(a) Prove that 𝐯1=((3)/(5), 0, (4)/(5))^T, 𝐯2=(-(4)/(13), (12)/(13), (3)/(13))^T, 𝐯3=(-(48)/(65)

step 1 We're given a set S of vectors in our 4. So S contains vectors U1, with components 1 -1 -1 -1, a

(a) Prove that 𝐯1=((3)/(5), 0, (4)/(5))^T, 𝐯2=(-(4)/(13),...

Key Concepts

Coordinate Representation

Expressing a vector in terms of a basis involves writing it as a linear combination of the basis vectors. The coefficients, or coordinates, in this expansion are unique and reflect how much of each basis direction is present in the vector. When the basis is orthonormal, these coefficients can be computed directly and efficiently using the dot product, making the coordinate representation particularly elegant.

Orthonormal Basis

An orthonormal basis for a vector space is a set of vectors that are both mutually orthogonal and each of unit length with respect to the chosen inner product (commonly the dot product in ?^n). This property greatly simplifies many computations in linear algebra, such as expressing any vector in the space uniquely as a linear combination of the basis vectors without the need for solving systems of equations.

Dot Product

The dot product is a binary operation that takes two vectors and returns a scalar. In ?^3, it is defined as the sum of the products of corresponding entries. This operation provides a way to measure the angle between two vectors, with a zero dot product indicating orthogonality. It is also used in calculating lengths and projections, which are key in verifying unit length and orthogonality in an orthonormal basis.

Projection (Expansion) Formula

When a set of vectors forms an orthonormal basis, any vector in the space can be reconstructed as the sum of its projections onto the basis vectors. This is given by the formula v = (v · v1)v1 + (v · v2)v2 + (v · v3)v3 in ?^3. This projection formula simplifies the computation of coordinates and verifies that the decomposition is unique and exact, exemplifying the power of using an orthonormal basis.

Transcript

00:02 We're given a set s of vectors in our 4.

00:07 So s contains vectors u1, with components 1 -1 -1 -1, as well as u2, with components 1 -1 -1 -1 -1 -1 -1.

00:24 U3, which is 1 -1 -1 -1 -1 -1, and u -4 is 1 -9 -1 -1 -1.

00:37 I guess i'll just let everyone know with that real one.

00:41 In part a, we're asked to show that s is orthogon and a basis of r4.

00:52 Well, let's compute all the inner products.

00:55 So the interproduct of u1 with u2 is 1 plus 1 minus 1 minus 1, which is 0.

01:02 The inner product of u1 with u3 is 1 minus 1 plus 1 minus 1, which is 0.

01:10 The inner product of u1 with u4 is equal to 1 minus 1 minus 1 minus 1.

01:16 1 plus 1, which is 0.

01:19 Likewise, the inner product of u2 with u3, this is 1 negative 1, negative 1, which is 0.

01:34 The inner product of u2 with u4 is 1 minus 1, which is 0.

01:47 And finally the inner product of u3 with u4.

01:54 This is 1 minus, say 1 plus 1 minus 1 minus 1, which is 0.

02:04 Since all of these vectors are orthogonal to each other, it follows up a set s is an orthogonal set.

02:15 All right.

02:31 To determine if s is in fact a basis, we check if the vectors are linear -bein -dependent.

02:41 This actually isn't too hard.

02:46 Well, suppose that a -u -1 plus b -u2 plus c -u -3 plus d -4 equals 0.

03:01 This gives us a plus b.

03:08 Plus c plus d okay gives us a system a plus b plus c plus d equals zero as well as a plus b minus c minus d equals zero a minus b plus c minus d equals zero and finally a minus d minus c plus c plus d equals zero and finally a we add the first two equations.

04:14 We get 2a plus 2b equals 0.

04:21 If we add the last two equations, we get 2a minus 2b equals 0.

04:26 And then adding these two equations, we get 2a minus 2b equals 0.

04:30 And then adding these two equations, we get 4a equals 0, and therefore a equals 0.

04:38 Plugging this back into one of the previous equations, we get b equals 0.

04:46 Okay.

04:50 So we have a and b are 0.

04:51 This gives us, we have c plus d equals 0, and also c minus d equals 0.

05:00 Therefore, 2c equals 0, so it's c equals 0, plugging this all back into the first equation, we get the d equals 0 as well...