As a warm-up to the subject of this blog post, consider the problem of how to classify matrices up to change of basis in both the source () and the target (). In other words, the problem is to describe the equivalence classes of the equivalence relation on matrices given by

.

It turns out that the equivalence class of is completely determined by its rank . To prove this we construct some bases by induction. For starters, let be a vector such that ; this is always possible unless . Next, let be a vector such that is linearly independent of ; this is always possible unless .

Continuing in this way, we construct vectors such that the vectors are linearly independent, hence a basis of the column space of . Next, we complete the and to bases of in whatever manner we like. With respect to these bases, takes a very simple form: we have if and otherwise . Hence, in these bases, is a block matrix where the top left block is an identity matrix and the other blocks are zero.

Explicitly, this means we can write as a product

where has the block form above, the columns of are the basis of we found by completing , and the columns of are the basis of we found by completing . This decomposition can be computed by row and column reduction on , where the row operations we perform give and the column operations we perform give .

Conceptually, the question we’ve asked is: what does a linear transformation between vector spaces “look like,” when we don’t restrict ourselves to picking a particular basis of or ? The answer, stated in a basis-independent form, is the following. First, we can factor as a composite

where is the image of . Next, we can find direct sum decompositions and such that is the projection of onto its first factor and is the inclusion of the first factor into . Hence every linear transformation “looks like” a composite

of a projection onto a direct summand and an inclusion of a direct summand. So the only basis-independent information contained in is the dimension of the image , or equivalently the rank of . (It’s worth considering the analogous question for functions between sets, whose answer is a bit more complicated.)

The actual problem this blog post is about is more interesting: it is to classify matrices up to *orthogonal* change of basis in both the source and the target. In other words, we now want to understand the equivalence classes of the equivalence relation given by

.

Conceptually, we’re now asking: what does a linear transformation between finite-dimensional *Hilbert spaces* “look like”?

**Inventing singular value decomposition**

As before, we’ll answer this question by picking bases with respect to which is as easy to understand as possible, only this time we need to deal with the additional restriction of choosing orthonormal bases. We will follow roughly the same inductive strategy as before. For starters, we would like to pick a unit vector such that ; this is possible unless is identically zero, in which case there’s not much to say. Now, there’s no guarantee that will be a unit vector, but we can always use

as the beginning of an orthonormal basis of . The question remains which of the many possible values of to use. In the previous argument it didn’t matter because they were all related by change of coordinates, but now it very much does because the length may differ for different choices of . A natural choice is to pick so that is as large as possible (hence equal to the **operator norm ** of ); writing , we then have

.

is called the **first singular value** of , is called its **first right singular vector**, and is called its **first left singular vector**. (The singular vectors aren’t unique in general, but we’ll ignore this for now.) To continue building orthonormal bases we need to find a unit vector

orthogonal to such that is linearly independent of ; this is possible unless , in which case we’re already done and is completely describable as ; equivalently, in this case we have

.

We’ll pick using the same strategy as before: we want the value of such that is as large as possible. Note that since , this is equivalent to finding the value of such that is as large as possible. Call this largest possible value and write

.

At this point we are in trouble unless ; if this weren’t the case then our strategy would fail to actually build an orthonormal basis of . Very importantly, this turns out to be the case.

**Key lemma #1:** Suppose is a unit vector maximizing . Let be a unit vector orthogonal to . Then is also orthogonal to .

*Proof.* Consider the function

.

The vectors are all unit vectors since are orthonormal, so by construction (of ) this function is maximized when . In particular, its derivative at is zero. On the other hand, we can expand out using dot products as

.

Now we can compute the first-order Taylor series expansion of this function around , giving

so setting the first derivative equal to gives as desired.

