If $A\vec{v}=\lambda\vec{v}$, then does $A=\lambda$? 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?
 A: As the comments have pointed out, you cannot infer that $A=\lambda$ because not only are they different types of variables (matrix vs. scalar), but different types of multiplications were used.

As a side note, the "cancellation" rule:

If $x \cdot z = y \cdot z$, then $x=y$.

holds true if $z$ has a multiplicative inverse $z^{-1}$ so that:
$$ \begin{array}{rrl}
&x \cdot z &=& y \cdot z \\
\implies& x \cdot z \cdot z^{-1} &=& y \cdot z \cdot z^{-1} \\
\implies& x&=&y
\end{array} $$
Note that we can't apply this cancellation rule since $\vec{v}$ doesn't have a multiplicative inverse (using either matrix multiplication or scalar multiplication).
A: This is an unfortunate misconception stemming from using matrix multiplication to model the action of a linear function.  Linear maps are functions, and as such, it should be clearer that to talk about $A(v) = \lambda v \implies A = \lambda$ is no more sensical than saying $\sin(\theta) = 2\theta \implies \sin = 2$.
A: 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 $A = \lambda I$.
A: Absolutely not - a matrix of size larger than 1 cannot equal a scalar.  However - the eigenvalue-eigenvector relationship says that for special vectors, called eigenvectors, the matrix $A$ will act like scalar multiplication - i.e. for eigenvectors, the act of multiplying that vector by the matrix $A$ will yield the same result as if we multiplied the vector by the scalar $\lambda$.
A: We simplify things when we are thinking about scalar operations and simply cross things from one side to the other like (for example):

$ 4x = 4(3w -1) $

the 4's on each side are cancelled out and the expression becomes:

$x = 3w -1 $

But we are actually doing the real operations of:

$ 4x = 4(3w -1) $
$ 4^{-1} 4x = 4^{-1} 4(3w -1) \iff \frac{1}{4}  4x = \frac{1}{4} 4(3w -1)  \iff x = 3w -1  $

What you want to do in your case is:

$A\vec{v}=\lambda\vec{v} \iff A\vec{v}\vec{v}^{-1}=\lambda\vec{v}\vec{v}^{-1}$

But the inverse of a vector is not defined because such a definition of an inverse would have limited use, since:

$\vec{v}\vec{v}^{-1} \ne \vec{v}^{-1}\vec{v}$

For example (as they would have different dimensions, and so on..).
A: Let
$$A = \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix}, \quad v = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \quad \lambda = 5.$$
Then,
$$Av = \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 5 \end{bmatrix} = 5 \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \lambda v,$$
even though $A \ne \lambda$.
As for the reason of confusion, this is the same problem that I have explained here: vectors do not have inverses, which you need in order to get from $A v = \lambda v$ to $A = \lambda$. Just like $2x = 2y$ means that $x = y$ because $2$ has an inverse (so you multiply both sides by $2^{-1}$), while $0x = 0y$ doesn't mean that $x = y$, because $0$ doesn't have an inverse.
