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2

Intuitively it means that, when combined with the other equations, they provide the same information. More rigorously, it means that there is some "linear combination" of the equations which gives $0=0$. This is easiest to see when, as in your case, you have $n$ equations and $n-1$ of them are redundant; this means that each one is a scalar multiple of the ...


2

Since $\det(A-\lambda I) = 0$, $A-\lambda I$ is singular and therefore not full (row) rank. Therefore some equations can be written into a linear combination of others, and those equations are considered to be "redundant", as they give no additional information about the solutions.


1

Let $V$ be a complex vector space equipped with an inner product $\langle\, , \rangle$. There is an induced inner product on $(p,q)$-forms which I will also denote by $\langle\, , \rangle$. The Hodge dual is defined intrinsically as follows: for $(p, q)$-forms $\alpha$ and $\beta$, $$\alpha\wedge\overline{\star\beta} = \langle\alpha, \beta\rangle dV.$$ ...


1

Not sure what there is to explain about 1 and 3. I will say that the second equation follows by block-matrix multiplication: $$ \pmatrix{e_1'\\e_2'} \pmatrix{x_1\\x_2} = \pmatrix{e_1'\\e_2'} (x) = \pmatrix{e_1'x\\e_2'x}= \pmatrix{y_1\\y_2} $$


1

I don't have a complete answer, but I think I can add some information. In the spirit of you last comment, the way toweards characterizing the cone of simple tensors is looking at maximal linear subspaces contained in it. Each line in $V$ determins a linear subspace of dimension $dim(W)$ contained in the cone and vice versa. These two families are exactly ...



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