Page 185 - Intro to Tensor Calculus

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which when expanded is a cubic equation of the form

2
3
f(λ)= −λ + I 1 λ − I 2 λ + I 3 =0, (2.1.21)
where I 1 ,I 2 and I 3 are invariants defined by the relations

I 1 = T ii
1 1
I 2 = T ii T jj − T ij T ij (2.1.22)
2 2
I 3 = e ijk T i1 T j2 T k3 .
¯
When T ij is subjected to an orthogonal transformation, where T mn = T ij ` im ` jn ,then

¯ ¯
` im ` jn (T mn − λδ mn )= T ij − λδ ij and det (T mn − λδ mn )= det T ij − λδ ij .

Hence, the eigenvalues of a second order tensor remain invariant under an orthogonal transformation.
If T ij is real and symmetric then

• the eigenvalues of T ij will be real, and
• the eigenvectors corresponding to distinct eigenvalues will be orthogonal.

Proof: To show a quantity is real we show that the conjugate of the quantity equals the given quantity. If
(2.1.19) is satisfied, we multiply by the conjugate A i and obtain

A i T ij A j = λA i A i . (2.1.25)

The right hand side of this equation has the inner product A i A i which is real. It remains to show the left
hand side of equation (2.1.25) is also real. Consider the conjugate of this left hand side and write

A i T ij A j = A i T ij A j = A i T ji A j = A i T ij A j .

Consequently, the left hand side of equation (2.1.25) is real and the eigenvalue λ can be represented as the
ratio of two real quantities.
ˆ ˆ
Assume that λ (1) and λ (2) are two distinct eigenvalues which produce the unit eigenvectors L 1 and L 2
with components ` i1 and ` i2 ,i =1, 2, 3 respectively. We then have

T ij ` j1 = λ (1) ` i1 and T ij ` j2 = λ (2) ` i2 . (2.1.26)

Consider the products
λ (1) ` i1 ` i2 = T ij ` j1 ` i2 ,
(2.1.27)
λ (2) ` i1 ` i2 = ` i1 T ij ` j2 = ` j1 T ji ` i2 .

and subtract these equations. We find that

[λ (1) − λ (2) ]` i1 ` i2 =0. (2.1.28)

By hypothesis, λ (1) is different from λ (2) and consequently the inner product ` i1 ` i2 must be zero. Therefore,
the eigenvectors corresponding to distinct eigenvalues are orthogonal.