# How can show the matrix is rank one?

I am civil engineering student , and I don't have any information about positive definite, rank, and conjugate . Please help me if you can.

I have an algorithm that finds the minimum.

I have a matrix $H$, and $H$ be an $n\times n$ symmetric matrix, and $$f(x)= c^Tx+\frac12x^THx.$$ Suppose $D_1$ is a $n×n$ positive definite symmetric matrix, let $x_1$ be an initial point. For $j=1,..,n$ let $\lambda_j= \arg\min_{λ\ge0} f(x+ \lambda d_j)$, and $x_{j+1}=x_j+ \lambda_jd_j$ where $d_j=-D_j\nabla f(x_j)$ and $D_{j+1}= D_j+ a_j$

$$a_j=\frac{(p_j-D_jq_j)(p_j-D_jq_j)^T}{q_j^T(p_j-D_jq_j)}$$

$p_j=x_{j+1}-x_j$ and $q_j=Hp_j$.

• In this algorithm, why is the matrix $a_j$ of rank $1$?

• If $D_j$ be a positive definite, is necessarily $D_{j+1}$ positive definite too?

• Are the direction $d_1,...,d_n$ necessarily conjugate?

Thanks

• Did you already consult wikipedia for information about "posiive definite", "rank" etc.? Jun 4, 2018 at 11:01
• Yes , but I can't proof Jun 4, 2018 at 11:03
• It's very difficult for me, is it simple? Jun 4, 2018 at 11:03
• How is $a_j$ a matrix? The denominator is of the form $u^Tv$, so it's a scalar, but the numerator is of the form $v^2$ for a vector $v$. Presumably it was meant as $vv^T$ which is a square matrix, with each column being a multiple of $v$, hence of rank $\le1$. Jun 4, 2018 at 11:35
• In my book $a_j= \frac{(p_j-D_jq_j)(p_j-D_jq_j)^T} {q_j^T(p_j-D_jq_j)}$ . Jun 4, 2018 at 11:48

The matrix $a_j$ is rank $1$ since it is a matrix of the form $uv^T$, where $u$ and $v$ are column vectors. Note that in such a matrix, every row will be a multiple of $v^T$.
The matrix $a_j$ is symmetric, and it will also be positive semidefinite since if the number on bottom is positive. In this case, we can say that $a_j$ is positive semidefinite since it can be written in the form $a_j = MM^T$. In particular, we have $$a_j = \left[\frac{p_j - D_j q}{\sqrt{q_j^T(p_j - D_jq_j)}}\right]\left[\frac{p_j - D_j q}{\sqrt{q_j^T(p_j - D_jq_j)}}\right]^T$$
If both $D_j$ and $a_j$ are positive semidefinite, then their sum $a_j + D_j$ will also be positive semidefinite.
• to show $d_1,..,d_n$ are conjugate , must show $d_iHd_j=0 , \forall i\neq j$ ?? Jun 4, 2018 at 13:07