Proof: If $\lambda$ is an eigenvalue of $A$ and $x$ is a corresponding eigenvector, then $sAx=s\lambda x$ for every scalar $s$.

Question

It's fairly straight forward to multiple $s$ on both sides of $Ax=\lambda x$, for $x$ is not the $0$ vector. How do you go about showing that $sAx=s\lambda x$ holds true for every $s$ scalar?

I'm thinking of 3 scenarios, when $s=0, s<0, and s>0$.

• When $s=0, 0Ax = 0\lambda x$, this is valid since $x$ is not the $0$ vector to hold true.

• When $s>0, sAx = s\lambda x$, which is scalar multiplication and is valid.

• When $s<0, -sAx = -s\lambda x$ is equivalent to $sAx = s\lambda x$.

Is this enough to show the above statement is true for every scalar $s$?

Thank you

Answer 1

Since $\lambda$ is an eigenvalue with corresponding eigenvector $x$ we have $$Ax = \lambda x$$

Note that $Ax$ and $\lambda x$ are identical vectors and scalar multiplication is a well defined operator therefore for every scalar $s$ you get $sAx=s\lambda x$

More interesting is the fact that $$A(sx)=\lambda (sx) $$ which is part of the proof for eigen space generated by $x$