We proved a theorem stating that for a linear mapping α : U → V , we havedim Kerα + dim Imα = dimU

This theorem is known as the Rank-Nullity Theorem in Linear Algebra. Here's a step-by-step proof:

1. Let's start by defining our terms. In this context, U and V are vector spaces, and α is a linear map from U to V. The Kernel (Kerα) of a linear map is the set of all vectors in U that α maps to the zero vector in V. The Image (Imα) of α is the set of all vectors in V that α maps to from U. The dimension (dim) of a vector space is the number of vectors in its basis.

2. We know that the Kernel and Image of α are both subspaces of their respective vector spaces. This means they each have a basis, and therefore a dimension.

3. The Rank-Nullity Theorem states that the dimension of the Kernel of α plus the dimension of the Image of α is equal to the dimension of U. In other words, dim Kerα + dim Imα = dim U.

4. To prove this, we can start by choosing a basis for Kerα. Let's call this basis {u1, u2, ..., un}, where n = dim Kerα.

5. Since α is a linear map, it preserves linear combinations. This means that if we take any vector in U and express it as a linear combination of the basis vectors, the image of this vector under α will be the same linear combination of the images of the basis vectors.

6. Now, let's extend our basis for Kerα to a basis for U. We can do this by adding vectors to our set until it spans U. Let's call this extended basis {u1, u2, ..., un, un+1, ..., un+m}, where m = dim U - dim Kerα.

7. The vectors {un+1, ..., un+m} form a basis for Imα. This is because any vector in Imα can be expressed as a linear combination of these vectors, and they are linearly independent.

8. Therefore, we have dim Kerα + dim Imα = n + m = dim U, which is what we wanted to prove.