Question

look at vectors
1 1 0
u= 0 v= 1 w = 1
1 0 -1

what is the dimension of the space spanned by these three vectors ?

1
is the vector -1 in tyhe span of vectors u , v , and w ?
4

Answer 1

Conditions

look at vectors

$$\begin{array}{c c c c c c} & 1 & & 1 & & 0 \\ u = & 0 & & v = & 1 & w = \\ & 1 & & & 0 & - 1 \end{array}$$

what is the dimension of the space spanned by these three vectors?

is the vector 1 in tyhe span of vectors u, v, and w?

4

Please explain

Solution

In mathematics, the dimension of a vector space $V$ is the cardinality (i.e. the number of vectors) of a basis of $V$. In linear algebra, a basis is a set of linearly independent vectors that, in a linear combination, can represent every vector in a given vector space or free module, or, more simply put, which define a "coordinate system" (as long as the basis is given a definite order). In more general terms, a basis is a linearly independent spanning set.

Given a basis of a vector space, every element of the vector space can be expressed uniquely as a finite linear combination of basis vectors. Every vector space has a basis, and all bases of a vector space have the same number of elements, called the dimension of the vector space.

Let's check, if $\mathbf{u},\mathbf{v},\mathbf{w}$ is linear independent.

$$\left( \begin{array}{c c c} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & - 1 \end{array} \right)$$

As we see, the $1^{\text{st}}$ element of $\mathbf{w}$ is 0, the $2^{\text{nd}}$ element of $\mathbf{u}$ is 0 and the $3^{\text{rd}}$ element of $\mathbf{v}$ is 0. This means that there is no exist a linear combination, which could transform one of these vectors to another. That means that $\mathbf{u}, \mathbf{v}, \mathbf{w}$ is a basis, and the dimension space, spanned by these 3 vectors is:

$$(x, y, z) = c _ {1} (1, 0, 1) + c _ {2} (1, 1, 0) + c _ {3} (0, 1, - 1) = \left(c _ {1} + c _ {2}, c _ {2} + c _ {3}, c _ {1} - c _ {3}\right)$$

Let's check, if the vector $(1, -1, 4)$ is in the span of $u, v, w$ ?

For this let's find $c_{1}, c_{2}, c_{3}$ for which the previous equation is correct.

$$(1, - 1, 4) = \left(c _ {1} + c _ {2}, c _ {2} + c _ {3}, c _ {1} - c _ {3}\right)$$ $$\left\{ \begin{array}{l} c _ {1} + c _ {2} = 1 \\ c _ {2} + c _ {3} = - 1 \\ c _ {1} - c _ {3} = 4 \end{array} \right.$$ $$c _ {1} = 1 - c _ {2}$$ $$c _ {2} = - 1 - c _ {3}$$ $$1 + 1 + c _ {3} - c _ {3} = 4$$

No solution exist.

That means, that the vector $(\mathbf{1}, -1, 4)$ is not from a subspace, spanned by $u, v, w$.