Inverses and pseudoinverses. Numerical issues, speed, symmetry.

Category: benchmarking
#inv #matrix inverse #numerical analysis #numerical methods #pinv #pinvh #positive semidefinite #pseudoinverse #symmetric #benchmarking #python

The matrix inverse is a cornerstone of linear algebra, taught, along with its applications, since high school. The inverse of a matrix $latex A$, if it exists, is the matrix $latex A\^{-1}$ such that $latex AA\^{-1} = A\^{-1}A = I_n$. Based on the requirement that the left and right multiplications should be equal, it follows that it only makes sense to speak of inverting square matrices. But just the square shape is not enough: for a matrix $latex A$ to have an inverse, $latex A$ must be full rank.

The inverse provides an elegant (on paper) method of finding solutions to systems of $latex n$ equations with $latex n$ unknowns, which correspond to solving $latex Ax = b$ for $latex x$. If we’re lucky and $latex A\^{-1}$ exists, then we can find $latex x = A\^{-1}b$. For this to work, it must be the case that:

  • We have exactly as many unknowns as equations
  • No equation is redundant, i.e. can be expressed as a linear combination of the others

In this setting, there is a unique solution for $latex x$.

The Moore-Penrose pseudoinverse

What if we have more equations than unknowns? It is most likely the case that we cannot satisfy all the equations perfectly, so let’s settle for a solution that best fits the constraints, in the sense of minimising the sum of squared errors. We solve $latex \operatorname{arg\,min}_x ||b - Ax||$.

And how about the other extreme, where we have a lot of unknowns, but just a few equations constraining them. We will probably have an infinity of solutions, how can we choose one? A popular choice is to take the one of least $latex \ell_2$ norm: $latex \operatorname{arg\,min}_x ||x|| \operatorname{s.t.} Ax = b$. Is there a way to generalize the idea of a matrix inverse for this setting?

The pseudoinverse of an arbitrary-shaped matrix $latex A$, written $latex A\^{+}$, has the same shape as $latex A\^{T}$ and solves our problem: the answer to both optimization methods above is given by $latex x = A\^{+}y$.

The theoretical definition of the pseudoinverse is given by the following conditions. The intuitive way to read them is as properties of $latex AA\^+$ or $latex A\^+A$:

  • $latex AA\^+A = A$
  • $latex A\^+AA\^+ = A\^+$
  • $latex (AA\^+)\^T = AA\^+$
  • $latex (A\^+A)\^T = A\^+A$

These conditions do not however give us a way to get our hands on a pseudoinverse, so we need something else.

How to compute the pseudoinverse on paper

The first time I ran into the pseudoinverse, I didn’t even know its definition, only the expression of the closed-form solution of such a problem, and given as:

$latex A\^+ = (A\^T A)\^{-1}A\^T$

What can we see from this expression:

  • It gives us a way to compute the pseudoinverse, and hence to solve the problem
  • If $latex A$ is actually invertible, it means $latex A\^T$ is invertible, so we have $latex A\^+ = A\^{-1}(A\^T)\^{-1}A\^T = A\^{-1}$
  • Something bad happens if $latex A\^TA$ is not invertible.

The pseudoinverse is still defined, and unique, when $latex A\^TA$ is not invertible, but we cannot use the expression above to compute it.

Numerical issues

Before going on, we should clarify and demystify some of the urban legends about numerical computation of least squares problems. You might have heard the following unwritten rules:

  1. Never compute $latex A\^{-1}$, solve the system directly
  2. If you really need $latex A\^{-1}$, use pinv and not inv

The first of these rules is based on some misguided beliefs, but is still good advice. If your goal is a one-shot answer to a system, there’s no use in explicitly computing a possibly large inverse, when all you need is $latex x$. But this paper shows that computing the inverse is not necessarily a bad thing. The key to this is conditional accuracy, and as long as the inv function used has good conditional bounds, you will get as good results as with a least squares solver.

