Why Initialization Matters
In overparameterized linear regression, there may be many coefficient vectors that fit the observed responses exactly. Gradient descent does not generally choose among them without reference to its starting value. The key geometric fact is that every gradient is in the row space of the design matrix. Consequently, updates can change the row-space component of the coefficients but cannot change the component in the design nullspace.
This tutorial derives that dependence for full-batch gradient descent on squared loss. The result separates two parts of the limit: a particular interpolator determined by the data, and the nullspace component inherited from initialization. We focus on exact interpolation, not on the spectral time course of the iterates.
Setup and Definitions
Let \(X\in\mathbb R^{n\times p}\) be a fixed design matrix and \(y\in\mathbb R^n\) a response vector. Assume \(n\leq p\) and \(X\) has full row rank \(n\). These assumptions ensure that \(XX^\top\) is invertible and that every response vector \(y\) can be interpolated. For \(\beta\in\mathbb R^p\), use the squared loss
The factor \(1/2\) makes its gradient \(\nabla\mathcal L(\beta)=X^\top(X\beta-y)\). Full-batch gradient descent with step size \(\eta>0\) and initialization \(\beta_0\) is the sequence
A coefficient vector in \(\mathcal N(X)\) contributes zero to every fitted value on the observed design, because \(Xz=0\). The projector formulas follow from full row rank: \(P_{\mathcal R}\) is symmetric and idempotent, its image is contained in \(\mathcal R(X^\top)\), and \(P_{\mathcal R}X^\top u=X^\top u\) for every \(u\in\mathbb R^n\). Thus it is the orthogonal projector onto that row space.
The Limit from a Nonzero Initialization
The minimum-Euclidean-norm interpolator is
Indeed, it interpolates because \(X\widehat\beta_{\min}=y\), and it lies in the row space. Any other interpolator differs from it by a vector in the nullspace, orthogonal to the row space, so adding a nonzero such vector increases the squared norm.
Proof. Write \(r_k=X\beta_k-y\) for the residual vector. Multiplying the gradient descent update by \(X\) and subtracting \(y\) gives
The matrix \(XX^\top\) is symmetric positive definite. Its eigenvalues \(\lambda\) satisfy \(0<\lambda\leq\lambda_{\max}(XX^\top)\). The step-size condition implies \(-1<1-\eta\lambda<1\) for each eigenvalue. By diagonalizing \(XX^\top\), every eigencomponent of \(r_k\) is multiplied at each step by a factor with absolute value less than one. Therefore \(r_k\to0\), so the fitted values converge to \(y\).
For the coefficient limit, repeatedly substitute the update to obtain
The same eigenvalue condition ensures that the matrix geometric sum converges. Since \(XX^\top\) is invertible,
For example, multiplying the finite sum on the left by \(XX^\top\) gives \(I-(I-\eta XX^\top)^k\), which tends to \(I\); invertibility then gives the stated limit. Substituting this limit into the expression for \(\beta_k\) yields
Finally, each update is a multiple of a vector in \(\mathcal R(X^\top)\). Hence \(P_{\mathcal N}\beta_{k+1}=P_{\mathcal N}\beta_k\), and induction gives \(P_{\mathcal N}\beta_k=P_{\mathcal N}\beta_0\) for every \(k\). This proves both claims. \(\square\)
The theorem gives a precise qualification to the minimum-norm conclusion: gradient descent reaches the minimum-norm interpolator if and only if \(P_{\mathcal N}\beta_0=0\). A zero initialization is one way to ensure this, but it is not necessary; any initialization lying in the row space also suffices. An arbitrary nonzero initialization does not, in general, lead to the minimum-norm solution.
