Finding the maximum area of isosceles triangle

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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