Supposing that a matrix A has an eigenvalue lambda, show that for any induced matrix norm, $||A|| \geq |\lambda|$.
I attempted the solution, but I am not sure if it is valid to cancel the norm of matrices on both sides of the equation. Is it valid to cancel them out?
Attempt at solution: If $A$ is a square $n\times n$ matrix we have $A v=\lambda v$, where $v$ is a $n\times 1$ vector. Taking the norm, $||Av|| = ||\lambda v||=|\lambda|~ ||v||$ because $\lambda$ is a real number.
Then, $||Av|| \leq ||A||~||v||$ implies that $||A||~||v|| \geq |\lambda|~ ||v||$. Cancelling $||v||$ on both sides, we see that $||A|| \geq |\lambda|$.
You cancelled by $\|v\|$ which is a nonnegative real number, moreover, as $v$ intends to be an eigenvector, $v\ne 0$, so $\|v\|\ne 0$, you can cancel it out.