2023-03-01 Singular Value Decomposition¶

Last time¶

• Householder QR
• Composition of reflectors

Today¶

• Comparison of interfaces
• Profiling
• Cholesky QR
• Matrix norms and conditioning
• Geometry of the SVD

Composition of reflectors¶

\begin{align} (I - 2 v v^T) (I - 2 w w^T) &= I - 2 v v^T - 2 w w^T + 4 v (v^T w) w^T \\ &= I - \Bigg[v \Bigg| w \Bigg] \begin{bmatrix} 2 & -4 v^T w \\ 0 & 2 \end{bmatrix} \begin{bmatrix} v^T \\ w^T \end{bmatrix} \end{align}

This turns applying reflectors from a sequence of vector operations to a sequence of (smallish) matrix operations. It's the key to high performance and the native format (QRCompactWY) returned by Julia qr().

This works even for very nonsquare matrices¶

This is known as a "full" (or "complete") QR factorization, in contrast to a reduced QR factorization in which $Q$ has the same shape as $A$.

• How much memory does $Q$ use?

Compare to numpy.linalg.qr¶

• Need to decide up-front whether you want full or reduced QR.
• Full QR is expensive to represent.

Cholesky QR¶

$$ R^T R = (QR)^T QR = A^T A$$

so we should be able to use $L L^T = A^T A$ and then $Q = A L^{-T}$.

Can we fix this?¶

Note that the product of two triangular matrices is triangular.

How fast are these methods?¶

Profiling¶

Condition number of a matrix¶

We may have informally referred to a matrix as "ill-conditioned" when the columns are nearly linearly dependent, but let's make this concept for precise. Recall the definition of (relative) condition number:

$$ \kappa = \max_{\delta x} \frac{|\delta f|/|f|}{|\delta x|/|x|} . $$

We understood this definition for scalar problems, but it also makes sense when the inputs and/or outputs are vectors (or matrices, etc.) and absolute value is replaced by vector (or matrix) norms. Consider matrix-vector multiplication, for which $f(x) = A x$.

$$ \kappa(A) = \max_{\delta x} \frac{\lVert A (x+\delta x) - A x \rVert/\lVert A x \rVert}{\lVert \delta x\rVert/\lVert x \rVert} = \max_{\delta x} \frac{\lVert A \delta x \rVert}{\lVert \delta x \rVert} \, \frac{\lVert x \rVert}{\lVert A x \rVert} = \lVert A \rVert \frac{\lVert x \rVert}{\lVert A x \rVert} . $$

There are two problems here:

• I wrote $\kappa(A)$ but my formula depends on $x$.
• What is that $\lVert A \rVert$ beastie?

Matrix norms induced by vector norms¶

Vector norms are built into the linear space (and defined in term of the inner product). Matrix norms are induced by vector norms, according to

$$ \lVert A \rVert = \max_{x \ne 0} \frac{\lVert A x \rVert}{\lVert x \rVert} . $$

• This equation makes sense for non-square matrices -- the vector norms of the input and output spaces may differ.
• Due to linearity, all that matters is direction of $x$, so it could equivalently be written

$$ \lVert A \rVert = \max_{\lVert x \rVert = 1} \lVert A x \rVert . $$

The formula makes sense, but still depends on $x$.¶

$$\kappa(A) = \lVert A \rVert \frac{\lVert x \rVert}{\lVert Ax \rVert}$$

Consider the matrix

$$ A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} . $$

• What is the norm of this matrix?
• What is the condition number when $x = [1,0]^T$?
• What is the condition number when $x = [0,1]^T$?

Condition number of the matrix¶

The condition number of matrix-vector multiplication depends on the vector. The condition number of the matrix is the worst case (maximum) of the condition number for any vector, i.e.,

$$ \kappa(A) = \max_{x \ne 0} \lVert A \rVert \frac{\lVert x \rVert}{\lVert A x \rVert} .$$

If $A$ is invertible, then we can rephrase as

$$ \kappa(A) = \max_{x \ne 0} \lVert A \rVert \frac{\lVert A^{-1} (A x) \rVert}{\lVert A x \rVert} = \max_{A x \ne 0} \lVert A \rVert \frac{\lVert A^{-1} (A x) \rVert}{\lVert A x \rVert} = \lVert A \rVert \lVert A^{-1} \rVert . $$

Evidently multiplying by a matrix is just as ill-conditioned of an operation as solving a linear system using that matrix.

Least squares and the normal equations¶

A least squares problem takes the form: given an $m\times n$ matrix $A$ ($m \ge n$), find $x$ such that $$ \lVert Ax - b \rVert $$ is minimized. If $A$ is square and full rank, then this minimizer will satisfy $A x - b = 0$, but that is not the case in general because $b$ is not in the range of $A$. The residual $A x - b$ must be orthogonal to the range of $A$.

