TITLE: If $A\vec{v}=\lambda\vec{v}$, then does $A=\lambda$?
QUESTION [5 upvotes]: When my professor started teaching eigenvectors and eigenvalues the other day, the very first thing I noticed was the fact that $A\vec{v}=\lambda\vec{v}$ (assuming that the equation is satisfied under the given variables) implies that $A=\lambda$... But given that A is an $n\times n$ matrix, and $\lambda\in\mathbb{R}$, how can a matrix equal a real number? What is the consequence of this?

REPLY [3 votes]: Well, the first observation is that $A$ and $\lambda$ are of different type: $A$ is a matrix, while $\lambda$ is a scalar. However, it's easy to "fix" that be noting that for any vector $\vec v$, $I\vec v=\vec v$ where $I$ is the $n\times n$ unit matrix. This gives us a natural mapping $\lambda\mapsto\lambda I$. So we can reformulate the question:
Does $A\vec v = \lambda I\vec v$ imply $A=\lambda I$?
Well, using the vector space properties and the linearity of the multiplication, we can reformulate yet another time, by bringing both terms on the same side and factoring out $\vec v$:
Does $(A - \lambda I)\vec v = 0$ imply $\vec v=0$?
Since we can add $\lambda I$ to any matrix $X$ we like, and then of course will find that subtracting it again recovers $X$, we can simplify this question further to:
Does $X\vec v=0$ imply $X=0$?
Now remember that one way to construct a matrix is to take a vector $\vec w$ and calculate the "outer product" $\vec w\vec w^T$. If $\vec w\ne0$, then $\vec w\vec w^T\ne 0$. Now assume that there exists a $\vec w$ which is orthogonal to $\vec v$, that is, $\vec w^T\vec v=0$. Then by multiplication with $w$ you get $\vec w\vec w^Tv=0$. But that means $X=\vec w\vec w^T$ is a non-zero matrix with $X\vec v=0$.
In other words, if there exist orthogonal vectors to $v$, then from $X\vec v=0$ it does not follow that $X=0$. But orthogonal vectors exist in any vector space of dimension larger than $1$.
So if we collect all the bits together, the answer is:
If the dimension of the vector space is larger than $1$, $A\vec v = \lambda\vec v$ does not imply $A = \lambda I$.