This is the technical heart of singular value decomposition, so it’s worth understanding in some detail. Michael Nielsen has a very nice interactive demo / explanation of this. Geometrically, the points trace out an ellipse centered at the origin, and by hypothesis describes the semimajor axis of the ellipse: the point furthest away from the origin. As we move away from , to first order we are moving slightly in the direction of , and so if were not orthogonal to it would be possible to move slightly further away from the origin than by moving either in the positive or negative direction, depending on whether the angle between and is greater than or less than . The only way to ensure that moving in the direction of does not, to first order, get us further away from the origin is if is orthogonal to .

Note that this gives a proof that the semiminor axis of an ellipse – the point closest to the origin – is always orthogonal to its semimajor axis. We can think of key lemma #1 above as more or less being equivalent to this fact, also known as the principal axis theorem in the plane, and which is also closely related to but slightly weaker than the spectral theorem for real matrices.

Thanks to key lemma #1, we can continue our construction. With as before, we inductively produce orthonormal vectors such that is maximized subject to the condition that for all , and write

where is the maximum value of on all vectors orthogonal to ; note that this implies that

.

The are the **singular values** of , the are its **right singular vectors**, and the are its **left singular vectors**. Repeated application of key lemma #1 shows that the are an orthonormal basis of the column space of , so the construction stops here: is identically zero on the orthogonal complement of , because if it weren’t then it would take a value orthogonal to . This means we can write as a sum

.

This is one version of the **singular value decomposition** (SVD for short) of , and it has the benefit of being as close to unique as possible. A more familiar version of SVD is obtained by completing the and to orthonormal bases of and (necessarily highly non-unique in general). With respect to these bases, takes, similar to the warm-up, a block form where the top left block is the diagonal matrix with entries and the remaining blocks are zero. Hence we can write as a product

where has the above block form, has columns given by , and has columns given by .

So, stepping back a bit: what have we learned about what a linear transformation between Hilbert spaces looks like? Up to orthogonal change of basis, we’ve learned that they all look like “weighted projections”: we are almost projecting onto the image as in the warmup, except with weights given by the singular values to account for changes in length. The only orthogonal-basis-independent information contained in a linear transformation turns out to be its singular values.

Looking for more analogies between singular value decomposition and the warmup, we might think of the singular values as a quantitative refinement of the rank, since there are of them where is the rank, and if some of them are small then is close (in the operator norm) to a linear transformation having lower rank.

Geometrically, one way to describe the answer provided by singular value decomposition to the question “what does a linear transformation look like” is that the key to understanding is to understand what it does to the unit sphere of . The image of the unit sphere is an -dimensional ellipsoid, and its principal axes have direction given by the left singular vectors and lengths given by the singular values . The right singular vectors map to these.

**Properties**

Singular value decomposition has lots of useful properties, some of which we’ll prove here. First, note that taking the transpose of a singular value decomposition gives another singular value decomposition

showing that has the same singular values as , but with the left and right singular vectors swapped. This can be proven more conceptually as follows.

**Key lemma #2:** Write . Then for every , the left and right singular vectors maximize the value of subject to the constraint that for all , that is orthogonal to , and that is orthogonal to . This maximum value is .

*Proof.* At the maximum value of subject to the constraint that is orthogonal to and is orthogonal to , it must also be the case that if we fix then takes its maximum value at . But for fixed , uniquely takes its maximum value when is proportional to (if possible), hence must in fact be equal to ; moreover, this is always possible thanks to key lemma #1. So we are in fact maximizing

subject to the above constraints and we already know the solution is given by .

**Left-right symmetry:** Let be the singular values, left singular vectors, and right singular vectors of as above. Then are the singular values, left singular vectors, and right singular vectors of . In particular, .

*Proof.* Apply key lemma #2 to , and note that is the same either way, just with the roles of and switched.

**Singular = eigen:** The left singular vectors are the eigenvectors of corresponding to its nonzero eigenvalues, which are . The right singular vectors are the eigenvectors of corresponding to its nonzero eigenvalues, which are also .

*Proof.* We now know that and that , hence

and

.

