True or false: The unique least norm solution to $Ax=b$ is the orthogonal projection of b onto $R(A)$

So, first isn't this the definition of least squares? The only incorrect thing I can think of is the solution may not be unique.

Also, the detailed answer states: The minimal solution of $Ax = b$ is in $R(A^T)$

Second, shouldn't it be in the $R(A)$, since it is the projection of $b$ onto the column space of $A$?

  • $\begingroup$ I believe you are right, it is unique if $A$ is a full rank matrix. $\endgroup$
    – Chee Han
    Oct 13, 2016 at 15:29
  • $\begingroup$ Suppose $A$ is $m \times n$. Then the least norm solution to $Ax = b$ belongs to $\mathbb R^n$, but the projection of $b$ onto $R(A)$ belongs to $\mathbb R^m$. So no, they are not equal. In fact they don't even belong to the same space. $\endgroup$
    – littleO
    Aug 23, 2020 at 23:11

1 Answer 1


Starting with $$ \mathbf{A}x = b, $$ where the system matrix of rank $\rho$ has $m$ rows, and $n$ columns: $$ \mathbf{A} \in \mathbb{C}^{m\times n}_{\rho}, \quad x \in \mathbb{C}^{n}, \quad b \in \mathbb{C}^{m}. $$ The general least squares problem is defined as $$ x_{LS} = \left\{ x \in \mathbb{C}^{n} \colon \lVert \mathbf{A}x - b\rVert_{2}^{2} \text{ is minimized}\right\} $$ and the solution is $$ x_{LS} = \color{blue}{\mathbf{A}^{\dagger} b} + \color{red}{\left( \mathbf{I}_{n} - \mathbf{A}^{\dagger}\mathbf{A} \right)y}, \quad y\in\mathbb{C}^{n}. $$ Range space components are in blue, nullspace in red.

The set of least squares minimizers is an affine space (dashed red line) passing through the range space of $\mathbf{A}$ at the point $x_{LS} = \mathbf{A}^{\dagger}b,$ as seen in the figure. enter image description here

Every point on the dashed line is a least squares minimizer. Which point has minimum length? That is, which point is closest to the origin? $$ \lVert x_{LS}(y) \rVert_{2}^{2} = \lVert \color{blue}{\mathbf{A}^{\dagger} b} + \color{red}{\left( \mathbf{I}_{n} - \mathbf{A}^{\dagger}\mathbf{A} \right)y} \rVert_{2}^{2} = \lVert \color{blue}{\mathbf{A}^{\dagger} b} \rVert_{2}^{2} + \lVert \color{red}{\left( \mathbf{I}_{n} - \mathbf{A}^{\dagger}\mathbf{A} \right)y} \rVert_{2}^{2} $$ We can control the nullspace term in red by selecting the vector $y=0$; this choice is the least squares minimizer of minimum norm, $\color{blue}{\mathbf{A}^{\dagger} b}.$

Notice that the nullspace is trivial $\color{red}{\mathcal{N}\left( \mathbf{A} \right)}=\left\{ \mathbf{0}\right\}$ when $n=\rho$. Therefore, the set of minimizers is a point, and the solution is unique.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.