Let’s say I have a set of vectors We called the group B And let’s say I have vectors V1, V2, ………. to Vk Let’s consider that this set of vectors is not any set of vectors There are some important things about these vectors The first thing is that each vector has a length of 1 So we can say that the length of the vector Vi is 1, since i equals a value between 1 and k or that i is equal to 1,2 down to k Each of these vectors has a length of 1 Alternatively, the square of the length of each vector is equal to 1 The square of the length of Vi, and therefore its length of 1, or the product of the point multiplication of vi equals 1, where i is any of these values \u200b\u200bwhether it is 1, 2, 3 ….. and even k This is the first interesting thing about it. I will write it in regular words. All vectors in B are 1 long. Or – in another way – all vectors were normalized. It is another way of saying that they are all normal. Or they are all unit vectors. The vector vectors are vectors whose length is 1 Or converted to unit vectors. All vectors printed So this is the first interesting thing about Group B. And then the second interesting thing about B is that all vectors are perpendicular to each other. So if you multiply it by itself, the resulting length is equal to one unit. But if you multiply the vector point by another vector, if you multiply the vector vi with the vector vj point, if you multiply v1 point by v2 then the result is equal to 0 if i is not equal to j. All of these vectors are orthogonal I will write it here All of these vectors are perpendicular to each other Of course, they are not perpendicular to themselves because they are all equal to 1. Thus the product of point multiplication is 1, while the product of point multiplication between one and the other vector in the same set is equal to 0. Maybe I can write it like this. The product of vi of vj or (vi.vj) is 0, where i does not equal j So if we multiply these vectors by themselves, the length is equal to 1 because i and j are equal So I have a special case, all of these vectors are deliberate with each other and the length of each is equal to 1 All of them are converted to unit vectors and all are orthogonal. We have a special name for this case, it is called the standard orthogonal group. Natural print All vectors are perpendicular All of these vectors are orthogonal to each other And all vectors converted to unit vectors All vectors are length (vector length means the sum of the vector and is converted to a non-vector value) Now, the first interesting thing about the standard orthogonal group is that it forms a linearly independent group Thus if B is orthogonal calibrated, B will be linearly independent How can I show you that? Well, suppose it is not linearly independent So we take vi and vj which belong to the group, and suppose i is not equal to j Now, we know the group is orthogonal calibrated Thus, vi times dot vj is equal to 0 It is orthogonal These vectors belong to the group Let’s consider it linearly independent I want to prove that it is linearly independent, and I proved it assuming that it is not linearly independent and then we come to terms with the hypothesis. So suppose vi and vj are not linearly independent. Well, this means that I can represent one of these vectors as the multiplication vector of a vector with a non-vector constant value. I can pick it up anyway. So let’s say that I can represent vi – let’s say that vi equals a constant product of vj. This is what linear dependency or linear autonomy means One of these vectors represents the other multiplied by a constant. If this is true, I can substitute this value for vi So what will I get? I get c times vj which is another image to represent vi, because I assumed it was linearly followed. The product of that in vj must be 0 That is vi And this vj They are perpendicular to each other. But here this is equal to c times vj times vi dot, which is equal to vj squared in c and equal to 0. It is orthogonal and therefore equal to 0 What knows us, the length of vj should equal 0 If we assume that this is a nonzero multiplication, it is impossible for this to be a multiplication of 0 – I must write it there – so c is not equal to 0 Why shouldn’t this be a non-plait operation? Because both are nonzero vectors. This is a nonzero vector So this cannot be zero The length of this vector is equal to 1 So if this vector is not 0, I can’t put a zero in here Because if I put 0, I have a vector zero. Thus, c is not equal to 0 If c is not 0, then this vector is 0 Thus we conclude that this vector is equal to 0 Which we know is not true The length of vj is equal to 1 This is a calibrated orthogonal group The length of all elements or vectors within Group B is 1 Thus we reached a contradiction This is a set aside the hypothesis. vj is not a zero vector, the length of vj is equal to 1. opposes So if we had a set of orthogonal vectors that were not zero, they must be linearly independent. Really interesting. So, if I had this group, the linearly independent group here, it is a set of linearly independent vectors, therefore it could be a subspace base in addition to being orthogonal calibrated so that it is partly space So let’s say B is some subspace base, say v Or we can say that v is equal to the range v1, v2 and even vk Then we can call B – if it is just a group – we might call it a calibrated orthogonal group, but we can consider it as a standard orthogonal norm that represents a partial space. So we can write – we can say that B is the orthogonal basis for a standardized v. Now, everything I’ve done is just so far so simple, I’m going to show you with a quick example, so that you can understand how a calibrated base would look like real numbers. So let’s say I have two vectors Let’s say I have the vector v1 which is about – let’s say we’re dealing with vectors in R3 – the values \u200b\u200bare 1/3, 2/3, 2/3, and 2/3 Let’s say I have another vector, which is v2, which is equal to 2/3, 1/3 and minus 2/3. Let’s say that B is a group consisting of v1 and v2 So the first question is: What are these vector lengths? Let’s find the lengths of the vectors (non-vectors or summed). The length of v1 square is the product of v1 with itself Which is just 1/3 squared which equals 1 by 9 Plus 2/3 squared equals 4/9, plus 2/3 squared equals 4/9, and the sum is 1 So the square of length is equal to 1, and this tells us that the length of the first vector is equal to 1. If the square of length is 1, then you can find the square root of 1 and equal 1 What about vector number 2? The length of the second vector squared equals the product of point v2 For a square, 2/3 squared equals 4/9, square 1/3 equals 1/9, and square 2/3 equals 4/9. The sum equals 1 So the length of the second vector is equal to 1 Thus we know that these vectors are unit vectors. We can call it a set of unit vectors However, is it a calibrated orthogonal group? Are they perpendicular to each other? To test this, we multiply both vectors together by a point. Times v1 with v2, 1/3 times 2/3 equals 2/9, plus 2/3 negative 2/3 – equals 4/9 – 2 plus 2 minus 4 equals 0 So the result is 0 Thus these vectors are actually orthogonal, indicating that B is an orthogonal group standardized And I have a subspace, let’s say that B equals the range of both v1 and v2, then we can say that the base of V, or we can say that B is a orthogonal base calibrated to V

Algebra