fourier jr
Oct20-04, 07:32 PM
we're doing tensor products in my algebra class. here's an example the prof gave us.
problem: Can the polynomial 1+xy+x^2y^2 be written in the form a(x)b(y)+c(x)d(y) for polynomials a, b, c, d?
sol: \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 polynomials in x & y whose degree is no greater than 2.
Let \Phi: \mathcal{P} \rightarrow \mathbb{R}^3. (which is an isomorphism) Then \Phi(1+xy+x^2y^2) = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) = I_3
\Phi(a(x)b(y)) = a matrix with rank 1
\Phi(c(x)d(y)) = a matrix with rank 1
The matrix I_3 has rank 3, but the sum of ranks is subadditive, ie. the sum of ranks of \Phi(a(x)b(y)) & \Phi(c(x)d(y)) can be 0, 1, 2 but not 3, so it is impossible.
problem: Can the polynomial 1+xy+x^2y^2 be written in the form a(x)b(y)+c(x)d(y) for polynomials a, b, c, d?
sol: \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 polynomials in x & y whose degree is no greater than 2.
Let \Phi: \mathcal{P} \rightarrow \mathbb{R}^3. (which is an isomorphism) Then \Phi(1+xy+x^2y^2) = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) = I_3
\Phi(a(x)b(y)) = a matrix with rank 1
\Phi(c(x)d(y)) = a matrix with rank 1
The matrix I_3 has rank 3, but the sum of ranks is subadditive, ie. the sum of ranks of \Phi(a(x)b(y)) & \Phi(c(x)d(y)) can be 0, 1, 2 but not 3, so it is impossible.