# proof of matrix singularity

If anyone can help me with the next question I would appreciate it a lot.

Let $A$ and $B$ be $n*n$ matrices and let $C=A-B$. Show that if $Ax_0=Bx_0$ and $x_0$ is not zero, then $C$ must be singular.

The first thing I don't get is the notation, what do $Ax_0$ and $Bx_0$ mean?

if $x_0 \neq 0$

$$Ax_0 = B x_0$$

then we have

$$(A-B) x_0=0$$

that is we have $x_0 \neq 0$ such that $Cx_0=0$ which means $C$ is singular.

• I understand till $Cx_0=0$ but why is C then singular? – Amaluena Nov 28 '16 at 22:22
• suppose $C$ is nonsingular, then I can multiply $C^{-1}$ on both sides of $Cx_0=0$ and obtain $x_0=0$ which is a contradiction. – Siong Thye Goh Nov 28 '16 at 22:29

Presumably, $x_0$ is any non zero vector of the space where the matrices apply, and $Ax_0$, $Bx_0$ are the products of these matrices and this vector.

If $A$ and $B$ map the vector $x_0 \ne 0$ to the same value then $$0 = A x_0 - B x_0 = (A-B) x_0 = C x_0 \quad (*)$$ by definition of matrix multiplication and definition of $C$.

This means $C$ has not only the zero vector as solution to $Cx=0$ but $x_0$ as well, which means that it is singular. (If there was an inverse $C^{-1}$ you would have both $0$ and $x_0$ among the choices for $C^{-1} 0$, but an inverse needs to be unique)

• Sorry but if $Cx_0 =0$ then why is $x_0$ a solution? – Amaluena Nov 29 '16 at 11:04
• $x_0$ fullfils the equation $(*)$, so it is called a solution. – mvw Nov 29 '16 at 11:58
• Oh of course! Don't know how i could miss that Thanks – Amaluena Nov 29 '16 at 14:37