The second rule comes from numerical stability, and will definitely bite you if misunderstood. If $latex A$ is a square matrix with a row full of zeros, it’s clearly not invertible, so an algorithm attempting to compute the inverse will fail and you will be able to catch that failure. But what if the row is not exactly zero, but the sum of several other rows, and a slight loss of precision is propagated at every step?

Numerical rank vs. actual rank

The rank of a matrix $latex A$ is defined as the number of linearly independent rows (or equivalently, columns) in $latex A$. In other words, the number of non-redundant equations in the system. We’ve seen before that if the rank is less than the total number of rows, the system cannot have a unique solution anymore, so the matrix $latex A$ is not invertible.

The rank of a matrix is a computationally tricky problem. On paper, with small matrices, you would look at minors of decreasing size, until you find the first non-zero one. This is unfeasible to implement on a computer, so numerical analysis has a different approach. Enter the singular value decomposition!

The SVD of a matrix $latex A$ is $latex A = USV\^{T}$, where $latex S$ is diagonal and $latex U, V$ are orthogonal. The elements on the diagonal of $latex S$ are called the singular values of $latex A$. It can be seen that to get a row full of zeros when multiplying three such matrices, a singular value needs to be exactly zero.

The ugly thing that could happen is that one (or usually more) singular values are not exactly zero, but very low values, due to propagated imprecision. Why is this a problem? By looking at the SVD and noting its properties, it becomes clear that $latex A\^{-1} = VS\^{-1}U\^{T}$ and since $latex S$ is diagonal, its inverse is formed by taking the inverse of all the elements on the diagonal. But if a singular value is very small but not quite zero, its inverse is very large and it will blow up the whole computation of the inverse. The right thing to do here is either to tell the user that $latex A$ is numerically rank deficient, or to return a pseudoinverse instead. A pseudoinverse would mean: give up on trying to get $latex AA\^+$ to be the identity matrix, simply aim for a diagonal matrix with approximately ones and zeroes. In other words, when singular values are very low, set them to 0.

How do you set the threshold? This is actually a delicate issue, being discussed on the numeric Python mailing list.

Scipy implementations

Scipy exposes inv, pinv and pinv2. inv secretly invokes LAPACK, that ancient but crazy robust code that’s been used since the 70s, to first compute a pivoted LU decomposition that is then used to compute the inverse. pinv also uses LAPACK, but for computing the least-squares solution to the system $latex AX = I$. pinv2 computes the SVD and transposes everything like shown above. Both pinv and pinv2 expose cond and rcond arguments to handle the treatment of very small singular values, but (attention!) they behave differently!

The different implementations also lead to different speed. Let’s look at inverting a random square matrix:

[sourcecode lang=”python”]
In [1]: import numpy as np

In [2]: from scipy import linalg

In [3]: a = np.random.randn(1000, 1000)

In [4]: timeit linalg.inv(a)
10 loops, best of 3: 132 ms per loop

In [5]: timeit linalg.pinv(a)
1 loops, best of 3: 18.8 s per loop

In [6]: timeit linalg.pinv2(a)
1 loops, best of 3: 1.58 s per loop

Woah, huge difference! But do all three methods return the “right” result?

[sourcecode lang=”python”]
In [7]: linalg.inv(a)[:3, :3]
array([[ 0.03636918, 0.01641725, 0.00736503],
[-0.04575771, 0.03578062, 0.02937733],
[ 0.00542367, 0.01246306, 0.0122156 ]])

In [8]: linalg.pinv(a)[:3, :3]
array([[ 0.03636918, 0.01641725, 0.00736503],
[-0.04575771, 0.03578062, 0.02937733],
[ 0.00542367, 0.01246306, 0.0122156 ]])

In [9]: linalg.pinv2(a)[:3, :3]
array([[ 0.03636918, 0.01641725, 0.00736503],
[-0.04575771, 0.03578062, 0.02937733],
[ 0.00542367, 0.01246306, 0.0122156 ]])

