Can the common linear transformations be defined using the characteristic equation? In textbooks, five different types of linear transformations are usually referred to: Rotation, Reflection, Scaling, Shear, Projection.
Would I be right in saying that one of the things which set these types of linear transformations apart has something to do with their eigenvalues/vectors?
I'm not sure if this is completely true so I wanted to ask here if the following definitions can be used if this is the case:
Transformation represented by $n\times n$ matrix $A$ has the following properties:
Rotation
$A$ is orthogonal and has characteristic equation
$$(\lambda^2+k\lambda+1)(\lambda - 1)^{n-2}\,\,\,,\,\,k\in]-2,2[$$
Reflection
$A$ is orthogonal symmetric and has characteristic equation
$$(\lambda + 1)(\lambda - 1)^{n-1}$$
Scaling
$A$ is symmetric and has characteristic equation
$$\prod^n_{i=1}(\lambda - a_i)\,\,\,,\,\,a_i\in\mathbb{R^+}$$
Shear
Not too sure...
Projection
$A$ has characteristic equation
$$\lambda^k(\lambda - 1)^{n-k}\,\,\,,\,\,k\in\mathbb{Z}^+$$where $k$ is the dimension of the null space of the transformation. The geometric and algebraic multiplicities of $\lambda=1$ are equal.

*As a small additional question, I've seen the characteristic polynomial defined to be $\det(\lambda I - A)$ and $\det(A - \lambda I)$ - which is it?
 A: Rotation
A rotation matrix $A$ need not be symmetric. Rather, $A^T=A^{-1}$. As for the eigenvalues, Wikipedia states

For even dimensions ($n$ even), the eigenvalues of a rotation matrix occur as pairs of complex conjugates which are roots of unity and may be written $ e^{\pm i\theta _{k}}$. Therefore, there may be no set of vectors which are unaffected by the rotation, and thus no axis of rotation. If there are any real eigenvalues, they will equal unity and will occur in pairs, and the axis of rotation will be an even dimensional subspace of the whole space. For odd dimensions, there will be an odd number of such eigenvalues, with at least one eigenvalue being unity, and the axis of rotation will be an odd dimensional subspace of the whole space.

So the characteristic polynomial need not take the form you wrote.
Shearing
Again, from Wikipedia, a typical shear matrix looks like
$$A=\begin{pmatrix} 1 & 0 & 0 & \lambda & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}$$
i.e., the identity matrix with the addition of some nonzero $\lambda$ in an off-diagonal entry. Its only eigenvalue is $1$, and the dimension of the eigenspace is $n-1$, so the characteristic polynomial will be $\pm(\lambda-1)^n$.
Projection
A projection matrix need not be symmetric. $k$ could be zero (if the projection is the identity transformation, i.e., the projection onto the entire space). I think you meant to write that $k$ is the dimension of the null space. 
Characteristic Polynomial
Note that 
$$\text{det}(\lambda I - A) = \text{det}(-(A-\lambda I)) = (-1)^n \text{det}(A-\lambda I)$$
where $A$ is an $x \times n$ matrix. Thus it does not matter which way you define the characteristic polynomial. If you compute it the opposite way, you either get the same answer (when $n$ is even) or the negative (when $n$ is odd), but typically we only care about the roots of the characteristic polynomial.
