I was recently reading Stephen Tu’s blog and stumbled upon his matrix inverse equality post which brough back some memories. Here’s the equality: let $X$ be an $n \times d$ matrix, let $\lambda > 0$, and let $I_k$ represent the $k \times k$ identity matrix. Then

$X(X^TX + \lambda I_d)^{-1}X^T = XX^T(XX^T + \lambda I_n)^{-1}$

Note that both sides are left-multipled by $X$, so for the purposes of checking equality, there’s no need to have it there, so instead consider

$(X^TX + \lambda I_d)^{-1}X^T = X^T(XX^T + \lambda I_n)^{-1}$

I was asked to prove this as part of a homework assignment in my CS 281A class with Professor Ben Recht a while back. Stephen Tu already has an excellent proof based on Singular Value Decomposition, but I’d like to present an alternative proof that seems even simpler to me. I also want to explain why this equality is important.

First, start with $\lambda X^T = \lambda X^T$. Then left-multiply the left hand side (LHS) by $I_d$ and right-multiply the right hand side (RHS) by $I_n$ to obtain

$\lambda I_d X^T = \lambda X^TI_n$

Note that this doesn’t change the equality, because multiplying a matrix with the identity will not change it, regardless of the multiplication ordering. As a sanity check, the LHS is $(d \times d)\times (d \times n)$ and the RHS is $(d\times n)\times (n\times n)$ and both result in a $(d \times n)$-dimensional matrix so the dimensions match up. Next, add $X^TXX^T$ to both sides and get

$X^TXX^T + \lambda I_d X^T = X^TXX^T + \lambda X^TI_n$

We now “distribute out” an $X^T$, though the particular $X^T$ to use is different on both sides. We obtain

$(X^TX + \lambda I_d)X^T = X^T(XX^T + \lambda I_n)$

Next, left-multiply both sides by $(X^TX + \lambda I_d)^{-1}$ and right-multiply both sides by $(XX^T + \lambda I_n)^{-1}$:

$(X^TX + \lambda I_d)^{-1}(X^TX + \lambda I_d)X^T(XX^T + \lambda I_n)^{-1} = (X^TX + \lambda I_d)^{-1}X^T(XX^T + \lambda I_n)(XX^T + \lambda I_n)^{-1}$ $X^T(XX^T + \lambda I_n)^{-1} = (X^TX + \lambda I_d)^{-1}X^T$

Thus proving the identity.

Now, you can leave it like that and in some cases you’ll be OK. But strictly speaking, you should show that the two inverses above actually exist before using them. When I was writing up this for my homework, I said “the problem implies the inverses exist” because the inverse was already written in the problem statement. In retrospect, however, that was a mistake. Fortunately, the TA didn’t take off any points, though he marked “No, but w/e …”

To make things extra clear, we now show that the inverses exist. Consider $(X^TX + \lambda I_d)$. We claim its inverse exist because it is positive definite (and all positive definite matrices are invertible). This is straightforward because for any nonzero $z \in \mathbb{R}^d$, we have

$z^T(X^TX + \lambda I_d)z = z^TX^TXz + \lambda zI_dz = (Xz)^T(Xz) + \lambda z^Tz = \|Xz\|_2^2 + \lambda \|z\|_2^2 > 0$

where we use the standard $w^Tw = \|w\|_2^2$ trick, which is everywhere in machine learning and optimization. If you don’t believe me, take EE 227B as I did and you’ll never be able to forget it. By the way, the above shows why we needed $\lambda > 0$, because $\|Xz\|_2^2$ is not necessarily greater than zero; it could be zero! This is because $X^TX$ could be positive semi-definite, but not positive definite.

A similar proof applies for the other inverse, which uses $\|X^Ty\|_2^2$ for $y \in \mathbb{R}^n$.

Having proved the equality, why do we care about it? The reason has to do with ridge regression and duality. The problem formulation for ridge regression is

${\rm minimize}_w \quad \|Xw - y\|_2^2 + \lambda \|w\|_2^2$

where $y$ is the $n$-dimensional vector of targets and $\lambda > 0$. Taking the gradient we get

\begin{align*} \nabla_w ((Xw - y)^T(Xw - y) + \lambda w^Tw) &= \nabla_w (w^TX^TXw - 2y^TXw + \lambda w^Tw) \\ &= 2X^TXw - 2X^Ty + 2 \lambda w \\ \end{align*}

Setting to zero, we find $w^*$ as

\begin{align*} 2X^TXw^* - 2X^Ty + 2\lambda w^* = 0 &\iff (X^TX + \lambda I_d)w^* - X^Ty = 0 \\ &\iff (X^TX + \lambda I_d)w^* = X^Ty \\ &\iff w^* = (X^TX + \lambda I_d)^{-1}X^Ty. \end{align*}

Gee, does this formulation of $w^*$ look familiar? Yep! The above is one way of finding $w^*$, but another way is by invoking the equality to get what is known as the dual form of $w^*$:

$w^* = (X^TX + \lambda I_d)^{-1}X^Ty = X^T\underbrace{(XX^T + \lambda I_n)^{-1}y}_{\alpha^*}$

Why do we like this new perspective? First, like any dual solution, it provides us with an alternative method if the “other” way (technically called the primal solution) is harder to compute. In this case, however, I don’t think that’s the main reason why (because this is fairly straightforward either way). I think we like this representation because $w^*$ is a linear combination of the rows of $X$, with coefficients specified by the vector $\alpha^* \in \mathbb{R}^n$! The rows of $X$ are the elements in our data $\{x_i\}_{i=1}^n$, so this gives the intuitive interpretation of the optimal solution being some combination of the data elements.