When does this matrix have one solution, no solution, or infinitely many solutions?

Question

Suppose I have an m x n matrix A is:

Under which conditions will the (A$^{T}$A) $x$ = B have one solution, no solution, infinitely many solutions?

Note: m x n can be anything. As in, we can have m>n, m<n, or m=n...etc. And B has dimensions m x 1

Answer 1

If $m>n$ then $C=A^TA$ will be a square matrix of order $n\times n$. Next, if $B$ is $n \times 1$. It may have unique, many or no solution. $\det|C|$ may or may not be zero.

But if $m<n$, then $D=A^T A$ will be singular matrix ($\det|D| = 0$) of order $n\times n$ and if $B$ is of order $n \times 1$. In this case the eq. $DX=B$ will have no unique solution (no solution or many solution are possible).

EDIT:

See the example ($m <n$): take $2 \times 3$ matrix as $A=\begin{pmatrix} a & c &e\\ b & d &f \end{pmatrix}.$ Then check that $A^T A=\begin{pmatrix} a & b \\ c & d\\ e & f \end{pmatrix} \begin{pmatrix} a & c &e\\ b & d &f \end{pmatrix} =\begin{pmatrix} a & b & 0 \\ c & d & 0\\ e & f &0 \end{pmatrix}. \begin{pmatrix} a & c &e\\ b & d &f \\ 0 & 0 & 0 \end{pmatrix} $ Hence, $\det| A^T A|=0,$ irrespective of the values of $a,b,c,d,e,f.$

Answer 2

The matrix $M = A^TA$ is a square matrix. It follows that either $M$ is invertible (i.e. there is a matrix $M^{-1}$ for which $MM^{-1} = M^{-1}M = I$ where $I$ is the identity matrix) or $M$ is not invertible. It turns out that $M$ will be invertible if and only if the columns of $A$ are linearly independent.

If $M$ is invertible, then $Mx = b$ will have a unique solution for every column-vector $b$.

If $M$ is not invertible, then $Mx = b$ will have either no solution or infinitely many solutions. There will be no solution if $b$ is outside the span of the rows of $A$, and there will be infinitely many solutions otherwise.