Stability and Numerical Aspects of Least Squares

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Assignment 5 and Midterm 2:

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Last time:

• Error analysis of least squares

Today

• Stability of least squares
• Numerical considerations

Reading: lecture notes 14/15

What if we observe \(\vecy = \matA\vecx_0+\vece\) and we apply the pseudo inverse? \(\hat{\vecx} = \matA^+\vecy\)

We can separate the error analysis into two components \[ \hat{\vecx}-\vecx_0 = \underbrace{\matA^+\matA\vecx_0-\vecx_0}_{\text{null space error}} + \underbrace{\matA^+\vece}_{\text{noise error}} \]

We will express the error in terms of the SVD \(\matA=\matU\boldsymbol{\Sigma}\matV^\intercal\) With

• \(\set{\vecv_i}_{i=1}^r\) orthobasis of \(\text{row}(\matA)\), augmented by \(\set{\vecv_i}_{i=1}^{r+1}\in\ker{\matA}\) to form an orthobasis of \(\bbR^n\)
• \(\set{\vecu_i}_{i=1}^r\) orthobasis of \(\text{col}(\matA)\), augmented by \(\set{\vecu}_{i=1}^{r+1}\in\ker{\matA^\intercal}\) to form an orthobasis of \(\bbR^m\)

The null space error is given by \[ \norm[2]{\matA^+\matA\vecx_0-\vecx_0}^2=\sum_{i=r+1}^n\abs{\dotp{\vecv_i}{x_0}}^2 \]

The noise error is given by \[ \norm[2]{\matA^+\vece}^2=\sum_{i=1}^r \frac{1}{\sigma_i^2}\abs{\dotp{\vece}{\vecu_i}}^2 \]

How do we mitigate the effect of small singular values in reconstruction? \[ \hat{\vecx} = \matV\boldsymbol{\Sigma}^{-1}\matU^\intercal\vecy = \sum_{i=1}^r\frac{1}{\sigma_i}\dotp{\vecy}{\vecu_i}\vecv_i \]

Truncate the SVD to \(r'<r\) \[ \matA_t\eqdef \sum_{i=1}^{r'}\sigma_i\vecu_i\vecv_i^\intercal\qquad\matA_t^+ = \sum_{i=1}^{r'}\frac{1}{\sigma_i}\vecu_i\vecv_i^\intercal \]

Reconstruct \(\hat{\vecx_t} = \sum_{i=1}^{r'}\frac{1}{\sigma_i}\dotp{\vecy}{\vecu_i}\vecv_i=\matA_t\)

Error analysis: \[ \norm[2]{\hat{\vecx}_t}^2 = \sum_{i=r+1}^n\abs{\dotp{\vecx_0}{\vecv_i}}^2+\sum_{i=r'+1}^r\abs{\dotp{\vecx_0}{\vecv_i}}^2+\sum_{i=1}^r'\frac{1}{\sigma_i^2}\abs{\dotp{\vece}{\vecu_i}}^2 \]

Regularization means changing the problem to solve \[ \min_{\vecx\in\bbR^n}\norm[2]{\vecy-\matA\vecx}^2+\lambda\norm[2]{\vecx}^2\qquad\ \lambda>0 \]

The solution is \[ \hat{\vecx} = (\matA^\intercal\matA+\lambda\matI)^{-1}\matA^\intercal\vecy = \matV(\boldsymbol{\Sigma}^2+\lambda\matI)^{-1}\boldsymbol{\Sigma}\matU^\intercal\vecy \]

We have seen several solutions to systems of linear equations \(\matA\vecx=\vecy\) so far

• \(\matA\) full column rank: \(\hat{\bfx} = (\matA^\intercal\matA)^{-1}\matA^\intercal\bfy\)
• \(\matA\) full row rank: \(\hat{\bfx} = \matA^\intercal(\matA\matA^\intercal)^{-1}\bfy\)
• Ridge regression: \(\hat{\bfx} = (\matA^\intercal\matA+\delta\matI)^{-1}\matA^\intercal\bfy\)
• Kernel regression: \(\hat{\bfx} = (\matK+\delta\matI)^{-1}\bfy\)
• Ridge regression in Hilbert space: \(\hat{\bfx} = (\matA^\intercal\matA+\delta\matG)^{-1}\matA^\intercal\bfy\)

Extension: constrained least-squares \[ \min_{\vecx\in\bbR^n}\norm[2]{\vecy-\matA\vecx}^2\text{ s.t. } \vecx=\matB\vecalpha\text{ for some }\vecalpha \]

• The solution is \(\hat{\bfx} = \matB(\matB^\intercal\matA^\intercal\matA\matB)^{-1}\matB^\intercal\matA^\intercal\bfy\)

All these problems involve a symmetric positive definite system of equations.

• Many methods to achieve this based on matrix factorization

• \(\bfA\in\bbR^{n\times n}\) invertible and diagonal
• \(O(n)\) complexity

• \(\bfA\in\bbR^{n\times n}\) invertible and orthogonal
• \(O(n^2)\) complexity

• \(\bfA\in\bbR^{n\times n}\) invertible and lower diagonal
• \(O(n^2)\) complexity