(a) Show that, $ δ_ijδ_ij = 3 $
(b) Show that, $ V_i = δ_{ij}V_j = δ_{ik}V_k $
(c) Simplify $ δ_{ij}δ_{ik}δ_{jk} $ and state if it is a scalar, vector, or 2nd order tensor.
1 (a)
$ δ_{ij}=1 $ $ if \ i = j $
$ δ_{ij}=0 $ $ if \ i \neq j $
$$ δ_{ij}δ_{ij} = δ_{1j}δ_{1j} + δ_{2j}δ_{2j} + δ_{3j}δ_{3j} $$$$ δ_{ij}δ_{ij} = δ_{11}δ_{11} + δ_{12}δ_{12} + δ_{13}δ_{13} + δ_{21}δ_{21} +δ_{22}δ_{22} + δ_{23}δ_{23} + δ_{31}δ_{31} + δ_{32}δ_{32} + δ_{33}δ_{33} $$$$ δ_{ij}δ_{ij} = δ_{11}δ_{11} + δ_{22}δ_{22} +δ_{33}δ_{33} = 1 + 1 + 1 = 3 $$Vector Notation: $$ δ_{ij}δ_{ij} = \left[ \begin{array}{ccc} δ_{11} & δ_{12} & δ_{13} \\ δ_{21} & δ_{22} & δ_{23} \\ δ_{31} & δ_{32} & δ_{33} \\ \end{array} \right] \left[ \begin{array}{ccc} δ_{11} & δ_{12} & δ_{13} \\ δ_{21} & δ_{22} & δ_{23} \\ δ_{31} & δ_{32} & δ_{33} \\ \end{array} \right] = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right] \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right] = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right] = 1 + 1 + 1 = 3$$
2 (b)
$ V_i=δ_{ij}V_j = δ_{ik}V_k $
$$ V_i = δ_{ij}V_j = \left[ \begin{array}{ccc} δ_{ij} & δ_{ij} & δ_{ij} \\ δ_{ij} & δ_{ij} & δ_{ij} \\ δ_{ij} & δ_{ij} & δ_{ij} \\ \end{array} \right](V_j) = \left[ \begin{array}{ccc} δ_{11} & δ_{12} & δ_{13} \\ δ_{21} & δ_{22} & δ_{23} \\ δ_{31} & δ_{32} & δ_{33} \\ \end{array} \right](V_j) = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right](V_j) = V_1 + V_2 + V_3 $$3 (c) Simplify $ δ_{ij}δ_{ik}δ_{jk} $ and state if it is a scalar, vector, or 2nd order tensor.
$$ δ_{ij}δ_{ik}δ_{jk} = δ_{ij}δ_{ij} = δ_{ij} = 3 $$ $$ SCALAR $$Number of Free Indices = 0 Therefore it is a scalar
The Alternator: (a) Show that $ e_{ijk}e_{ijk} = 6 $ (b) Let A be a matrix with components aij (e.g. a stress tensor), show that $ det|A| = e_{ijk}a_{1i}a_{2j}a_{3k} $
2 (a) Show that $ e_{ijk}e_{ijk} = 6 $
$ e_{ijk}= 1 $ if ijk is an even permutation
$ e_{ijk}= -1 $ if ijk is an odd permutation
$ e_{ijk}= 0 $ if two indices are repeated
$$ e_{ijk}e_{ijk} = δ_{ii}δ_{jj}-δ_{ij}δ_{ji} = (3*3) - δ_{ii} = 9 - 3 = 6 $$2 (b) Let A be a matrix with components $a_{ij}$ (e.g. a stress tensor), show that $ det|A| = e_{ijk}a_{1i}a_{2j}a_{3k} $
when i = 1 $$ e_{ijk}a_{1i}a_{2j}a_{3k} = e_{123}a_{11}a_{22}a_{33} +e_{132}a_{11}a_{23}a_{32} = a_{11}a_{22}a_{33}- a_{11}a_{23}a_{32} = a_{11}(a_{22}a_{33}-a_{23}a_{32}) $$
when i = 2 $$ e_{ijk}a_{1i}a_{2j}a_{3k} = e_{213}a_{12}a_{21}a_{33} +e_{231}a_{12}a_{23}a_{31} = -a_{12}a_{21}a_{33}+ a_{12}a_{23}a_{31} = -a_{12}(a_{21}a_{33}-a_{23}a_{31}) $$
when i = 3 $$ e_{ijk}a_{1i}a_{2j}a_{3k} = e_{312}a_{13}a_{21}a_{32} +e_{321}a_{13}a_{22}a_{31} = -a_{13}a_{21}a_{32}+ a_{13}a_{22}a_{31} = -a_{13}(a_{21}a_{32}-a_{22}a_{31}) $$
Using index notation, prove the vector identity: $ a × (b × c) = (c · a)b − (b · a)c $, where a,b,c are vector quantities. [hint: note that, $ d = (b × c)_i = e_{ijk}b_jc_k $]