Let L: U →V be a linear function, and let W ⊂U be a subspace of the domain space. (a) Prove that

Let L: U →V be a linear function, and let W ⊂U be a subspace...

Key Concepts

Dimension Inequality Under Linear Transformation

The principle that the dimension of the image of a subspace under a linear transformation is less than or equal to the dimension of the original subspace is fundamental in linear algebra. This fact is often demonstrated by constructing a basis for the subspace and observing that their images under the transformation are either linearly independent or collapse, hence not exceeding the original dimension.

Dimension of a Vector Space

The dimension of a vector space is the number of vectors in any basis for the space, which serves as a measure of its size or degrees of freedom. Understanding dimension is important for comparing the sizes of different subspaces and for proving that linear transformations do not increase dimension.

Image of a Linear Transformation

The image of a linear transformation is the set of all outputs produced by the transformation, forming a subspace of the codomain. This concept is crucial in understanding how the transformation maps entire subspaces and in proving properties like dimension inequalities.

Linear Transformation

A linear transformation is a function between two vector spaces that preserves addition and scalar multiplication. This property assures that the transformation maps subspaces to subspaces, which is essential for understanding the behavior of images of sets under such mappings.

Subspace

A subspace is a subset of a vector space that is itself a vector space under the same operations of addition and scalar multiplication. Recognizing subspaces allows us to analyze parts of vector spaces that still maintain structural properties, which is critical when studying images of linear transformations.

Transcript

0:00 Hello there.

00:01 So for this exercise we need to prove some a couple of things.

00:05 So the first is to the first thing that we need to do is well we need to consider a subspace of the finite dimensional inner product space v so the first thing that we need to take into account of this information that a part at w is a subspace of b is that the finite dimensionality of b implies that w is also finite dimensional.

00:43 So we need to take into account this.

00:46 The next thing is that now we are going to define a transformation that will go from b to w.

00:56 And this transformation is defined as a is defined as a projection of b onto w.

01:02 So we need to prove that this transformation is linear.

01:06 So the first thing that we need to do is how we're going to define this projection well the fact that w is a finite dimensional subspace of this inner product space means that we can consider a basis for w so we are going to consider the basis v1 v2 up to vr and just to give you numbers due to v is a finite dimensional we're going to say that the dimension of b it's going to be n and in this case we're considering r vectors for w that means that the dimension of w is equal to r where clearly r will need to be less than n now so we have here our vectors this form a basis but we are going to impose something extra and is that this is an orthogonal basis so for any subspace we can found this orthogonal basis we just need to first pick a set of linearly dependent, linearly independent vectors, and then they apply the granite schmid procedure.

02:27 So we can get this basis.

02:32 Then let's consider a vector v in the subspace v.

02:38 And when we apply the transformation to this, that means that we need to calculate the projection of this vector v on the subspace w.

02:49 But we can do this easily because we have a basis for double that means that this transformation of the vector b on the subspace w is given by the summation i'm going to write it in the simplified notation the summation from i equals to one up to r of the inner product of b uh well i'm going to put here another notation because to avoid confusions i'm going to put this basis as alpha 1 alpha 2 alpha r okay and these are you need unitary vectors so the inner product of b weight alpha 1 times alpha 1 remember that these are unitary vectors so we we don't need to divide by the square of the norm so for any vector b the the transformation is defined in this way now so having this definition now we can pick we need we know we can prove the linearity and to prove the linearity we need to pick two vectors v1, v2 in v, such that t, for any vector, v1 and v2, we need to show that the transformation of the summation of these two vectors is equals to the summation of the transformation of the two vectors.

04:19 In other words, we need to prove this equality here.

04:22 So let's start with the left -hand side.

04:26 So the left -hand side of t v1 plus v2 is equal to the summation from i equals to 1 up to r of the inner product of v1 plus v2 with alpha i times alpha and look here we can apply the linearity of the first term of the first component of the inner product so that means that this becomes the inner the summation from i equals to the first term of the first component of the inner product.

04:58 1 up to r of the inner product of b1 with alpha i alpha i plus the the inner product of b2 with alpha i alpha i and we can distribute this sum operation onto each of the components in the brackets so we obtain that this is equal to the summation from i equals to 1 up to r of of the inner product of b i with v1 with alpha i plus the summation from i equals to one up to r of the inner product of the second vector with alpha i's alpha i and look this is just the transformation of the vector v1 plus the transformation of the vector v2 so this property is satisfied.

06:03 And the second thing that we need to show is that for any scalar, in this case, any real value, we have that the application of this transformation to alpha b is equal to alpha t of b.

06:27 So again, we're going to start from the left hand side.

06:31 So we have the t alpha b is equal to.

06:34 To the summation from i equals to 1 up to r of the inner product of alpha b uh sorry here i'm going to put gamma to avoid confusions of gamma times the vectors alpha i and look here by the property of the inner product we can take out this gamma outside because it's just and scalar and that gamma will multiply to each of the components in summation so we can take it out from the summation just multiplying the whole summation so we obtained that gamma times the summation from i equals to one up to r of the inner product of b with alpha i times alpha i and this part here you can observe is just gamma t that is what we wanted to prove so right these t is a linear transformation so t is a linear now we need to see what's going to be to the for the second part of this exercise we need to find first the range of t and second the kernel of t so let's by calculating the range so basically the range is defined for any transformation is defined as the vectors w in the the the the all the the set of vectors w that are obtained by yeah well first let me put here the the definition of t first okay so t let's remember is a function is a is a is a transformation from from from the inner product space v to the inner product space w where w is containing b so the range of t is defined as all the set as all the vectors w in w such that w is equals to t of b for some v in v okay so it's basically to which set we are mapping the vectors so we can prove these in two ways one is really easy and the second one is a we need to work a little bit so i'm going to do it in the two ways so let's start with the hardest one so in this case actually going to show and this kind in this sense is not quite obvious but this t will we can say that the range of t is actually actually the subspace w.

10:06 So let's see if this is true.

10:07 So we're going to prove that the range of t is equals to w.

10:13 So to prove the equality in this kind of exercises, we need to first prove that the range of t is contained in w, and then we need to prove that w is contained in the range of t.

10:31 If these two properties hold, then we have the equality so let's start by proving like that the range of t is contained in w so that means this is equivalent to say that if we pick some vector in the range of t let's say some vector v is in the range of t then this implies that the vector b is also in dobby...

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