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Download Algebra and analysis: problems and solutions by Lefort G. PDF

By Lefort G.

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Amm Bn1 A1m B12 A1m B22 . . A1m Bn2 . . . Amm B12 Amm B22 .. Amm Bn2 ··· ··· .. ··· ··· ··· .. ··· A1m B1n ⎞ A1m B2n ⎟ ⎟ . ⎟ . ⎟ . ⎟ A1m Bnn ⎟ ⎟ ⎟ . ⎟ .. ⎟ ⎟. ⎟ . ⎟ . ⎟ Amm B1n ⎟ ⎟ Amm B2n ⎟ ⎟ ⎟ .. ⎠ . Amm Bnn Now let A : V → V and B : W → W be linear maps and let v ∈ V and w ∈ W . The tensor product map A ⊗ B : V ⊗ W → V ⊗ W is naturally defined by (A ⊗ B)(v ⊗ w) = Av ⊗ Bw, extended by linearity. Pick a basis {e1 , . . , em } for V and { f 1 , . . , f n } for W .

The same idea works in general. For example, the components of u ⊗ v ⊗ w are just u i v j wk and so on. It is perhaps worth observing that a tensor of the form v ⊗ w for some vectors v and w is not the most general order-2 tensor. The reason is that the most general order-2 tensor has nine algebraically independent components, whereas vi w j has only six algebraically independent components (three from each vector). For example, v2 w1 = (v1 w1 )(v2 w2 )/v1 w2 . 1 Some examples of tensors you may have already encountered include the inertia tensor and the electromagnetic field strength tensor.

62) becomes σ = cg(e J , e J )σ. It follows that c = g(e J , e J ), so we may write e I = g(e J , e J )e J . 63) A similar argument shows that e J = bg(e I , e I )e I for some constant b, determined as follows. We have (−1) p(n− p) σ = (−1) p(n− p) e I ∧ e J = e J ∧ e I = g( e J , e I )σ = g(e I , e J )σ, so that (−1) p(n− p) = g(e I , e J ) = b[g(e I , e I )]2 = b and thus e J = (−1) p(n− p) g(e I , e I )e I . 55). 61). 4 Let V = M4 and let (0, 0, 1, 0), and e3 = (0, 0, 0, 1). Thus ⎛ −1 ⎜ 0 (gi j ) = ⎜ ⎝ 0 0 e0 = (1, 0, 0, 0), e1 = (0, 1, 0, 0), e2 = 0 1 0 0 0 0 1 0 ⎞ 0 0⎟ ⎟.

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