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Theorem. Let $B_k$ be a symmetric matrix. Let $B_{k+1} = B_k+C$ where $C \neq 0$ is a matrix of rank one. Assume that $B_{k+1}$ is symmetric, $B_{k+1}s_{k} = y_k$ and $(y_{k}-B_{k}s_{k})^{T}s_{k} \neq 0$. Note that $s_k = x_{k+1}-x_k$. Then $$C = \frac{(y_{k}-B_{k}s_{k})(y_{k}-B_{k}s_{k})^{T}}{(y_{k}-B_{k}s_{k})^{T}s_{k}}$$

We know that $C = \gamma ww^{T}$ where $\gamma$ is a scalar and $w$ is a vector of norm $1$. Ultimately I get to the step that $$\gamma(w^{T}s_{k})w = y_{k}-B_{k}s_{k}$$

If $w \neq 0$, why does this imply that $w = \theta(y_{k}-B_{k}s_{k})$ where $$\theta = \frac{1}{\|y_{k}-B_{k}s_{k}\|}$$

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1 Answer 1

$C$ is rank one symmetric so certainly we can write $C = \gamma w w^T$. As you noted, the equation $B_{k+1} s_k = y_k$ yields $$ \gamma (w \cdot s_k) w = y_k - B_k s_k. $$ Note that $w \cdot s_k := w^T s_k$ is just a number. So we can divide both sides by $\gamma (w \cdot s_k)$ to get $$ w = \frac{1}{\gamma (w \cdot s_k)} \left( y_k - B_k s_k \right). $$ We want $w$ to have norm $1$, and we can do this by choosing $\gamma$ to be $$ \gamma = \frac{\|y_k - B_k s_k\|}{w \cdot s_k}, $$ which gives $$ w = \frac{y_k - B_k s_k}{\|y_k - B_k s_k\|}. $$ By your assumption $(y_k - B_k s_k)^T s_k \neq 0$, we have that $\|y_k - B_k s_k\| \neq 0$, and so this expression for $w$ actually makes sense. Since $w$ is a nonzero multiple of $y_k - B_k s_k$, this also implies that $w \cdot s_k \neq 0$, and so our expression for $\gamma$ also makes sense.

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