1. This is a problem from least square approximation, where we solve the equation $A^TAx = A^Tb$ when $Ax = b$ is unsolvable.

  2. The case I am dealing with is when A has dependent columns, i.e. A is an m by n matrix where the rank r is smaller than n. In this case $A^TA$ is a singular n by n matrix with dependent columns, the rank of which is also r (Rank($A^TA$)=Rank($A$)).

  3. Now in the book Introduction to Linear Algebra of the legendary Gilbert Strang, he says and I quote, when A is singular, $A^TA$ is also singular, and the equation $A^TAx = A^Tb$ had infinitely many solutions, the pseudoinverse gives us a way to choose a "best solution" $x^+=A^+b$.

  4. I understand why the equation has infinitely many solutions geometrically:
    Because what the equation asserts geometrically is to find the projection of b, denoted by p, in the column space of A, then solve the new equation $A\hat x$ = p. Because we can always project b onto the column space of A, whether it's singular or not, we know there must be a solution to the equation $A\hat x$=p and if there is a solution, there are infinitely many because A is singular.

  5. My question is how do we know that the equation have infinitely many solution algebraically, to make it clearer, I don't understand why the equation has at least one solution. I do understand that once it has at least one solution, it has infinitely many.

  6. Algebraically, I understands that $A^Tb$ will take us to $C(A^T)$, and it will take away the part of b that lies in $N(A)$. But what does it has to do with $C(A^TA)$ ? My hypothesis is there is some formula regarding $C(A^TA)$ and $C(A^T)$ that I am not aware of. For example, if $C(A^TA) = C(A^T)$, then my problem is solved.

  7. Also, I found this How come least square can have many solutions?, I know what $\hat x^TA\hat x$ in sums will looks like, but I don't know where it came from, but assuming that this is actually correct, I understand the arguments made in that thread.

  8. Any instruction will be appreciated.

Let $A$ be an $n \times m$ matrix.

Consider $$T_{A^T} : \mathbb R^n \to \mathbb R^m , T_{A^T}=A^Tx \\ T_{A} : \mathbb R^m \to \mathbb R^n , T_{A}=Ax \\ T_{A^TA} : \mathbb R^m \to \mathbb R^m , T_{A^TA}=A^TAx $$

Since $T_{A^TA}=T_{A^T}\circ T_A$ we have $$C(A^TA)=range(T_{A^TA}) \subset range(T_{A^T})= C(A^T)$$

If we can show that these two spaces have the same dimension we are done.

But this comes for free from the rank nullity theorem and the fact that $$ker(T_{A^TA})=\ker(T_A)$$

Indeed, the $\supset$ inclusion is obvious. For $\subset$ let $x \in ker(T_{A^TA})$, then $$A^TAx=0 \Rightarrow x^TA^TAx=0 \Rightarrow (Ax)^T(Ax)=0 \Rightarrow (Ax)\cdot(Ax)=0 \Rightarrow \| Ax\|^2=0 \Rightarrow Ax=0$$

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  • $\begingroup$ Could you elaborate on or point me for further instructions regarding $C(A^TA)=range(T_{A^TA}) \subset range(T_{A^T})= C(A^T)$, I only have a vague idea of range of transformation, so I don't really understand why. $\endgroup$ – Sam May 3 '17 at 2:41
  • $\begingroup$ I kinda understand it now from the matrix point of view, since $ABx$ could be rewritten as $Ac$ with $Bx = c$, we can see that $ABx$ is a linear combination of columns of A, and that's why $C(A^TA) \subset C(A^T)$. Just in case anyone might need this. $\endgroup$ – Sam May 3 '17 at 3:32


First and foremost, we must have existence before we can talk about uniqueness. To have existence, we require that the data vector $b$ is not in the $\color{red}{null}$ space: $$ b\notin\color{red}{\mathcal{N} \left( \mathbf{A} \right)} $$


When the matrix $\mathbf{A}$ has dependent columns, the $\color{red}{null}$ space $\color{red}{\mathcal{N} \left( \mathbf{A} \right)}$ is nontrivial. Let $\color{red}{z}$ be any vector in the $\color{red}{null}$ space, and let $x_{LS}$ be a known least squares minimizer.

$$ \mathbf{A} \left( x_{LS} + \color{red}{z} \right) = \mathbf{A} x_{LS} + \mathbf{A} \color{red}{z} = \mathbf{A} x_{LS} + \mathbf{0} = \mathbf{A} x_{LS} $$

Under the action of $\mathbf{A}$, the solutions $\left( x_{LS} + \color{red}{z} \right)$ and $x_{LS}$ are equivalent.

