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Linear least squares minimizes $\Vert Ax-b{\Vert }^2$ over a vector $x$. Its normal equations for linear least squares are $A^{T}Ax=A^{T}b$; geometrically $Ax$ is the orthogonal projection of $b$ onto the range of $A$.

This is an orthogonal projection interpretation of linear least squares. The best affine predictor has slope covariance divided by predictor variance and intercept chosen to match means.

A finite-dimensional linear subspace $W$ of an inner product space has a unique closest vector to any given vector $x$. This is its orthogonal projection $P_{W}x$. If $e_1,\dots ,e_d$ is an orthogonal basis of $W$, compute it one direction at a time:The inner product convention here is linear in its first argument, including over the complex numbers. For an orthonormal basis, each denominator is one. When $W$ consists only of the zero vector, the empty sum gives $P_{W}x=0$. The finite-dimensional formula is described in Francis Narcowich, Inner Products and Norms: minimization problems.

Take $W=\mathrm{span}\{(1,2{)}^T\}$ and $x=(3,1{)}^T$, using the dot product. With $e=(1,2{)}^T$,The orthogonal projection residual has $r\cdot e=2-2=0$. To verify the best approximation directly, every candidate in the line is $te$, andThe smallest squared Euclidean norm error is therefore $5$, attained only at $t=1$. The diagram connects the perpendicular residual to this one-variable quadratic polynomial.

The general mechanism is the Pythagorean theorem in an inner-product space. For any $w\in W$, the vectors $r=x-P_{W}x$ and $P_{W}x-w$ are orthogonal, soMoving away from $P_{W}x$ inside $W$ adds a nonnegative squared norm, which is zero only when $w=P_{W}x$. This proves both optimality and uniqueness.

The useful feature is separation of preparation from evaluation. Construct an orthonormal basis once with the Gram-Schmidt process, then obtain each new best approximation using inner products instead of a fresh optimization problem. When $W$ is the column space of a matrix, this gives linear least squares; when $W$ is spanned by orthogonal polynomials, it gives a least-squares polynomial in an orthogonal-polynomial basis. The ambient inner product space need not be a Hilbert space for this finite-dimensional construction. In an infinite-dimensional target linear subspace, existence requires additional hypotheses, supplied by orthogonal decomposition by a closed subspace.

Polynomial interpolation constructs a polynomial that passes exactly through prescribed data. Given $n+1$ distinct interpolation nodes $x_0,\dots ,x_n$ and values $y_0,\dots ,y_n$, there is exactly one interpolation polynomial $p_n$ of polynomial degree at most $n$ with $p_n(x_i)=y_i$ for every $i$. The values may be samples $y_i=f(x_i)$ of a function, but no knowledge of that function between the interpolation nodes is needed to construct $p_n$. The Lagrange polynomial and Newton polynomial are two representations of this same answer.^[ref]

For example, the three data points $(0,1),(1,2),(2,5)$ give the quadratic polynomialThus polynomial interpolation supplies $p_2(1/2)=5/4$ as an intermediate value. Whether this approximates an unknown underlying function well is a separate question about polynomial interpolation error.

Both hypotheses in the uniqueness statement matter. The Lagrange cardinal polynomials prove existence constructively. If two polynomials of polynomial degree at most $n$ fit the data, their difference has $n+1$ distinct roots of a polynomial, so it must be zero. Repeated interpolation nodes with conflicting values have no solution, and repeated equal values provide no additional independent condition; prescribed derivatives instead lead to Hermite interpolation. Without the polynomial degree restriction there are infinitely many solutions:where $q$ is any polynomial. By the factor theorem, this describes every polynomial fitting the same data. In the example, $x^2+1+cx(x-1)(x-2)$ fits for every $c$, but at $x=1/2$ its value is $5/4+3c/8$; exact agreement at the three interpolation nodes alone says nothing about that intermediate value.

The usefulness of polynomial interpolation is that data become an explicit function whose values, derivatives, and integrals can be calculated. Differentiating or integrating the Lagrange cardinal polynomials produces differentiation weights from nodal evaluations or a quadrature rule. Exact passage through observations is the defining constraint. In polynomial regression, linear least squares instead chooses a polynomial degree and minimizes a sum of squared residuals, generally without passing through every observation. Interpolating noisy values reproduces their errors at the interpolation nodes and can amplify them between interpolation nodes; polynomial interpolation does not itself smooth measurement noise. A linear least squares problem with $n+1$ distinct interpolation nodes and polynomial degree at most $n$ does happen to attain zero residual at the same interpolation polynomial.