Let A be an m ×n matrix of rank r. Suppose 𝐯1, …, 𝐯n are a basis for ℝ^n such that 𝐯r+1, …, 𝐯n f

Let A be an m ×n matrix of rank r. Suppose 𝐯1, …, 𝐯n are a...

Key Concepts

Basis

A basis of a vector space is a set of vectors that are both linearly independent and span the space. This concept is crucial because expressing elements of vector spaces in terms of a basis allows for unique coordinate representations, which simplifies analysis and computation in linear algebra.

Kernel of a Linear Transformation

The kernel (or null space) of a linear transformation is the set of all vectors from the domain that map to the zero vector in the codomain. This subspace is key in understanding the structure of the transformation, particularly in relation to solutions of homogeneous equations and in applying the rank-nullity theorem.

Image of a Linear Transformation

The image (or range) of a linear transformation is the set of all vectors in the codomain that can be produced by applying the transformation to vectors from the domain. It is a subspace that reveals important aspects of the transformation such as surjectivity and its effective dimensionality, as measured by the rank.

Rank of a Matrix

The rank of a matrix, which represents a linear transformation, is the dimension of its image or column space. This concept provides a measure of the non-degeneracy of the transformation, indicating how many dimensions of the target space are actually reached.

Rank-Nullity Theorem

The rank-nullity theorem states that for any linear transformation from a finite-dimensional vector space, the sum of the dimensions of the kernel (nullity) and the image (rank) equals the dimension of the domain. This theorem is a fundamental tool in understanding the balance between the dimensions of the kernel and the image, and it underpins many arguments about the structure and behavior of linear transformations.

Transcript

00:01 Hi there.

00:02 For the following exercise we need to prove we need to work with a proof.

00:06 Okay, so first we're going to consider an inner transformation ta defined with an invertible matrix and further we're going to consider that you want you to until you and form a basis for the euclidean space and dimensional.

00:28 Okay, so we need to show we need to prove actually that the set of transformed vectors u u uh with respect to the this transformation t is also a basis for rn okay so let's start by the the fact that this set form a basis for r n okay so if we know that you u1, u2 until un form a basis for rn, okay? this is equivalent to say that this set u1 until un is linearly independent and it span rn.

01:40 Okay that's the conditions to be a basis so on one hand the linearly independence can be represented by the sum from i equals to one up until n of c i ui equals to zero if and only if all these cis are equals to zero with i equals to one until n okay this c i is are just constants okay but these the linear independence means that these linear combinations of the basis elements are equals to zero if and only if all these constants are equals to zero and on the other hand we need to show that the span is that for a generic vector v in our n then there exists well then this vector v can be represented uniquely from c i equals to 1 until n of c i as a linear combination of the basis elements ui okay so there is of course there is a unique unique set of c i's that satisfy this equation okay so we can represent any vector as a linear combination of this basis elements okay so now you can observe that for the linearly independent condition let me put it in a different color the linearly independent condition if we apply the transformation t a we have here the linear transformation applied to the sum from i equals to one up until n of c i u i equals to the linear transformation applied to the here is the vector zero i forgot to mention here we have the zero vector the zero vector so we have this we know and this is inherited this equation is inherited from the previous exercise so basically this is saying that this is equals to zero if all the cis are equals to zero okay for all the i's okay we just apply this transformation to this this this equation here okay so we have that this is equals to this we apply just the linear transformation to the previous equation and we know that the a linear transformation map the zero vector to the zero vector.

05:02 In this case we're map into the same space so we obtain the same zero vector and then we by linearity property of the operation we have that this is equals to the sum from i equals to one until n of c i t a ui equals to zero...

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