Computing the image and kernel of a linear transformation

Let V be an n-dimensional vector space with basis B = \{e_i\}_{i=1}^n and let W be an m-dimensional vector space with basis E = \{f_i\}_{i=1}^m. Let \varphi : V \rightarrow W be a linear transformation, and let A = M^E_B(\varphi). (That is, the jth column of A is the coordinates of e_j in terms of the ordered basis E. Let A^\prime be the reduced row echelon form of A.

  1. Prove that the dimension of \mathsf{im}\ \varphi is the row rank (i.e. number of nonzero rows) of A^\prime. Prove moreover that the columns of A corresponding to the pivotal columns of A^\prime form a basis of \mathsf{im}\ \varphi.
  2. Note that the kernel of \varphi is precisely the solution of the equation A^\prime X = 0. We saw how to solve such equations in this previous exercise. Prove that \varphi is injective if and only if A^\prime has row rank n. If \varphi is not injective, construct a basis for \mathsf{ker}\ \varphi.

  1. Write A = [A_1 | \cdots | A_n] as a column matrix. Then A[x_1 | \cdots | x_n]^\mathsf{T} = \sum A_i x_i. In particular, the image of \varphi is the span of the columns of A. In particular, as we showed previously, the columns of A corresponding to the pivotal columns of A^\prime form a basis of \mathsf{im}\ \varphi. In particular, the dimension of this image is the row rank of A^\prime.
  2. Note that \varphi is injective if and only if \mathsf{im}\ \varphi has dimension n, if and only if (by part 1) A^\prime has row rank n. We showed previously how to construct the solutions of A^\prime X = 0; specifically, for each nonpivotal column of A^\prime, we fix one corresponding entry in X to be 1 and the other (nonpivotal) entries 0. The other entries of X are then determined uniquely. These vectors are certainly linearly independent, and generate \mathsf{ker}\ \varphi.
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