Possible confusion, the inertia of something yields a tensor? (trying to understand an example) I was reading the text by Dan Fleisch titled a A Student’s Guide to Vectors and on first pages he says:

An example of a tensor is the inertia that relates the angular velocity of a > rotating object to its angular momentum. Since the angular velocity vector
  has a direction and the angular momentum vector has a (potentially different)
  direction, the inertia tensor involves multiple directions. 

I can't seem to visualize this example, so intuitively the inertia is usually defined as the resistance of any physical object to any change in its state of motion, so if I push something it resists, it thus has inertia, the angular velocity is the speed (in angles unit over time, example radians x seconds-1) at which something rotates + the direction of rotation since it is a vector, we can write it as 
$$\boldsymbol\omega = \frac{d\phi}{dt}\boldsymbol u $$ 
and the angular momentum is how much mass moves through a rotation 
$$ \mathbf{L}=rmv\mathbf{\hat{u}}$$ 
so what does daniel mean when he states "the inertia that relates the angular velocity .. and the angular momentum"? 
Another question, what do we mean when we state that a higher-order tensor  is the mathematical representation of a physical entity that may be
characterized by magnitude and multiple directions? what do we mean by multiple directions? Can't we instead of writing a tensor of rank 2 as
$$\begin{pmatrix}x_{11} & x_{12} & x_{13}\\x_{21}&x_{22}&x_{23}\\x_{31}&x_{32}&x_{33}\end{pmatrix}$$ 
Why don't we just write it as a vector in 9 dimensions like this? $$\begin{pmatrix}x_{11} \\ x_{12} \\ x_{13}\\x_{21}\\x_{22}\\x_{23}\\x_{31}\\x_{32}\\x_{33}\end{pmatrix}$$
Thanks very much in advance for any type of help.
 A: The angular momentum is a vector quantity whose most important properties is that it is conserved and therefore absorbs torques linearly.
The angular velocity is a vector quantity whose magnitude specifies a rate at which an object is rotating, and whose direction specifies the axis about which it is rotating.
The specific expression you are using is valid for one particle rotating about a center -- boring. It turns out you can always define an angular momentum for many-particle systems. Furthermore, when you are dealing with rigid bodies (masses of particles whose relative distances are fixed) you can still define their motion as a composition of a translation (as witnessed e.g. by the center of mass) and a rotation -- no other linear transforms (besides reflections) preserve relative distances.
So you still have an axis of rotation, but in this more complicated scenario there is no guarantee that the axis of rotation is the same as the axis about which the angular momentum points. A simple example is the moments of inertia for a disc which can rotate either about an axis along the disc ($I = \frac14 M R^2$) or the axis which goes through the disc ($I = \frac12 M R^2$). When a disc is not rotating about any single one of these axes but is instead in a "wobbling" rotation combining the two effects, they will not be on the same axis, rather we will have to describe its angular-velocity as precessing about the angular momentum, if we look very carefully at its dynamics. In general this motion is called "tumbling".
To see these with your eyes, try this YouTube video, which uses a computer to compute which way the object is rotating about, giving you a nice visual of where the $\omega$ is. Of course, you might not believe me that this is a real effect, so here is a video of someone doing something similar in zero-G. The dramatic reversals due to having a lot of mass on one axis, almost no mass on a second, and a low mass on the third, really drive home the fact that the rotations of this object are crazy -- but the way that the heavy side always seems to keep rotating in the same direction even as the low-mass prong oscillates from side to side, really drives home the fact that the angular momentum is remaining constant.
As for treating matrices as big vectors, yes, you can do that! My Master's thesis had a ton of Kronecker products and the easiest way to program it was to say "hey, I'm going to treat these matrices as if they were vectors." However the reason that we write these things as a matrix is to get the benefits of matrix algebra, which lets us write $v_i = \sum_j M_{ij} u_j.$  This looks like transposing the vector $\vec u$ and overlaying with each row of $\mathbf M$, taking the product of each of the overlayed numbers, and then summing across the row. This reveals for instance that the first column of $\mathbf M$ is the response of the matrix to the unit vector $\hat e_1$ and so forth: these are the "outputs" to the basis vector "inputs".
A: This is a very formalistic answer, because I have recently reviewed the first chapter of Marion and Thornton in preparation for teaching from the text. It may not be suitable for beginners in the sense of following the math, but the answer that is reached might be helpful even so.
I'm going to use the notation $\vec{x}$ (normal weight with arrow overset) for a vector and $\mathbf{X}$ (bold face and usually capital letters) for a 2nd rank tensor, and will write the elements of either a vector or tensor (as expressed in some coordinate system) with an un-decorated symbol and one or two indices: $x_i$ and $X_{ij}$.
This particular development follows the approach of Marion and Thornton thought I've stripped out most of the detail.
A final note: there is nothing that requires the use of vector spaces as I'm doing here and doing without them has become increasingly popular in recent years, but I still think of that as an advanced approach.

Consider that the angular momentum $\vec{L}$ of an extended body is related to its rotation $\vec\omega$ (where the direction of the vector tells you the orientation of the axis of rotation and the magnitude tell you the speed of rotation) by the matrix multiplication
$$ \vec{L} = \mathbf{I} \vec\omega \,,$$
using the rules for multiplication or vectors and matrices from linear algebra. (This is easiest if you take rotations about the center of mass, thought the formalism can be made to work out around other points.)
Now, we written both the vectors and the tensor in a form that assumes a particular set of coordinate axis in order to take advantage of linear algebra in this form.
So ask what happens if you re-write them in a new set of axis that have the same origin as the original set but may be oriented in another way. I'll use primed symbols when things are expressed in terms of these rotated axis.
The physics hasn't changed, so we should still be able to use $\vec{L}' = \mathbf{I}' \vec\omega'$ to express this relationship, but this will only work if the transformations
\begin{align*}
\vec{L} &\longrightarrow \vec{L}'\\
\vec{\omega} &\longrightarrow \vec{\omega}'\\
\mathbf{I} &\longrightarrow \mathbf{I}'
\end{align*}
are compatible with one another.
You should be able to convince yourself that the transformation for the vectors can be written in terms of a matrix multiplication and we're going to call the matrix $\mathbf{\lambda}$ whose elements $\lambda_{ij}$ are the direction cosines between the new and old coordinate axis (with 1,2,3 corresponding to the $x$,$y$,and $z$ axis now often written. as $\hat{x}_1$ through $\hat{x}_3$ or $\hat{e}_1$ through $\hat{e}_3$ In that framework we write the transformation rule for the vectors as
$$ v_i' = \sum_{j=1}^3 \lambda_{ij} v_j \,.$$
The corresponding rule for transforming the inertial tensor is
$$ I_{ij}' = \sum_{k=1}^3 \sum_{l=i}^3 \lambda_{ik} I_{kl} \lambda_{jl} \,. \tag{*}$$
(note: because the rotation matrix $\mathbf{\lambda}$ is orthogonal this is actually a similarity transform.) 
It is the structure of equation (*) that makes the organization into rows and columns more than just a convenience: the rows and the columns are subject to vector-like transformation rules (though simultaneously).

A set of exercises that might be helpful even without working out the nasty detail.


*

*Show that $\sum_{k=1}^3 \lambda_{ik} \lambda_{kj} = \delta^i_j$.

*Show that the transposition of $\mathbf{\lambda}$ is also it's inverse.

*Show explicitly that the transformation $\vec{L} \longrightarrow \vec{L}'$ follows from the transformations of $\omega$ and $\mathbf{I}$ using the transformation rules indicated.

