Let 𝐯1=( 1 0 2 ), 𝐯2=( 3 -1 1 ), 𝐯3=( 2 -1 -1 ),

Let 𝐯1=( 1 0 2 ), 𝐯2=( 3 -1 ...

Let $\mathbf{v}_1=\left(\begin{array}{l}1 \\ 0 \\ 2\end{array}\right), \mathbf{v}_2=\left(\begin{array}{r}3 \\ -1 \\ 1\end{array}\right), \mathbf{v}_3=\left(\begin{array}{r}2 \\ -1 \\ -1\end{array}\right), \mathbf{v}_4=\left(\begin{array}{r}4 \\ -1 \\ 3\end{array}\right)$. (a) Do $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ span $\mathbb{R}^3$ ? Why or why not? (b) Are $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ linearly independent? Why or why not? (c) Do $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ form a basis for $\mathbb{R}^3$ ? Why or why not? If not, is it possible to choose some subset that is a basis? (d) What is the dimension of the span of $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ ? Justify your answer.

Let $\mathbf{v}_1=\left(\begin{array}{l}1 \\ 0 \\ 2\end{array}\right), \mathbf{v}_2=\left(\begin{array}{r}3 \\ -1 \\ 1\end{array}\right), \mathbf{v}_3=\left(\begin{array}{r}2 \\ -1 \\ -1\end{array}\right), \mathbf{v}_4=\left(\begin{array}{r}4 \\ -1 \\ 3\end{array}\right)$.
(a) Do $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ span $\mathbb{R}^3$ ? Why or why not?
(b) Are $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ linearly independent? Why or why not? (c) Do $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ form a basis for $\mathbb{R}^3$ ? Why or why not? If not, is it possible to choose some subset that is a basis?
(d) What is the dimension of the span of $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ ? Justify your answer.

Key Concepts

Span

Span is the collection of all possible linear combinations of a set of vectors. It represents the subspace generated by those vectors, indicating whether they cover an entire space (such as ?³) or just a part of it.

Linear Independence

A set of vectors is linearly independent if none of the vectors can be expressed as a linear combination of the others. This concept is vital because it ensures that each vector adds a unique direction to the space, and it is fundamental in determining the minimum number of vectors needed to span a space.

Basis

A basis of a vector space is a set of linearly independent vectors that also span the space. A basis provides the simplest and most efficient way to represent every element in the space, and any vector in the space can be uniquely written as a linear combination of basis vectors.

Dimension

The dimension of a vector space or subspace is defined as the number of vectors in its basis. It indicates the number of independent directions in the space and helps in understanding the structure and complexity of the space.

Transcript

00:02 Okay, suppose i have a linear transformation, okay, that goes from r3 to r3, where the linear transformation is represented by the matrix, okay, a, which is equal to, let's say, 1 ,1, 0, 1, negative 3, and negative 1, 2, and 0.

00:23 Okay, so, and i want to see whether according to the standard basis, which we will denote as e1, e2, and a3, okay, if i were to transform, okay, the standard basis by this ta, so we consider ta applied to e1, ta applied to e2, and ta applied to e3, we want to see whether these three vectors will form a linear, whether this.

00:53 These three vectors is linearly independent, okay, in r3.

00:58 Okay, so how would we go about doing this? well, first of all, the trick, okay, is to note that ta applied to any of these coordinate vectors is basically just this matrix a multiplied by the corresponding coordinate vectors.

01:17 Okay, so for example, e1, that's one zero, okay? therefore, t .a of e1 is just this matrix multiplied by this vector.

01:25 Again, the same goes for tae2 and t8e3.

01:27 Okay, and the special thing about coordinate vectors like this is that basically they just pick out the corresponding column, okay, of the matrix.

01:38 So, for example, this matrix a times one zero zero, that's going to give us one zero and negative one, because that's the first column of the matrix a, okay? and tae2 is going to give us 112, and tae3 is going to give us 1 negative 3, 0.

01:57 Okay? so basically what they're asking us then is whether these three vectors are linearly independent.

02:04 Well, there's only one way to find out, which is to put them back together into the original matrix, a, and then try to see whether the determinant is non -zero.

02:16 Right if the determinant is non -zero then they're linearly independent on the other hand if the determinant is zero then they're linearly dependent okay so how do we calculate the determinant of this thing we have one times zero minus negative three times two okay and then i have minus one times a zero minus minus one minus three okay and then last last and not least, i have plus 1 times 0 minus minus 1 basically.

02:54 So let me switch to another color and do the sum cancellation in arithmetic.

02:59 So this thing becomes 6.

03:03 This thing becomes negative 3.

03:07 But then there's minus 1 in front, so this whole thing becomes plus 3...

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