where $r_{ii}$ are the diagonal entries of $R$. I'm not sure if this implies that the singular values of $R$ are the same as the singular values of $A$. If thats true, it might be possible and less expensive for my application to compute $SVD(R)$ rather than $SVD(A)$.

2. Oct 24, 2015

### jasonRF

The singular values of $A$ are the nonzero eigenvalues of $A^H A$ or $A A^H$. The largest eigenvalue of a matrix can usually be computed using the power method with just a few iterations. The smallest eigenvalue of a full-rank matrix is the largest eigenvalue of the inverse, so again the power method can work; if you have hte QR the inverse is cheap to compute. If not full rank ... I'm not sure what clever methods there are.

jason

3. Oct 24, 2015

### vibe3

The problem is that my matrix $A$ is very ill-conditioned, so computing eigenvalues from $A^T A$ is unreliable (hence the reason I'm computing the QR decomposition instead). I'm looking for a numerically stable way to estimate the singular values from the QR decomposition for ill-conditioned matrices.

Suppose we have $A = QR = U \Sigma V^T$. Then suppose the SVD of $R$ is $R = U_r \Sigma_r V_r^T$.

Because $Q$ is orthogonal, $A = Q U_r \Sigma_r V_r^T$ which implies $U = Q U_r$ and $V = V_r$. This would then mean that $\Sigma = \Sigma_r$ and so the singular values of $R$ and $A$ are the same. Is this correct logic?

Last edited: Oct 24, 2015
4. Nov 4, 2015

### rs1n

Did you meant to write that the singular values of A are the square root of the eigenvalues of $A^HA$ ?

5. Nov 4, 2015

### rs1n

Yes, this is mathematically sound. A and R would have the same singular values.

6. Nov 4, 2015

oops. yep.