Why is the determinant necessary to find out the eigenvalues of a matrix? Say I have a $2\times2$ matrix $A$:
$$A = \begin{bmatrix}1&2\\4&3\\ \end{bmatrix}.$$
To find the eigenvalues, I have to solve
$$Au = \lambda u,$$ where $u$ is a non-zero vector. Solving this I get
$$0 = \lambda u -Au \Leftrightarrow \\
0 = (\lambda*I_n -A)u.$$
Since $u$ is non-zero, $(\lambda I_n-A) = 0$. Why can't I then find the values of $\lambda$ for which this yields the null matrix? Why do I have to do
$$\det(\lambda I_n -A)=0$$
instead?
I think that to get a null vector you don't have to multiply it by a null matrix necessarily, so I figure it has something to do with that, but I don't understand why I have to use the determinant.
 A: You partially answered your own question in saying that "I think that to get a null-vector you don't have to multiply it by a null-matrix necessarily".
The other part is that you don't need to use the determinant.
If you find an eigenvalue however you do, that's good. You can try to solve $Av = \lambda v$, you can do it by inspection, you can do it by heavenly inspiration (provided you check, ha!).
The fact of matter is, that for an $n\times n$ matrix $A$, $\lambda$ is an eigenvalue of $A$ if and only if $\det(\lambda I - A) = 0$.
This is because if an $n\times n$ matrix $M$ has a nonzero vector $v$ in its kernel, then that $v$ is an eigenvector associated to the eigenvalue $0$.
It follows that $\det M$, as the product of $M$'s eigenvalues, is $0$.
The thing is, this also goes in reverse: if $\det M = 0$, then some eigenvalue must be $0$, and so there must be nonzero vector in $M$'s kernel.
A: You wanted to say $(\lambda I_n - A)u=0$ and $u\ne 0 \implies \lambda I_n-A=0$, 
but that's not true, as the example  $\pmatrix{2 & 2\\4 & 4}\pmatrix{1\\-1}=\pmatrix{0\\0}$, where $\lambda=-1$,  shows.
What is true is that $\det(\lambda I_n-A)=0$.
A: You are not after the null matrix. The eigenvalues of $A$ are $5$ and $-1$ but neither $A-5\operatorname{Id}$ nor $A+\operatorname{Id}$ are the null matrix. The numbers $5$ and $-1$ are the numbers $\lambda$ for which the equation $(A-\lambda\operatorname{Id}).\left[\begin{smallmatrix}x\\y\end{smallmatrix}\right]=\left[\begin{smallmatrix}0\\0\end{smallmatrix}\right]$ has non-null solutions. And those $\lambda$'s are precisely the solutions of the equation $\det(A-\lambda\operatorname{Id})=0$.
A: Actually, for numerical work with matrices  much bigger than $2 \times 2$, you do not want to compute or solve the characteristic polynomial.  It is numerically unstable.  There are much more efficient and stable numerical methods, e.g. the QR algorithm. 
