small question I'm having a disagreement with a classmate over - it's a question in a book:
Let $A$ be a square matrix $n \times n$
Prove or disprove:
If the system $A\underline{x}=\underline{b}$ has infinite solutions then $A^2\underline{x}=\underline{b}$ also has infinite solutions.
I say it's true, because if $A\underline{x}=\underline{b}$ has infinite solutions then $A$ must be singular and therefore contains at least one row of zeroes somewhere, so its square must also contain a row of zeroes in the same places and so the second system must also have infinite solutions.
The book says it's false and doesn't elaborate or supply a proof - my classmate agrees with the book. What do you guys think?
Thanks in advance!
The book is right. Take $A$ as the matrix with all entries $0$, except for the right upper corner, with entry $1$. Then $A$ is a nilpotent $n\times n$ matrix with $A\neq 0$ and $A^2=0$. Then let $x=(0,0,\ldots ,r)^t$ for $r\in \mathbb{R}$, $r\neq 0$, and $b=(r,0,0,\ldots ,0)^t$. Then $Ax=b$, with infinitely many $x$, but $A^2x=0\neq b$, so that $A^2x=b$ has no solution.
As to your argument. If $A$ is singular, it need not contain a row of zeroes somewhere. Take the matrix $A$ where every entry is equal to $1$. Then $\det(A)=0$ for every $n\ge 2$, but every row is $(1,1,\ldots ,1)$.