- #1

TheFerruccio

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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.

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.

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:

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.

[edit] 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.

**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.

[edit] 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: