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A least-squares solution minimizes $\Vert Ku-f\Vert$ over the source Hilbert space. It exists exactly when $f\in \mathcal{R}(K)\oplus \mathcal{R}(K{)}^{\perp }$, and then the solutions form a closed affine subspace with direction $\mathcal{N}(K)$.

The orthogonal projection of a prescribed point $u_0$ onto the affine subspace of least-squares solutions is unique. It retains $u_0$'s null space component and chooses the determined component from the Moore–Penrose inverse of an operator.

Use the Hilbert space interpretation of a linear inverse problem, with a bounded linear operator $K:\mathcal{U}\to \mathcal{V}$. A least-squares solution minimizes $\Vert Ku-f{\Vert }_{\mathcal{V}}$ over $u\in \mathcal{U}$. A minimum-norm least-squares solution additionally has the smallest $\mathcal{U}$-norm among all least-squares solutions. When the equation is consistent this is the smallest-norm exact solution.

The least-squares solutions form a nonempty closed affine subspace precisely whenThe closest point theorem in a Hilbert space then gives a unique nearest point to zero, hence a unique minimum-norm least-squares solution. Equivalently, it is the unique least-squares solution in $\mathcal{N}(K{)}^{\perp }$. If $\mathcal{R}(K)$ is closed, this exists for every datum; a nonclosed operator range can leave some data without any least-squares solution.

For example, on the l2 sequence space, take $(Ku{)}_n=u_n/n$ and $f_n=1/n$. The equation would require $u_n=1$ for every $n$, which is not square summable. However, setting the first $N$ coordinates of $u$ to one and the rest to zero givesThus the least-squares infimum is zero but is not attained, and no minimum-norm least-squares solution exists for these data.

The map $K^{\ast }K$ is a bounded linear operator, so the inverse image of the closed singleton $\{K^{\ast }f\}$ is closed. That inverse image is also a convex set: if $K^{\ast }Ku=K^{\ast }Kv=K^{\ast }f$, every convex combination has the same image.

The identity $\mathcal{N}(K^{\ast })=\mathcal{R}(K{)}^{\perp }$ givesConsequently such a $u$ exists exactly when $f$ is the sum of an element of $\mathcal{R}(K)$ and one of its orthogonal complement. ThusThese are precisely the least-squares solutions. Indeed, if $r=Ku-f$ and $K^{\ast }r=0$, then for every $h$ the Pythagorean identity givesConversely, differentiating the squared residual along any real direction $h$ forces $\mathrm{Re}⟨K^{\ast }r,h⟩=0$, hence the normal equation for a linear inverse problem. For any $u_{\ast }\in \mathbb{L}$, the difference of two solutions of the normal equation for a linear inverse problem lies in $\mathcal{N}(K)$, since $⟨K^{\ast }Kh,h⟩=\Vert Kh{\Vert }^2$. Thereforea closed affine subspace; the empty case is closed and convex as well.

Write $u^{†}=K^{†}f$. Every least-squares solution is $u^{†}+w$ for some $w\in \mathcal{N}(K)$, while $u^{†}\in \mathcal{N}(K{)}^{\perp }$. Decompose $u_0=P_{\mathcal{N}(K)}{u}_0+P_{\mathcal{N}(K{)}^{\perp }}{u}_0$ using orthogonal projections. Then the Pythagorean identity givesThe first term is independent of $w$, and the second vanishes at exactly one point. Hence the unique nearest least-squares solution isGeometrically this is the orthogonal projection of $u_0$ onto the closed affine subspace of least-squares solutions.

The Gaussian likelihood is proportional toso maximum likelihood estimation is equivalent to ordinary least squares. If $p\le n$ and $X$ has rank $p$, then $X^{T}X$ is invertible and the normal equations giveIf $p>n$, then $\mathrm{rank}X<p$ and $\ker X\ne \{0\}$. A least-squares minimizer always exists, but it is not unique: by rank-deficient ordinary least squares, the complete set isan affine subspace of dimension $p-\mathrm{rank}X>0$. Thus there are uncountably infinitely many likelihood maximizers. In the generic full-row-rank case their dimension is $p-n$.

If a design matrix $X$ has deficient rank, every ordinary least squares minimizer has the form $X^{+}Y+z$ with $z\in \ker X$, where $X^{+}$ is the Moore-Penrose inverse. Hence the fitted values are unique but the coefficients are not: the minimizers form an affine subspace of dimension $p-\mathrm{rank}X$.