Show that if W ⊂ℝ^3 is a subspace containing the vectors (1,2,-1)^T,(2,0,1)^T, (0,-1,3)^T, then

Show that if W ⊂ℝ^3 is a subspace containing the vectors...

Key Concepts

Subspaces and Dimension

A subspace is a subset of a vector space that is itself a vector space under the same operations. The dimension of a subspace is the number of vectors in its basis. Since R³ is a three-dimensional space, any subspace of R³ that contains three linearly independent vectors must necessarily be R³ itself.

Linear Independence

In linear algebra, a set of vectors is said to be linearly independent if none of the vectors can be written as a linear combination of the others. This concept is crucial because if a set of vectors in an n-dimensional space has exactly n linearly independent vectors, then they form a basis for the space. Demonstrating that the given vectors are linearly independent indicates that they can span the entire space when properly combined.

Basis and Span

A basis of a vector space is a set of linearly independent vectors that span the entire space, meaning every vector in the space can be expressed as a linear combination of the basis vectors. In this problem, showing that the provided vectors form a basis implies that they span R³, and consequently, any subspace containing these vectors must be the whole space.

Transcript

00:01 Show that the set of vectors, and i'm going to write it down here, does not span three -dimensional real space, but it does span the subspace of three -dimensional space, consisting of all vectors lying in the plane with equation x minus 2y plus z equals 0.

00:28 So i'm going to write down the vectors, 3, 4, 5, and it'll span this subset, this plane.

00:56 All right.

01:01 So if we write that any vector is going to be a constant times v1 plus a constant times v2, plus a constant times v3, i should note that this is v1, this is v2, and this is v3.

01:36 Then a constant times 1, 2, 3.

01:46 That's v1.

01:48 A constant times 3, 4, 5.

01:55 That's v2, and a constant times that, which is v3.

02:08 If i write down the x coordinates, it would be 1 times c1 plus 3 times c2 plus 4 times c3 is the x coordinate.

02:38 2 times c1 plus 4 times c2 plus 5 times c3 is the y coordinate.

02:52 And just to be clear, that's where i'm getting them from.

03:00 And to be more clear, this equals x, y, z, any vector in three -dimensional space.

03:11 Okay? also, three times c1 plus five times c2 plus six times c3 is z.

03:30 Now, we can write this as an augmented matrix.

03:36 1, 3, 4, 2, 4, 5, 3, 5, 6.

03:50 I got them from right here.

03:53 1, 3, 4, 2, 4, 5, 3, 5, 6.

04:01 And then the solution, not the solution, but the xyz matrix, is just x, y, z.

04:10 The solution matrix would be c1, c2, c3.

04:18 All right, we can reduce this matrix.

04:25 I'm going to take negative 2 times the first row and add it to the second row.

04:40 Oh my goodness, that is not what it says.

04:51 Okay, two times the first row and add it to the second row, that gives us 0...

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