Stanford's Linear Algebra Review
I’ll take a break from the recent discussion on graphical models to go over some old linear algebra concepts. While there are many references on linear algebra designed to help a student relearn what he or she may have forgotten, I found the handout from Stanford’s CS 229 (Machine Learning) class to be one of the best when considering the content, size, and clarity together. And despite being advertised as a review, it actually did teach me a lot. In college, I only took one linear algebra class, a basic introductory and proofsbased course^{1}. While I found it difficult when I was taking it, in retrospect, we did not go over that much material. Two things that really tricked me up when I first arrived in Berkeley were
 Positive Definite Matrices (and related to that, Singular Value Decomposition)
 Matrix Calculus (with gradients, Hessians, etc.)
These two concepts are used all the time in AI, and I really wish I could know them better. Thus, in this post, I’ll quickly go over some of the basic review concepts of linear algebra as presented in the Stanford handout, and then spend a little more time on those previously mentioned topics.
Basic Concepts and Matrix Multiplication
A few years ago, I kept getting confused between column and row vectors when I was reading machine learning literature, but now I know that all vectors are assumed to be columns by default. Also, when denoting row vectors, be careful that the transpose sign does not necessarily mean we are taking the corresponding column and then transposing it (as the notes say, the definitions of \(a_1\) and \(a_1^T\) are ambiguous).
Matrix multiplication is at the heart of linear algebra and what we do in AI. I find it easiest to reason about matrix multiplication by always checking the number of rows and columns of the matrices involved, and making sure they align correctly. But this handout made it easier for me to think of other ways to view how multiplication works. They describe four interesting ways of looking at matrix multiplication, of which the most interesting to me is when the product \(AB\) is expressed as the sum of outer products (between vectors).
Operations and Properties
Some interesting operations and properties of matrices:

Any square matrix \(A \in \mathbb{R}^{n\times n}\) can be represented as a sum of a symmetric matrix and an antisymmetric matrix, since \(A = .5 (A + A^T) + .5 (A  A^T)\).

The trace is the sum of the diagonal elements of a matrix. Despite its seeming simplicity, the trace operator will actually be very useful in matrix calculus (the “trace trick”) and in the study of eigenvalues and eigenvectors, since the sum of the eigenvalues of a matrix equals the trace, even though the individual components being summed up generally have no relation.

The norm of a vector \(\x\\) is like an informal measure of the distance. For some reason, these kept confusing me when I was trying to work on my undergraduate thesis, so it’s nice to see it formally treated here. There are several properties of norms, but in general, I tend to view norms of vectors as just the \(\ell_p\) norm. Also, when working with norms, make sure I know the CauchySchwarz inequality: \(x \cdot y  \le \x\ \y\\).

The rank of a matrix is equal to the number of its linearly independent columns (or rows). There are many implications of the rank; for me, the one I remember the best is that in order for an inverse to exist, a (square) matrix has to be full rank. If we generated a square matrix by filling in its values from a random number generator, then it will almost always have full rank.

A square matrix \(U\) is orthogonal if \(U^TU = I\). This requires the columns of \(U\) to be orthonormal^{2}. A nice property: \(\Ux\_2 = \x\_2\), which we can prove by squaring them, thus considering \((Ux)^T(Ux)\) and \(x^Tx\).

The range, i.e., column space, of a matrix \(A \in \mathbb{R}^{m \times n}\) is \(\mathcal{R}(A) = \{v \in \mathbb{R}^m \: : \: v = Ax, x \in \mathbb{R}^n \}\). The null space is \(\mathcal{N}(A) = \{x \in \mathbb{R}^n \: : \: Ax = 0\}\)^{3}. Vectors in the former set have length \(m\), those in the latter have length \(n\), and we can get the complement of each of those sets by considering \(\mathcal{R}(A^T)\) and \(\mathcal{N}(A^T)\). For instance, considering the sets \(\mathcal{R}(A^T)\) and \(\mathcal{N}(A)\), it turns out that the intersection of those two sets is empty, and that every vector in \(\mathbb{R}^n\) can be expressed as the sum of two vectors, one in each of those two sets. Thus, they are orthogonal complements. The two column spaces, combined with the two column spaces (using \(A\) and \(A^T\)) form the four fundamental subspaces, a term popularized by Gilbert Strang. I’m not sure I really get it, though.

The determinant of a square matrix \(A\), denoted as \(det A\) or \(A\), is a mysterious function that maps \(A\) to the real numbers. The three main properties it satisfies are:
 that the identity has determinant one
 that if we multiply a single row in \(A\) by a scalar \(t\), the determinant of the new matrix is \(tA\)
 that if we exchange any two (different) rows, then the determinant of the new matrix is \(A\)
All the other properties of determinants follow from this, including the cofactor expansion, which I will not list here. Determinants have an intuitive interpretation in terms of volume: if we take the span of the rows of \(A\), and restrict all the coefficients to be in \([0,1]\), then the determinant is the volume of this set. Also, determinants are useful for checking if \(A^{1}\) exists. Finally, the adjoint of a matrix, \({\rm adj}(A)\), is \(A^{1} = {\rm adj}(A) / A\).