Worked Example: A Three-Coefficient Interpolator
Take
First compute \(XX^\top\) by taking dot products of the rows of \(X\). The first row has squared length \(1^2+0^2+1^2=2\); the second has squared length \(0^2+1^2+1^2=2\); and their dot product is \(1\cdot0+0\cdot1+1\cdot1=1\). Thus
Multiplying the inverse by \(y\) gives \((XX^\top)^{-1}y=(0,1)^\top\), so the minimum-norm interpolator is
The vector \(v=(-1,-1,1)^\top\) is in the nullspace: \(Xv=(-1+1,-1+1)^\top=(0,0)^\top\). Its squared length is \(3\), and \(\beta_0^\top v=-1\); therefore the projection of \(\beta_0\) onto the nullspace is \((\beta_0^\top v/\|v\|_2^2)v=(-1/3)v=(1/3,1/3,-1/3)^\top\). The theorem gives
Checking the fit directly, the first row gives \(1/3+2/3=1\) and the second gives \(4/3+2/3=2\). The nonzero initialization has changed the interpolating coefficients, but not the training predictions.
Worked Consequences of Nullspace Dependence
Worked Example: Full Rank Leaves No Nullspace Component
Let \(X=\begin{pmatrix}2&0\\0&1\end{pmatrix}\), \(y=(4,-1)^\top\), and \(\beta_0=(3,4)^\top\). This square matrix is invertible, so its nullspace contains only zero. Solving \(X\beta=y\) gives the unique interpolator \((2,-1)^\top\), which is also the minimum-norm interpolator. With \(\eta=1/8\), the step condition holds because \(\lambda_{\max}(XX^\top)=4\) and \(0<1/8<2/4\).
The coordinate updates are \(\beta_{1,k+1}=\beta_{1,k}-\frac18\,4(\beta_{1,k}-2)\) and \(\beta_{2,k+1}=\beta_{2,k}-\frac18(\beta_{2,k}+1)\). Thus the deviations from the solution are multiplied by \(1/2\) and \(7/8\), respectively. Starting from deviations \(1\) and \(5\), the iterates are
There is no initialization dependence in the limit because there is no nonzero nullspace direction in which it could persist.
Worked Example: Same Training Fit, Different New Prediction
Consider one training row \(X=(1\ \ 1)\) and response \(y=2\). The minimum-norm interpolator is \(\widehat\beta_{\min}=(1,1)^\top\): it fits because \(1+1=2\). The nullspace consists of multiples of \(v=(1,-1)^\top\). From initialization \(\beta_0=(3,0)^\top\), the nullspace projection is
The limiting coefficients are therefore \((5/2,-1/2)^\top\), which still fit the training response since \(5/2-1/2=2\). Now consider a new design row \(x_{\mathrm{new}}^\top=(1\ \ 0)\). The minimum-norm interpolator predicts \(1\), whereas the limit from this initialization predicts \(5/2\). The difference is invisible on the training row, which is orthogonal to the nullspace direction, but it affects a new row that is not.
Common Mistake and Practical Meaning
A common mistake is to reason, “Gradient descent converges to an interpolator, so it must converge to the minimum-norm interpolator.” Interpolation alone does not imply minimum norm: when \(p>n\), a nullspace vector can be added to any interpolator without changing its training fit. The correction is to check initialization. Under the theorem's assumptions, minimum norm follows when the initial nullspace projection is zero; otherwise the projection remains in the limit.
This distinction matters when interpreting fitted coefficients and predictions. Two runs with the same design, responses, and step size can reach different interpolating coefficient vectors if their initializations have different nullspace components. Those vectors agree on the observed design but need not agree at new design points. Thus exact training fit does not by itself specify a unique fitted function away from the training data.
Check Your Understanding
Use the theorem and the row-space/nullspace decomposition to answer the following.
- Why is every gradient \(X^\top(X\beta-y)\) orthogonal to \(\mathcal N(X)\)?
- For which initializations does gradient descent converge to the minimum-norm interpolator?
- What does the step-size restriction ensure about the residual iteration?
- If \(z\in\mathcal N(X)\), why does adding \(z\) to an interpolator preserve its training fit?
- Can two interpolators with identical training predictions make different predictions at a new design point? State the geometric condition that allows this.