If $x$ is a solution of a linear system $Ax=b$ and $y$ is a vector orthogonal to $x$, is $x+y$ also a solution?

Question

This system has infinite solutions.

I think I found some examples where that doesn't hold but I'm not sure how to prove it.

If $x+y$ was a solution, then $A(x+y)=b$ should hold. So, it should be $Ay=0$. But I'm not sure how to use the orthogonality to prove/disprove this.

Answer 1

You cannot prove it, since it is false. Suppose that $A=\left[\begin{smallmatrix}1&-1\\1&-1\end{smallmatrix}\right]$. Then, if $x=\left[\begin{smallmatrix}1\\1\end{smallmatrix}\right]$, $A.x=\left[\begin{smallmatrix}0\\0\end{smallmatrix}\right]$. But, if you take $y=\left[\begin{smallmatrix}1\\-1\end{smallmatrix}\right]$, then $x$ and $y$ are orthogonal and $A.(x+y)\neq\left[\begin{smallmatrix}0\\0\end{smallmatrix}\right]$.

Answer 2

Clearly not. Let $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $x=\begin{pmatrix}-1&1\end{pmatrix}$, $y=\begin{pmatrix}1&1\end{pmatrix}$. Then $x\cdot y = -1+1=0$ but $Ax^T = \begin{pmatrix}0&1\end{pmatrix}$ and $A(x+y)^T = A\begin{pmatrix} 0&2\end{pmatrix}^T = \begin{pmatrix}2&2\end{pmatrix}$. In particular $Ay^T =\begin{pmatrix}2&1\end{pmatrix}\ne 0$.