Geometrically, the issue is shown below. The general least squares solution is the affine space represented by the red, dashed line.

  1. If the $\color{red}{null}$ space is trivial, $\color{red}{\mathcal{N} \left( \mathbf{A} \right)} = \mathbf{0}$, the least squares solution is the point $\color{blue}{\mathbf{A}^{+}b}$ in the $\color{blue}{range}$ space $\color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)}$.

  2. If the $\color{red}{null}$ space is not trivial, $\color{red}{\mathcal{N} \left( \mathbf{A} \right)} \ne \mathbf{0}$, the least squares solution is the affine space going through the point $\color{blue}{\mathbf{A}^{+}b}$ in $\color{blue}{range}$ space $\color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)}$ and extending through the $\color{red}{null}$ space.


Least squares solution

Given a matrix $\mathbf{A}\in\mathbb{C}^{m\times n}_{\rho}$, and a data vector $b\in\mathbb{C}^{m}$ such that $$ b\notin\color{red}{\mathcal{N} \left( \mathbf{A} \right)} $$ The least squares solution is defined as $$ x_{LS} = \left\{ x\in\mathbb{C}^{n} \colon \lVert \mathbf{A} x - b \rVert_{2}^{2} \text{ is minimized} \right\} $$ The least squares solution is computed using
$$ x_{LS} = \color{blue}{\mathbf{A}^{+} b} + \color{red}{ \left( \mathbf{I}_{n} - \mathbf{A}^{+} \mathbf{A} \right) y}, \quad y \in \mathbb{C}^{n} $$ In this form, it's clear that having a unique solution demands $$ \mathbf{A}^{+} \mathbf{A} = \mathbf{I}_{n} $$ which happens when $\color{red}{\mathcal{N} \left( \mathbf{A} \right)} = \mathbf{0}$. More details are in the subsequent links.

Explore Stack Exchange:

Existence and uniqueness of least squares solutions: Query about the Moore Penrose pseudoinverse method

Derivation of the SVD solution and conditions on existence and uniqueness: Singular value decomposition proof

Subspace decomposition for least squares: Singular Value Decomposition

How the two null spaces affect the least squares solution: Pseudo-inverse of a matrix that is neither fat nor tall?, What forms does the Moore-Penrose inverse take under systems with full rank, full column rank, and full row rank?

How full column rank changes the inverse: How to find the singular value decomposition of $A^{T}A$ & $\left( A^{T}A \right)^{-1}$

How null spaces affect the pseudoinverse: generalized inverse of a matrix and convergence for singular matrix, When pseudo inverse and general inverse of a invertible square matrix will be equal or not equal?

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Forgetting how we got the equation (by transposing and multiplying), and just looking at it in its own right, we get this:

$c = A^Tb$ is a column vector and $B = A^TA$ is a square matrix with linearly dependent columns. Lastly, we know that $Bx = c$ has at least one solution.

Algebraically, $Bx$ is a linear combination of the columns of $B$. There are infinitely many of these linear combinations that give $0$, so for each linear combination of columns there are infinitely many others that give the same result (just add one of the infinitely many linear combinations that give $0$). That includes an arbitrary linear combination that gives $c$, i.e. a solution to $Bx = c$.

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  • $\begingroup$ As a matter of fact, by looking at your answer I know that I haven't made my question clear enough. The part I am struggling with is that why $A^TAx=A^Tb$ has at least one solution, which is not answered in your post(The part is skipped by you saying that we know $Bx = c$ has at least one solution). $\endgroup$ – Sam May 2 '17 at 14:16

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