linear transformation's geometric meaning Linear transformations can be represented using matrix, like $$v = Au$$, which transforms vector $u$ into $v$. And my intuitive understanding about linear transformations is that, it rotates the vector $u$ by some degrees and meanwhile stretches it by some scales. But if $u$ is the eigenvector, only stretching without rotating.
Here comes my problem,


*

*So a linear transformation has just 2 effects on a vector, that is streching and rotating, right?

*Generally how a vector $u$ will be rotated? Someone tells me that general vector will be rotated as much as parallel with the eigenvector, what if there're more than one eigenvector pointing to different directions? 

*Does the determinant of $A$ reveals how much a vector will be stretched?
 A: Hopefully some examples will help. The overriding theme is that the action of a matrix is determined by its action on a basis  - in case you don't know what a basis is, it's a set of vectors such that every vector in the space can be written uniquely as a linear combination of those in the set. For the examples, I'll work in $\mathbb{R}^2$, and the most useful basis will be $e_1=\left(\begin{smallmatrix}1\\0\end{smallmatrix}\right)$ and $e_2\left(\begin{smallmatrix}0\\1\end{smallmatrix}\right)$.
The matrices I will be interested in are:
$$A_1=\begin{pmatrix}2&0\\0&2\end{pmatrix}\qquad A_2=\begin{pmatrix}2&0\\0&0\end{pmatrix}\qquad A_3=\begin{pmatrix}2&1\\0&2\end{pmatrix}$$
All of these matrices have eigenvector $e_1$ with eigenvalue $2$, but I aim to show that they have very different properties, by considering their action on the basis $\{e_1,e_2\}$ (or, to take a different point of view, the square they determine - this is the $1\times 1$ square with bottom-left corner at the origin).
The matrix $A_1$ is just twice the identity, so both of our basis vectors (and indeed every vector) are eigenvectors for $A_1$ with eigenvalue $2$. So this matrix stretches the plane by a factor of $2$ away from the origin. Our square has the lengths of all of its sides doubled, and so the area is scaled by a factor of $4=\det{A_1}$.
When we apply the matrix $A_2$, it doubles the length of $e_1$, but maps $e_2$ to zero. Our square is collapsed to a line of length $2$ in the direction of $e_1$, which has no area. So the area has been multiplied by a factor of $0=\det{A_2}$. This is very different from the behaviour of $A_1$, but we wouldn't have seen this difference if we only considered the vector $e_1$.
The matrix $A_3$ acts in a slightly more complicated way, doubling the length of $e_1$ as before, but mapping $e_2$ to $e_1+2e_2$. Our square is stretched in the $e_1$ direction, but also tilted away from the $e_2$ direction, to give a parallelogram (which must have area $4=\det{A_3}$). Unlike in the other two cases, $e_2$ is not an eigenvector, and indeed there is no basis of $\mathbb{R}^2$ consisting only of eigenvectors of $A_3$. (The vector $e_2$ is however a generalized eigenvector).
You may also be interested in Perron-Frobenius theory (although the wiki article is - at the time of writing - a little technical) which explains that if all the matrix entries are positive there is a unique largest eigenvalue, and the effect of multiplying by the matrix repeatedly is quite closely linked to this eigenvalue and its eigenvectors.
