Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $$\theta = \frac{1}{\|y_{k}-B_{k}s_{k}\|}$$

share|cite|improve this question

$C$ 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.