Hence are orthonormal eigenvectors of respectively. Moreover, these matrices have rank at most (in fact exactly) , so this exhausts all eigenvectors corresponding to nonzero eigenvalues.

This gives an alternative route to understanding singular value decomposition which comes from writing as

and then applying the spectral theorem to to diagonalize, but I think it’s worth knowing that there’s a route to singular value decomposition which is independent of the spectral theorem.

In addition to the above algebraic characterization of singular values, the singular values also admit the following variational characterization.

**Variational characterizations of singular values (Courant, Fischer):** We have

and

.

*Proof. *For the first characterization, any -dimensional subspace intersects nontrivially, hence contains a unit vector of the form

.

We compute that

and hence that

.

We conclude that every contains such that , hence . Equality is obtained when .

The second characterization is very similar. Any -dimensional subspace intersects nontrivially, hence contains a unit vector of the form

.

We compute that

and hence that

.

We conclude that every contains a vector such that , hence . Equality is obtained when .

The second variational characterization above can be used to prove the following important theorem.

**Low rank approximation (Eckart, Young):** If is the SVD of , let where has diagonal entries and all other entries zero. Then is the closest matrix to in operator norm with rank at most ; that is, minimizes subject to the constraint that . This minimum value is .

*Proof.* Suppose is a matrix of rank at most . Let be the nullspace of , which by hypothesis has dimension at least . By the second variational characterization above, this means that it contains a vector such that , and since this gives

and hence that . Equality is obtained when as defined above.

The variational characterizations can also be used to prove the following inequality relating the singular values of two matrices and of their sum, which can be thought of as a quantitative refinement of the observation that the rank of a sum of two matrices is at most the sum of their ranks.

**Additive perturbation (Weyl):** Let be matrices with singular values . Then

.

*Proof.* We want to bound in terms of the singular values of and . By the second variational characterization, we have

.

To give an upper bound on a minimum value of a function we just need to give an upper bound on some value that it takes. Let and be the subspaces of of dimensions respectively which achieve the minimum values of and respectively, and let be their intersection. This intersection has dimension at least , and by construction

.

Since has dimension at least , the above is an upper bound on the value of for any -dimensional subspace , from which the conclusion follows.

The slightly curious off-by-one indexing in the above inequality can be understood as follows: if and are both very small, this means that and are close to matrices of rank at most and respectively, and hence is close to a matrix of rank at most , hence also ought to be small.

Setting in the additive perturbation inequality we deduce the following corollary.

**Singular values are Lipschitz:** The singular values, as functions on matrices, are uniformly Lipschitz with respect to the operator norm with Lipschitz constant : that is,

.

*Proof.* Apply additive perturbation twice with , first to get

(remembering that is the operator norm), and second to get

(remembering that ).

This is very much not the case with eigenvalues: a small perturbation of a square matrix can have a large effect on its eigenvalues. This is explained e.g. in in this blog post by Terence Tao, and is related to pseudospectra.

Setting , or equivalently , in the additive perturbation inequality, we deduce the following corollary.

**Interlacing:** Suppose are matrices such that . Then

.

*Proof. *Apply additive perturbation twice, first to get

and second to get

.

Interlacing gives us some control over what happens to the singular values under a low-rank perturbation (as opposed to a low-norm perturbation; a low-rank perturbation may have arbitrarily high norm, and vice versa). For example, we learn that if all of the singular values are clumped together, then a rank- perturbation will keep most of the singular values clumped together, except possibly for either the largest or smallest singular values. We can’t expect any control over these, since in the worst case a rank- perturbation can make the largest singular values arbitrarily large, or make the smallest singular values arbitrarily small.

A particular special case of a low-rank perturbation is deleting a small number of rows or columns (note that a row or column which is entirely zero does not affect the singular values, so deleting a row or column is equivalent to setting all of its entries to zero), in which case the upper bound above can be tightened.