In [10]: np.testing.assert_array_almost_equal(linalg.inv(a), linalg.pinv(a))

In [11]: np.testing.assert_array_almost_equal(linalg.inv(a), linalg.pinv2(a))

Looks good! This is because we got lucky, though, and a was invertible to start with. Let’s look at its spectrum:

[sourcecode lang=”python”]
In [12]: _, s, _ = linalg.svd(a)

In [13]: np.min(s), np.max(s)
Out[13]: (0.029850235603382822, 62.949785645178906)

This is a lovely range for the singular values of a matrix, not too small, not too large. But what if we built the matrix in a way that would always pose problems? Specifically, let’s look at the case of covariance matrices:

[sourcecode lang=”python”]
In [14]: a = np.random.randn(1000, 50)

In [15]: a =, a.T)

In [16]: _, s, _ = linalg.svd(a)

In [17]: s[-9:]
array([ 7.40548924e-14, 6.48102455e-14, 5.75803505e-14,
5.44263048e-14, 4.51528730e-14, 3.55317976e-14,
2.46939141e-14, 1.54186776e-14, 5.08135874e-15])


a has at least 9 tiny singular values. Actually it’s easy to see why there are 950 of them:

[sourcecode lang=”python”]
In [18]: np.sum(s \< 1e-10)
Out[18]: 950

How do our functions behave in this case? Instead of just looking at a corner, let’s use our gift of sight:[]

The small eigenvalues are large enough that inv thinks the matrix is full rank. pinv does better but it still fails, you can see a group of high-amplitude noisy columns. pinv2 is faster and it also gives us a useful result in this case.

Wait, does this mean that pinv2 is simply better, and pinv is useless?

Not quite. Remember, we are now trying to actually invert matrices, and degrade gracefully in case of rank deficiency. But what if we need the pseudoinverse to solve an actual non-square, wide or tall system?

[sourcecode lang=”python”]
In [19]: a = np.random.randn(1000, 50)

In [20]: timeit linalg.pinv(a)
10 loops, best of 3: 104 ms per loop

In [21]: timeit linalg.pinv(a.T)
100 loops, best of 3: 7.08 ms per loop

In [22]: timeit linalg.pinv2(a)
10 loops, best of 3: 114 ms per loop

In [23]: timeit linalg.pinv2(a.T)
10 loops, best of 3: 126 ms per loop

Huge victory for pinv in the wide case! Hurray! With all this insight, we can draw a line and see what we learned.

  • If you are 100% sure that your matrix is invertible, use inv for a huge speed gain. The implementation of inv from Scipy is based on LAPACK’s *getrf + *getri, known to have good bounds.
  • If you are trying to solve a tall or wide system, use pinv.
  • If your matrix is square but might be rank deficient, use pinv2 for speed and numerical gain.

Improving the symmetric case

But wait a second, can’t we do better? $latex AA\^T$ is symmetric, can’t we make use of that to speed up the computation even more? Clearly, if $latex A$ is symmetric, in its SVD $latex A = USV\^T$, we must have $latex U = V$. But this is exactly the eigendecomposition of a symmetric matrix $latex A$. The eigendecomposition can be computed cheaper than the SVD using Scipy eigh, that uses LAPACK’s *evr. As part of my GSoC this year, with help from Jake VanderPlas, we made a pull request to Scipy containing a pinvh function that is equivalent to pinv2 but faster for symmetric matrices.

[sourcecode lang=”python”]
In [24]: timeit linalg.pinv2(a)
1 loops, best of 3: 1.54 s per loop

In [25]: timeit linalg.pinvh(a)
1 loops, best of 3: 621 ms per loop

In [26]: np.testing.assert_array_almost_equal(linalg.pinv2(a), linalg.pinvh(a))

[]: http://localhost:8001/wp-content/uploads/2012/08/pseudoinverses.png

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