# Show that the following tensors have the same principal values

1. Sep 16, 2013

### TheFerruccio

I apologize for the sheer volume of questions I am asking. I have never faced this with an assignment. I get 90% of the way then spent 8 hours on the last 10%. This is inefficient.

Problem Statement
If T is has a non-zero determinant and is second order, show that $\textbf{T}^\top \textbf{T}$ and $\textbf{T}T^\top$ have the same principal values.

Attempt at Solution

In order to determine principal values, the following equation must hold true for the tensor:

$\det{\textbf{A}-\lambda \textbf{1}}=0$

For a 2nd order tensor, this means solving for the following cubic equation:

$-\lambda^3+I_1(\textbf{A})\lambda^2+I_2(\textbf{A})\lambda+I_3(\textbf{A})=0$

Where I_1, I_2, and I_3 are the three scalar invariants for a second order tensor.

Since my A is $\textbf{T}^\top \textbf{T}$, I will redefine A, and define B, as such:
$A_{ij}=(T)_{ij}$
$B_{ij}=(T^\top)_{ij}$

I am assuming anyone here who knows how to assist knows what the scalar invariants are. Typing up this problem alone has been a huge headache, and I haven't even gotten to the part where I have to express the invariant values in index notation.

So, my short question is: Is there a simpler way to solve this?

My current method is: Brute force expand the entire cubic equation, write out each of the three scalar invariants in index notation (16 indices show up, so I have to make sure I match and contract the right ones while still staying "legal" with this index notation), then show that, in both the cases of A and B, that the scalar invariants are equal. If they are equal, then the principal values will be equal, since the coefficients of the cubic equation are equal.

Is there any faster way to do this without getting lost in a huge sea of index notation? It is driving me up the wall, right now. I have spent hours on this and I keep messing up the numerous indices that I have to match up. I do not want to waste a day on a single problem like I did last time.

 If it is indeed the case that I have to expand out the scalar invariants in index notation, how the heck do I do that while still preserving and collapsing the right indices? I'm just not seeing it. I fully expanded it out, and I am ending up with a mess that I cannot seem to resolve.

For instance, in the case of I_2, I expanded everything as such:

I_2:
in the case of $\textbf{T}^{\top} \textbf{T}$...
$\frac{1}{2}((B_{kj}A_{jk})(B_{ab}A_{ba})-(B_{ij}A_{jk})(B_{ka}A_{ai}))$

in the case of
$\textbf{T}\textbf{T}^{\top}$...
$\frac{1}{2}((A_{kj}B_{jk})(A_{ab}B_{ba})-(A_{ij}B_{jk})(A_{ka}B_{ai}))$

That is what I have so far, for I_2, and I see absolutely no way to contract it. Even then, I am pretty sure I used too many matching indices in each term, but I have no idea how to write the indices in such a way that the summations are all preserved, as well as the relationships between the terms.

Last edited: Sep 16, 2013
2. Sep 16, 2013

### Dick

Forget the index notation. That's no help. Can you show that if v is an eigenvector of $T^T T$ with eigenvalue λ (i.e $T^T Tv=\lambda v$), then $Tv$ is an eigenvector of $T T^T$ with eigenvalue λ?

3. Sep 16, 2013

### TheFerruccio

I am so glad that I do not need to do this mess with index notation.

However, wouldn't your suggestion lead me right back to using the characteristic polynomials and getting the coefficients (scalar invariants) to match up? I don't quite see how I would proceed beyond doing that.

4. Sep 16, 2013

### Dick

Just think abstractly. If $T^T Tv=\lambda v$ then what is $T T^T Tv$?? If you can show every principal value (or eigenvalue) of $T^T T$ corresponds to an principal value of $T T^T$ and vice versa then you are done. You know a principal value (or eigenvalue) of a matrix A is just a value of λ such that $A v=\lambda v$ for some vector v, right?

Last edited: Sep 16, 2013
5. Sep 17, 2013

### HallsofIvy

Staff Emeritus
You call these "tensors" but there is no use of general "tensor" properties. You are really doing a "matrix" problem. (Every tensor can be represented by a matrix in a given coordinate system.)

6. Sep 17, 2013

### TheFerruccio

I am just using the vocabulary taught during class. "Show that the tensors have the same principal values." That is precisely what the assignment states.

7. Sep 17, 2013

### TheFerruccio

Let's see...
if
$\textbf{T}^{\top}\textbf{T}\textbf{v}=\lambda\textbf{v}$
then
$\textbf{T}\textbf{T}^{\top}\textbf{T}\textbf{v}=\textbf{T} \lambda \textbf{v}$
then (this is where I am unsure)
$\textbf{T}\textbf{T}^\top\textbf{T}\textbf{T}^{-1}\textbf{v}=\textbf{T}\textbf{T}^{-1}\lambda\textbf{v}$
then
$\textbf{T}\textbf{T}^\top\textbf{v}=\lambda\textbf{v}$

So, both have the same principal values via this method. Am I doing this right?

8. Sep 17, 2013

### TheFerruccio

Actually, scratch all of that. I have to use index notation to solve this problem. That is the point of the exercise. So, I am not allowed to do it the short way. Thus, I still have the dilemma I wrote in the first post.

9. Sep 17, 2013

### vela

Staff Emeritus
Even though you're not supposed to solve the problem this way, you may be interested in seeing what Dick was getting at.
The idea here is to now rewrite it slightly as
$$\textbf{T}\textbf{T}^{\top}(\textbf{T}\textbf{v}) = \lambda (\textbf{T}\textbf{v}).$$ Can you see how this says that $\lambda$ is an eigenvalue of $\textbf{T}\textbf{T}^{\top}$?

10. Sep 17, 2013

### TheFerruccio

Yep! That makes sense. The result of Tv is simply a vector anyway, so, that shows that Tv is the corresponding eigenvector for TT^T for the given eigenvalue.

11. Sep 17, 2013

### TheFerruccio

Hey everyone. I ended up figuring it out. When I put it all in indicial notation, I can regroup the terms. I just had to be extremely careful with assigning the indices for each tensor. I had to first establish the relationship between the "T-T-transpose" and "T-transpose-T"