**Cauchy interlacing:** Suppose is a matrix and is obtained from by deleting at most rows. Then

.

*Proof.* The lower bound follows from interlacing. The upper bound follows from the observation that we have for all , then applying either variational characterization of the singular values.

Cauchy interlacing also applies to deleting columns, or combinations of rows and columns, because the singular values are unchanged by transposition. In particular, we learn that if is obtained from by deleting either a single row or a single column, then the singular values of interlace with the singular values of , hence the name.

In particular, if all of the singular values of are clumped together then so are those of , with no exceptions. Taking the contrapositive, if the singular values of are spread out, then the singular values of must be as well.

**Three special cases**

Three special cases of the general singular value decomposition are worth pointing out.

First, if has orthogonal columns, or equivalently if is diagonal, then the singular values are the lengths of its columns, we can take the right singular vectors to be the standard basis vectors , and we can take the left singular vectors to be the unit rescalings of its columns. This means that we can take to be the identity matrix, and in general suggests that is a measure of the extent to which the columns of fail to be orthogonal (with the caveat that is not unique and so in general we would want to look at the closest to ).

Second, if has orthogonal rows, or equivalently if is diagonal, then the singular values are the lengths of its rows, we can take the left singular vectors to be the standard basis vectors , and we can take the right singular vectors to be the unit rescalings of its rows. This means that we can take to be the identity matrix, and in general suggests that is a measure of the extent to which the rows of fail to be orthogonal (with the same caveat as above).

Finally, if is square and an orthogonal matrix, so that , then the singular values are all equal to , and an arbitrary choice of either the left or the right singular vectors uniquely determines the other. This means that we can take to be the identity matrix, and in general suggests that is a measure of the extent to which fails to be orthogonal. In fact it is possible to show that the closest orthogonal matrix to is given by , or in other words by replacing all of the singular values of with , so

is precisely the distance from to the nearest orthogonal matrix. This fact can be used to solve the orthogonal Procrustes problem.

In general, we should expect that the SVD of a matrix is relevant to answering any question about whose answer is invariant under left and right multiplication by orthogonal matrices. This includes, for example, the question of low-rank approximations to with respect to operator norm we answered above, since both rank and operator norm are invariant.

on March 18, 2017 at 11:17 am |RamsayBtw. In the proof of interlacing, I don’t see the first displayed equality: . What am I missing?

on March 18, 2017 at 2:14 pm |Qiaochu YuanOops, that’s a typo; it should just be on the RHS.

on March 18, 2017 at 11:14 am |RamsayThanks for that concise and clear introduction to the SVD. I do not understand why it is often not even touched in a first class in linear algebra. It seems to me that it would make sense to introduce it even before the spectral theorem.

Regarding “weighted projections”: up to a scale factor (i.e., a single weight , you can view a linear transformation

between Euclidean spaces as an orthogonal projection.

Specifically, if and are subspaces of of dimension and respectively, and is the orthogonal projection onto restricted to , then, the singular values of are the cosines of the principal angles

https://en.wikipedia.org/wiki/Angles_between_flats

If , then these singular values can take any value in $\latex [0,1]$.

So we see that any can be represented as for an appropriate choice of $\latex X,Y$ as subspaces of .

Also, regarding the best orthogonal transformation to represent , it is worth pointing out that you are talking about the orthogonal factor in the polar decomposition

https://en.wikipedia.org/wiki/Polar_decomposition

which is an immediate consequence of the SVD. We can always represent our matrix as a composition of an orthogonal matrix and a positive semidefinte matrix:

,

where and $R= V \Sigma V^T$ and $R’ = U\Sigma U^T$.

on March 15, 2017 at 2:57 am |Ammar HusainThinking of the QR algorithm and the Toda lattice, you gain when you swap and refactor repeatedly as dressing transformations. Haven’t thought about what if anything reasonable happens when you permute the factors in a square SVD.