Total Solutions Amount

Question

If we assume $Ax = 0$ only has ONE solution.

How many solutions will $(A^T)Ax$ = 0 have?

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Will there be 1 solution? $A^TA = I_n$, the identity matrix. Hence there is only one possible vector that gives the solution?

Answer 1

For arbitrary $A \in \mathbb{R}^{m\times n}$ and $x \in \mathbb{R}^n$ we have $$ x^\top A^\top A x = y^\top y = y^2 = \lVert y \Vert_2^2 \ge 0 $$ with $y = Ax$.

The norm $\lVert y \rVert_2$ vanishes only for $y = 0$.

As $Ax = 0$ has only the trivial solution, we know $x = 0 \iff y = 0$.

Which means the symmetric $n \times n$ matrix $A^\top A$ is positive definite.

This means it can be diagonalized with positive eigenvalues only and thus $$ \det(A^\top A) = \prod \lambda_i \ne 0 $$ So $A^\top A$ is bijective, with only $0$ as solution to $A^\top A x = 0$.

In general $A^\top A$ might be different from the identity matrix.

Answer 2

If it has only One solution, then it must be $\mathbf 0$, hence $\boldsymbol A$ is invertible, hence so is $\boldsymbol A ^{\mathsf T} \boldsymbol A$. Therefore the solution is only $\mathbf 0$.