When you find the least squares solution you solve $$A^TA = A\vec b$$ but to find the orthogonal projection into the "subspace" A, you multiply this result (the least squares solution) with the original matrix. Why is this?

If you use the analogy with the light shining orthogonally on to the subspace and the orthogonal projection is the shadow in the subspace, isn't this shadow also the least squares solution?

  • $\begingroup$ The least square solution is actually the orthogonal projection on the subspace formed by the columns of A, if this is what you're asking. $\endgroup$
    – Augustin
    May 25, 2015 at 16:22
  • $\begingroup$ @Augustin But what happens when you multiply the least squares solution with A? $\endgroup$
    – martin
    May 25, 2015 at 16:28
  • 5
    $\begingroup$ @Augustin A least squares solution of the system Ax = b is a vector x such that Ax is the orthogonal projection of b onto the column space of A. It is not the orthogonal projection itself. $\endgroup$
    – ivme
    Feb 20, 2019 at 21:25
  • $\begingroup$ With regard to what is written, the left-hand side is a square (Gram) matrix, whereas the right-hand side is a linear combination of the columns of $A.$ That is, the right-hand side is a vector in the column space of $A.$ Presumably, you wanted to write the normal equations $A^{t}Ax = A^{t}b,$ which follow from a geometric or calculus-type argument. $\endgroup$
    – sunspots
    Jul 3, 2021 at 3:59

3 Answers 3


A least squares solution is not the shadow you refer to in the shining light analogy. This shadow is the orthogonal projection of $b$ onto the column space of $A$, and it is unique. Call this projection $p$. A least squares solution of $Ax = b$ is a vector $x$ such that $Ax = p$. The vector $x$ need not be unique.

Consider the matrix

$$A = \begin{bmatrix} 4 &8 \\ 6 &12 \end{bmatrix}$$ and the vector

$$b = \begin{bmatrix} 5 \\ 1 \end{bmatrix} $$

which is not in $C(A) = \textrm{span} \left( \begin{bmatrix} 4 \\ 6 \end{bmatrix} \right)$. The orthogonal projection of $b$ onto $C(A)$ is given by

$$ p = \frac{\begin{bmatrix} 4 \\ 6 \end{bmatrix} \cdot \begin{bmatrix} 5 \\ 1 \end{bmatrix}}{\begin{bmatrix} 4 \\ 6 \end{bmatrix} \cdot \begin{bmatrix} 4 \\ 6 \end{bmatrix}} \begin{bmatrix} 4 \\ 6 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$$

A least squares solution of $Ax = b$ is a vector $x$ such that

$$Ax = p$$

This system has infinitely many solutions. The solution set is

$$x = \left\{ t \begin{bmatrix} -2 \\ 1 \end{bmatrix} + \begin{bmatrix} \frac{1}{2} \\ 0 \end{bmatrix}, t \in \mathbb{R} \right\} $$

Therefore, both $x_1 = \begin{bmatrix} \frac{1}{2} \\ 0 \end{bmatrix}$ and $x_2 = \begin{bmatrix} -\frac{7}{2} \\ 2 \end{bmatrix}$, for instance, are least squares solutions, because both $Ax_1 = p$ and $Ax_2 = p$. But neither of these solutions is the "shadow" you refer to in the shining light analogy. Rather, $p$ is the shadow, and $x_1$ and $x_2$ are simply vectors you could multiply $A$ by to get $p$.

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You want to minimize $||Ax-b||^2$. If $A$ is a full rank matrix, the unique solution is given by $$x=(A^TA)^{-1}A^Tb$$ So $Ax=A(A^TA)^{-1}A^Tb=Hb$, with $H=A(A^TA)^{-1}A^T$ the projection matrix on the subspace formed by the columns of $A$.

  • $\begingroup$ So when finding the least squares solution you're solving a full rank matrix, but when finding the projection you have to multiply by A to use the general projection matrix for the subspace? $\endgroup$
    – martin
    May 25, 2015 at 17:07
  • 1
    $\begingroup$ @martin. The vector projected onto $col(A)$ can be represented in coordinates either in the basis of $col(A)$, which is $x$, or in the basis of the ambient space, $Ax$. $\endgroup$
    – rych
    Sep 14, 2015 at 5:41
  • $\begingroup$ isn't the projection matrix just $A A^T$? How did you know H was a projection? $\endgroup$
    – ions me
    Jul 14 at 22:31

