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I have to solve the following problem (exact wording):


Given is a Matrix $A_{m \times n} \in \mathbb{R}^{m \times n}$ and $r = \mathrm{rank}(A)$. The goal is to find the vectors $V=\{v_1,v_2,\ldots,v_n\}$ of the orthogonal basis of $\mathrm{im}(A)$ by using the Gram Schmidt algorithm.

May $a_1,\ldots,a_n \in \mathbb{R}^m$ be the column vectors of A. We define $u_1,\dots,u_n \in \mathbb{R}^m$ by means of the Gram Schmidt algorithm,

$$ u_r = a_r - \sum_{s=1}^{r\;-1}\;\mathrm{proj}_{u_s}(a_r)\quad \text{for r = 1,...,n} $$

where

$$ \mathrm{proj}_{u_s}(a_r) = \begin{cases} \frac{u_s\cdot a_r}{u_s\cdot u_s}u_s & \text{for}~u_s \neq 0\\ \phantom| \\ \qquad0 & \text{for}~u_s=0 \end{cases} $$

a) What is the value of $u_r$, if $a_r\in \mathrm{span}(a_1,\dots,a_{r\,-1})$ with $r \leq n$? How can $\{v_1,\ldots,v_k\}$ be expressed in terms of $\{u_1,\ldots,u_n\}$?


What I don't understand is how $\mathrm{proj}_{u_s}(a_r)$ is defined, as far as I understand $u_s$ and $a_r$ are column vectors of size $m$, thus how can they be multiplied with one another?

My hunch is that $u_r$ is zero if $a_r \in \mathrm{span}(a_1,\dots,a_{r\,-1})$ as $a_r$ is not linearly independent in this case and can thus be expressed as sum of $(u_1,\dots,u_{r\,-1})$.

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1 Answer 1

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The vector product being described here is the dot product, which is an important way to take a product of vectors. For column vectors, the dot product can be computed as

$$ \mathbf a \cdot \mathbf b = \mathbf a^{\mathsf T} \mathbf b = a_1 b_1 + a_2 b_2 + \cdots + a_m b_m\;.$$

It is possible to show that $\mathbf a \cdot \mathbf b = \| \mathbf a \| \| \mathbf b \| \cos(\theta) $, where $ \| \mathbf v \|$ is the length of a vector $\mathbf v$ (e.g. the distance of the point $\mathbf v$ from the origin in Euclidean space), and $\theta$ is the angle between the two vectors $\mathbf a$ and $\mathbf b$ (i.e. between the line segment connecting the origin to the point $\mathbf a$, and the similar segment for $\mathbf b$). As a result, for any vector $\mathbf a$ and any unit vector $\mathbf u$, the vector $(\mathbf a \cdot \mathbf u) \; \mathbf u$ represents the projection of $\mathbf a$ onto the line passing through the origin and pointing in the direction of $\mathbf u$. This property is implicitly the basis on which the Gram-Schmidt process operates.

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