Given a square, symmetric matrix \(A\), a quadratic form is a function \(f(x) = x^TAx = \sum_{i=1}^n \sum_{j=1}^n A_{ij}x_ix_j\) mapping \(A\) and \(x\) to the reals. Our variables are \(x\), so even if the dimensions of \(A\) are large, the highest power of any \(x_i\) (i.e., a component of \(x\)) that can exist in \(f(x)\) is two, hence the intuitive name.
Something that is related is the concept of a positive semidefinite matrix, which is a symmetric matrix such that \(x^TAx \ge 0\) for all \(x\). We can also define the obvious analogues for definite, negative definite, etc., matrices. Given any, not necessarily square matrix \(A\), the matrix \(A^TA\) is always positive semidefinite; prove this by using \(\Ax\_2^2\).
Note: quadratic forms like these are the start of an introduction to what really happens in machine learning and AI, where we have to understand matrix representations.

For determining eigenvalues and their corresponding eigenvectors (note: eigenvectors are not unique), refer to \({\rm det}(A  \lambda I) = 0\) if we are going to be solving them by hand.
We tend to use \(\Lambda\) (capital lambda) to represent the matrix of eigenvalues, so the matrix equation will result in \(AS = S\Lambda\), implying that \(A = S\Lambda S^{1}\), hence \(A\) is diagonalizable.
Some interesting properties: the trace and determinant are the sum and product, respectively, of the eigenvalues. The eigenvalues of a triangular matrix are just the elements on the diagonal. If \(A\) is nonsingular, then \(1/\lambda_i\) is an eigenvalue of \(A^{1}\) with eigenvector \(x_i\).
If we have a symmetric matrix \(A\), then (1) all eigenvalues are real, and (2) its eigenvectors can be scaled to be orthogonal to each other, so that \(S\) above can turn orthogonal, and hence we have \(S^{1} = S^T\), a good thing because transposes are easier computationally than inverses. Unfortunately, I don’t currently know how to prove these off the top of my head.
Singular Value Decomposition
With a symmetric matrix \(A\), we can set \(A = U\Lambda U^T\) instead of using \(U^{1}\). It turns out we can generalize this to arbitrary (not even square!) matrices \(A\) by expressing them in singular value decomposition form, \(A = U\Sigma V^T\). The \(U\) and \(V\) matrices are square and orthogonal, and represent eigenvectors of \(AA^T\) and \(A^TA\), respectively.
The \(\Sigma\) matrix is diagonal, possibly nonsquare, and contains something called the singular values (not the eigenvalues!) of matrix \(A\). They fill the first \(r\) diagonal elements for a rank \(r\) matrix \(A\), and the rest of the elements are zero. The singular values are the square roots of the nonzero eigenvalues of both \(AA^T\) and \(A^TA\) (which are symmetric, so they have real eigenvalues).
For a positive definite matrix, \(U = V\).
There are many applications of SVD, and the ones I’m familiar with happen to be in computer vision. Some of these applications have to do with the fact that \(U\) and \(V\) give orthonormal bases for all four fundamental subspaces. For instance, to use SVD to compute the null space of \(A\), then look at the last \(nr\) columns of \(V\).
Matrix Calculus
I really wish I had completely digested this section before taking CS 281A here because I had no idea how matrix calculus worked until I reviewed this document. Given a function \(f : \mathbb{R}^{m \times n} \to \mathbb{R}\), the gradient of \(f\) with respect to \(A\) is a matrix (i.e., it’s the same size as the original matrix input to \(f\)) of partial derivatives of \(f(A)\).
The Hessian is the second derivative analogue to the gradient (well, almost). Here, we assume we are taking the Hessian with respect to a function \(f : \mathbb{R}^n \to \mathbb{R}\), where we take a vector as input (not a general matrix, even though that’s possible).
The Trace Trick
Here’s a neat trick I learned from reading Mike Jordan’s notes. Since the trace is such that \(tr[ABC]\) is the same no matter what the ordering of the three matrices, and since \(\frac{\partial}{ \partial A} tr[AB] = B^T\) (verify this by explicit computation) then we can claim:
\[\begin{align} \frac{\partial}{\partial A} x^TAx &= \frac{\partial}{\partial A} tr[x^TAx] \\ &= \frac{\partial}{\partial A} tr[xx^TA] \\ &= \frac{\partial}{\partial A} [xx^T]^T = xx^T \end{align}\]Intuitively, this makes sense since we just eliminated \(A\). In the past, when I tried computing derivatives like these, I would convert the entire expression to an enormous set of summations, and try to reason element by element. That is way too much work.
There is a related trick involving determinants: the derivative of \(\log A\) with respect to \(A\) is \(A^{T}\).
One can use these tactics to take the MLE of the covariance matrix \(\Sigma\) of a multivariate Gaussian distribution, since the log likelihood of that term has \(\log \Sigma\) and \((x_n\mu)^T\Sigma^{1}(x_n\mu)\) in it.

When I took the course (with Professor Elizabeth Beazley), it was called MATH 211. Now it’s called MATH 250 since the department put the core courses in the “50”s with their muchimproved numbering system. ↩

Watch out with the terminology! Some people might call such matrices \(U\) as orthonormal matrices, but I call them orthogonal. ↩

This is technically called the right null space, or the kernel, since we may wish to refer to the left null space, which uses \(A^Tx = 0 \Rightarrow x^TA = 0\). ↩