Problem 1: Consider the vectors u⃗ =

Problem 1: Consider the vectors $\vec{u} = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$ and $\vec{v} = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$. Determine if each vector below belongs to the span of $S = \{\vec{u}, \vec{v}\}$. (a) $\begin{bmatrix} 3 \\ 1 \\ 0 \end{bmatrix}$ (b) $\begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}$ (c) $\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

Transcript

00:01 We are going to determine whether each set spans the given vector space.

00:07 If yes, we prove it and if no, we give an example of a vector outside the span.

00:13 In part a, the set s is formed by three vectors, b1 equal 1 to negative 1, b2 equals 0 to 1 and v3 equal 1 -11, and the vector space is r3, the space r3.

00:32 And in part b, we have s equal two vectors, v1 equal negative 1 -1, and v2 equal to negative 2, and let the vector space vr2, the plane.

00:43 Okay, so let's start with part a, and here we are going to apply the fact that r3, we know, has dimension 3, that is, the maximum number of linear independent vector is 3.

00:56 And because we have three vectors in set s, if we prove that the set s is linearly independent, then it must be a base of r3.

01:12 So it will span that space.

01:16 So let us prove that s in part a, that is this set over here, is linearly.

01:31 Independent.

01:39 And for that, let alpha, beta, gamma, b any three real numbers.

01:54 Real numbers, such that the linear combination, alpha v1 plus beta, b2, plus gamma b3 is equal to zero vector zero zero zero.

02:18 If we prove that alpha, beta, and gamma are all of of them equal to 0 then this set s will be linearly independent so we write this expression here alpha times b1 is 1 negative 1 2 negative 1 plus coefficient beta times vector b2 is 0 to 1 plus gamma times 1 1 1 and that equal to 0 vector 0 0 0 0 0 0 0 okay so now we do the vector, scalar vector products and the sum of vectors and we will get, and then we apply the equality.

03:11 Let me do one step in the middle.

03:15 That is what we do here is that we do vector operations on the left, and we get a vector which is alpha plus gamma because beta will be multiplied by zero.

03:27 It's the first component of the result.

03:30 Then 2 alpha plus 2 beta plus gamma.

03:36 It's the second component.

03:37 And the third one is negative alpha plus beta plus gamma.

03:45 And that's the vector operation on the left.

03:48 And on the right we have zero vectors.

03:50 So now the quality of vectors implies that the components, corresponding components, get to be equal.

03:57 So we get a system of three equations...

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