eigen

nafems4.md

@@ -93,7 +93,7 @@ It looks (and works) like a regular $3 \times 3$ matrix. Some brief comments abo
     - $\tau_{yz} = \tau_{zy}$, and
     - $\tau_{zx} = \tau_{xz}$.
  * The elements of the tensor depends on the orientation of the coordinate system.
- * There exists a particular coordinate system in which the stress tensor is diagonal, i.e. all the shear stresses are zero. In this case, the three diagonal elements are called the principal stresses.
+ * There exists a particular coordinate system in which the stress tensor is diagonal, i.e. all the shear stresses are zero. In this case, the three diagonal elements are called the principal stresses, which happen to be the three [eigenvalues](https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors) of the stress tensor. The basis of the coordinate system that renders the tensor diagonal are the eigenvectors.
 
 
 What does this all have to do with mechanical engineering? Well, once we know what the stress tensor is for every point of a solid, in order to obtain the internal forces per unit area acting in a plane passing through that point and with a normal given by the direction $\vec{n}$, all we have to do is “project” the stress tensor through $\vec{n}$. In plain simple words:
@@ -599,7 +599,7 @@ So what is the source of this unexpected non-linear effect in an otherwise nice
 
 Recall that principal stresses are the eigenvalues of the stress tensor. And the fact that in a linear elastic formulation means that the stress tensor of case\ C above is the sum of the individual stress tensors from cases\ A and B does not mean that their eigenvalues can be summed (think about it!). Again, imagine the eigenvalues and eigenvectors of cases A & B. Got it? Good. Now imagine the eigenvalues and eigenvectors for case\ C. Should they sum up? No, they should not! Let us make another experiment, this time with matrices using [Octave](https://www.gnu.org/software/octave/) or whatever matrix-friendly program you want.
 
-First, let us create a 3$\times$3 random matrix $R$ and then multiply it by its transpose\ $R^T$ to obtain a symmetric matrix\ $A$ (recall that the stress tensor is symmetric):
+First, let us create a 3 $\times$ 3 random matrix $R$ and then multiply it by its transpose\ $R^T$ to obtain a symmetric matrix\ $A$ (recall that the stress tensor from [@sec:tensor] is symmetric):
 
 ```octave
 octave> R = rand(3); A = R*R'
@@ -610,7 +610,7 @@ A =
    1.31108   0.57570   1.09657
 ```
 
-Do the same to obtain another 3$\times$3 symmetric matrix\ B:
+Do the same to obtain another 3 $\times$ 3 symmetric matrix\ B:
 
 ```octave
 octave> R = rand(3); B = R*R'