• Is this the same as saying $A^T (A x - b) = 0$?
• If $QR = A$, is it the same as $Q^T (A x - b) = 0$?

In the quiz, we showed that $QQ^T$ is an orthogonal projector onto the range of $Q$. If $QR = A$, $$ QQ^T (A x - b) = QQ^T(Q R x - b) = Q (Q^T Q) R x - QQ^T b = QR x - QQ^T b = A x - QQ^T b . $$ So if $b$ is in the range of $A$, we can solve $A x = b$. If not, we need only orthogonally project $b$ into the range of $A$.

Solution by QR (Householder)¶

Solve $R x = Q^T b$.

• QR factorization costs $2 m n^2 - \frac 2 3 n^3$ operations and is done once per matrix $A$.
• Computing $Q^T b$ costs $4 (m-n)n + 2 n^2 = 4 mn - 2n^2$ (using the elementary reflectors, which are stable and lower storage than naive storage of $Q$).
• Solving with $R$ costs $n^2$ operations. Total cost per right hand side is thus $4 m n - n^2$.

This method is stable and accurate.

Solution by Cholesky¶

The mathematically equivalent form $(A^T A) x = A^T b$ are called the normal equations. The solution process involves factoring the symmetric and positive definite $n\times n$ matrix $A^T A$.

• Computing $A^T A$ costs $m n^2$ flops, exploiting symmetry.
• Factoring $A^T A = R^T R$ costs $\frac 1 3 n^3$ flops. The total factorization cost is thus $m n^2 + \frac 1 3 n^3$.
• Computing $A^T b$ costs $2 m n$.
• Solving with $R^T$ costs $n^2$.
• Solving with $R$ costs $n^2$. Total cost per right hand side is thus $2 m n + 2 n^2$.

The product $A^T A$ is ill-conditioned: $\kappa(A^T A) = \kappa(A)^2$ and can reduce the accuracy of a least squares solution.

Solution by Singular Value Decomposition¶

Next, we will discuss a factorization $$ U \Sigma V^T = A $$ where $U$ and $V$ have orthonormal columns and $\Sigma$ is diagonal with nonnegative entries. The entries of $\Sigma$ are called singular values and this decomposition is the singular value decomposition (SVD). It may remind you of an eigenvalue decomposition $X \Lambda X^{-1} = A$, but

• the SVD exists for all matrices (including non-square and deficient matrices)
• $U,V$ have orthogonal columns (while $X$ can be arbitrarily ill-conditioned).

Indeed, if a matrix is symmetric and positive definite (all positive eigenvalues), then $U=V$ and $\Sigma = \Lambda$. Computing an SVD requires a somewhat complicated iterative algorithm, but a crude estimate of the cost is $2 m n^2 + 11 n^3$. Note that this is similar to the cost of $QR$ when $m \gg n$, but much more expensive for square matrices. Solving with the SVD involves

• Compute $U^T b$ at a cost of $2 m n$.
• Solve with the diagonal $n\times n$ matrix $\Sigma$ at a cost of $n$.
• Apply $V$ at a cost of $2 n^2$. The total cost per right hand side is thus $2 m n + 2n^2$.

Activity: Geometry of the Singular Value Decomposition¶

Diagonal matrices¶

Perhaps the simplest transformation is a scalar multiple of the identity.

The diagonal entries can be different sizes.

$$ A = \begin{bmatrix} 2 & 0 \\ 0 & .5 \end{bmatrix}$$

The circles becomes an ellipse that is aligned with the coordinate axes

Givens Rotation (as example of orthogonal matrix)¶

We can rotate the input using a $2\times 2$ matrix, parametrized by $\theta$. Its transpose rotates in the opposite direction.

Reflection¶

We've previously seen that reflectors have the form $F = I - 2 v v^T$ where $v$ is a normalized vector. Reflectors satisfy $F^T F = I$ and $F = F^T$, thus $F^2 = I$.

Singular Value Decomposition¶

The SVD is $A = U \Sigma V^T$ where $U$ and $V$ have orthonormal columns and $\Sigma$ is diagonal with nonnegative entries. It exists for any matrix (non-square, singular, etc.). If we think of orthogonal matrices as reflections/rotations, this says any matrix can be represented as reflect/rotate, diagonally scale, and reflect/rotate again.

Let's try a random symmetric matrix.

• What visual features indicate that this is a symmetric matrix?
• Is the orthogonal matrix a reflection or rotation?
  • Does this change when the determinant is positive versus negative (rerun the cell above as needed).

Parts of the SVD¶

Putting it together¶

Observations¶

• The circle always maps to an ellipse
• The $U$ and $V$ factors may reflect even when $\det A > 0$