# linear algebra – if \$vec(W_{k-1}) = x_{k-1}otimes A^Tv_{k-1}\$ Then \$vec(W_k) = x_k otimes A^Tv_k\$ Answer

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linear algebra – if \$vec(W_{k-1}) = x_{k-1}otimes A^Tv_{k-1}\$ Then \$vec(W_k) = x_k otimes A^Tv_k\$

This query could be very unostentatious, however notation weighty. Bear with me.

We have a ceaseless matrix $$A in mathbb R^{n occasions d}$$ the place $$d geq n$$ and $$rank(A) = n$$.

We too have a ceaseless vector $$b in mathbb R^{n occasions 1}$$ and a scalar $$alpha in mathbb R^+$$.

We have two sequences, one is a sequence of matrices, the opposite is a sequence of vectors:

$${W_k}_{ok=0}^infty subset mathbb R^{d occasions d}, {x_k}_{ok=0}^{infty} subset mathbb R^{d occasions 1}$$. Where $$W_0, x_0$$ got to us. The relaxation of the weather of the sequences are outlined recursively:

$$W_k = W_{k-1} – alpha A^T(AW_{k-1}x_{k-1}-b)x_{k-1}^T$$

$$x_k = x_{k-1} – alpha W^TA^T(AW_{k-1}x_{k-1}-b)$$

We now want to show (or disprove) that if $$vec(W_{k-1}) = x_{k-1}otimes A^Tv_{k-1}$$ for some $$n occasions 1$$ vector $$v_{t-1}$$ then $$vec(W_k) = x_k otimes A^Tv_k$$ for some $$n occasions 1$$ vector $$v_t$$. Several numerical experiments display that is undoubted, however I’ve but to provide you with a convincing dispute.

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