we're doing tensor products in my algebra class. here's an example the prof gave us.(adsbygoogle = window.adsbygoogle || []).push({});

problem: Can the polynomial [tex]1+xy+x^2y^2[/tex] be written in the form [tex]a(x)b(y)+c(x)d(y)[/tex] for polynomials a, b, c, d?

sol: [tex]\mathbb{R}^3 \otimes \mathbb{R}^3 \cong \mathbb{R}^3 \cong \mathcal{P}_2(x) \otimes \mathcal{P}_2(y) \cong \mathcal{P} \cong[/tex] polynomials in x & y whose degree is no greater than 2.

Let [tex]\Phi: \mathcal{P} \rightarrow \mathbb{R}^3[/tex]. (which is an isomorphism) Then [tex]\Phi(1+xy+x^2y^2) = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) = I_3[/tex]

[tex]\Phi(a(x)b(y)) = [/tex] a matrix with rank 1

[tex]\Phi(c(x)d(y)) = [/tex] a matrix with rank 1

The matrix [tex] I_3 [/tex] has rank 3, but the sum of ranks is subadditive, ie. the sum of ranks of [tex]\Phi(a(x)b(y))[/tex] & [tex]\Phi(c(x)d(y))[/tex] can be 0, 1, 2 but not 3, so it is impossible.

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# Putnam 2003, prob B1 solved the hard way

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