Given the linear system $$ \mathbf{A} x = b $$ we look for solutions of the form $$ \mathbf{A} x - b = 0. $$ If the data vector $b$ is in the column space of $\mathbf{A}$, the the above equation has an exact solution, $$ x = \mathbf{A}^{-1}b. $$ If $b$ has a nullspace component, we can't satisfy the difference equation, so we relax the requirement and ask the $\mathbf{A}x - b$ be as small as possible. This leads to the definition of the least squares problem: $$ x_{LS} = \left\{ x\in\mathbb{C}^{n} \colon \lVert \mathbf{A} x_{LS} - b \rVert_{2}^{2} \text{ is minimized} \right\} $$ The general least squares solution is $$ x_{LS} = \mathbf{A}^{\dagger} b + \left( \mathbf{I}_{n} - \mathbf{A}^{\dagger}\mathbf{A} \right)y, \quad y \in\mathbb{C}^{n}. $$

The are multiple avenues for solution. For example, the normal equations which you allude to: $$ \mathbf{A}^{*} \mathbf{A} x = \mathbf{A}^{*} b $$ which offers the solution $$ x_{LS} = \left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} \mathbf{A}^{*} b $$ (Construction of the normal equations: Overdetermined System Ax=b)

The fundamental projectors are defined using the SVD in Least squares solutions and the orthogonal projector onto the column space. You are asking about the projector onto the range space of $\mathbf{A}$: $$ \mathbf{P}_\color{blue}{\mathcal{R}\left( \mathbf{A} \right)} = \mathbf{A}\mathbf{A}^{\dagger} $$ The operator projects $m$ vectors on the range space (column space, image).

If the classic inverse exists, then it is also the pseudoinverse Pseudo inverse of a singular value decomposition SVD is equal to its “real” inverse for a square matrix?

For your normal equations example, the inverse is also the pseudoinverse. Existence of a normal equations solution implies we have at least as many columns as rows $m\ge n$, and the matrix has full column rank $m=\rho$. The singular value decomposition is $$ \mathbf{A} = \mathbf{U}\, \Sigma \, \mathbf{V}^{*} = \underbrace{\left[ \begin{array}{cc} \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}} \end{array} \right]}_{m\times m} % \underbrace{\left[ \begin{array}{c} \mathbf{S}_{\rho \times \rho} \\\mathbf{0} \end{array} \right]}_{m\times n} % \underbrace{\left[ \begin{array}{c} \color{blue}{\mathbf{V}_{\mathcal{R}}} \end{array} \right]^{*}}_{n\times n} . $$ This implies the pseudoinverse is $$ \mathbf{A}^{\dagger} = \mathbf{V}\, \Sigma^{\dagger} \, \mathbf{U}^{*} = % \underbrace{\left[ \begin{array}{c} \color{blue}{\mathbf{V}_{\mathcal{R}}} \end{array} \right]}_{n\times n} % \underbrace{\left[ \begin{array}{cc} \mathbf{S}_{\rho \times \rho}^{-1} & \mathbf{0} \end{array} \right]}_{n\times m} % \underbrace{\left[ \begin{array}{cc} \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}} \end{array} \right]^{*}}_{m\times m} $$ The normal equations solution is $$ \begin{align} \left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} \mathbf{A}^{*} % &= % \left( \left( \mathbf{V}\, \Sigma^{\mathrm{T}} \mathbf{U}^{*} \right) \left( \mathbf{U}\, \Sigma \, \mathbf{V}^{*} \right) \right)^{-1} \mathbf{V}\, \Sigma^{\mathrm{T}} \mathbf{U}^{*} \\ % &= % \left( \mathbf{V}\, \mathbf{S}^{2} \mathbf{V}^{*} \right)^{-1} \mathbf{V}\, \Sigma^{\mathrm{T}} \mathbf{U}^{*} \\ % &= % \left( \mathbf{V}\, \mathbf{S}^{-2} \mathbf{V}^{*} \right) \mathbf{V}\, \Sigma^{\mathrm{T}} \mathbf{U}^{*} \\ % &= % \left( \mathbf{V}\, \Sigma^{\dagger} \mathbf{U}^{*} \\ % &= % \mathbf{A}^{\dagger} % \end{align} $$

  • $\begingroup$ what are dagger and asterisk in your notation? is dagger = conjugate transpose? is asterisk the conjugate non-transpose? $\endgroup$
    – ions me
    Jul 14 at 22:38
  • 1
    $\begingroup$ @ionsme the asterisk is the conjugate transpose, the dagger is the Moore-Penrose pseudoinverse (see linked question). $\endgroup$
    – Augustin
    Aug 8 at 16:16

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