Give (in terms of orthonormal basis ) a vector in a given Span

Question

I don't know how to do these "Give in terms of $(u_1,u_2, \ldots)$ a vector in Span (...) questions. I don't understand why $b$ and $c$ in the image below are done differently or why we have to set the linear combination to $0$ etc. Does this have anything to do with a change of basis? If someone could just give a very thorough explanation of the concepts behind $b$ and $c$ of this question, I would be eternally grateful !!

Image of question

Their Answers

Answer 1

A span is a set of all linear combination of some vectors. The paper says $Span(\left\{\vec{v}_i\right\})$ where $\left\{\vec{v}_i\right\}$ is a set of vectors $\vec{v}_1,\vec{v}_2,..$. That means the span contains any vector $\vec{v}=a_1\cdot \vec{v}_1+a_2\cdot \vec{v}_2+..$ for any values $a_1, a_2,..$

In terms of means 'using those terms' (referring to the unit vectors $\vec{u}_1,\vec{u}_2,..$ in the paper). So you use those vectors in your answer.

In problem (b), notice that the span lists just one vector $$\vec{v}_1=\vec{u}_1+2\vec{u}_2+2\vec{u}_3$$ that happens to be a linear combination of several unit vectors.

Then all you have to do is find $a_1 \cdot \vec{v}_1$ which satisfies the constraint, a unit vector. That means a vector whose length is 1. You can convert a vector into a unit vector of the same orientation by dividing it by its original length. Length is computed using $l^2$-norm.

In problem (c), notice that the span lists three vectors separated by commas. And also notice the perpendicular sign $\perp$. That means the span actually includes all vectors that are perpendicular (orthogonal) to any linear combinations of those three vectors.

For two non-zero vectors to be orthogonal, their dot product must be 0.

To find a vector that meets the criteria, define a non-zero vector $$\vec{v}=a_1\vec{u}_1+a_2\vec{u}_2+a_3\vec{u}_3+a_4\vec{u}_4$$, dot-product it with one vector of some arbitrary linear combination of the three vectors in the span list, and equate that to 0 to make it orthogonal. Then solve for $a_1,a_2,..$.

The paper is confusing because it interchangeably used $a,b,c$, $u_1,u_2,..$ (they are not vectors), and $c_1,c_2,..$ as vector component values, and used the term 'linearly independent', which is not exactly the same as orthogonal. Not sure if the matrix-vector computation is correct either, since the layout of the matrix and the dimension of the vector seems wrong, but I didn't bother checking further.