Let A be a 4 ×4 matrix and let U be its row echelon form....

Let $A$ be a $4 \times 4$ matrix and let $U$ be its row echelon form. (a) Suppose columns 1,2 , 4 of $U$ form a basis for its image. Do columns 1, 2, 4 of $A$ form a basis for its image? If so, explain why; if not, construct a counterexample. (b) Suppose rows 1, 2, 3 of $U$ form a basis for its coimage. Do rows $1,2,3$ of $A$ form a basis for its coimage? If so, explain why; if not, construct a counterexample. (c) Suppose you find a basis for $\operatorname{ker} U$. Is it also a basis for ker $A$ ? (d) Suppose you find a basis for coker $U$. Is it also a basis for coker $A$ ?

Key Concepts

Kernel Invariance under Row Operations

The kernel (or null space) of a matrix is the set of all vectors that are mapped to zero by the matrix. One key invariant under elementary row operations is the kernel, which means that the null space of the row echelon form is identical to the null space of the original matrix. Therefore, a basis for the kernel of the row echelon form is also a basis for the kernel of the original matrix.

Row Space (Coimage)

The row space, sometimes referred to as the coimage in this context, is the subspace spanned by the rows of a matrix. Since elementary row operations are simply invertible linear combinations of the rows, they do not change the row space. Consequently, a basis formed by the rows of the row echelon form of a matrix is also a basis for the row space of the original matrix.

Cokernel and Its Non-Invariance

The cokernel of a matrix is defined as the quotient of the codomain by the image of the matrix. Unlike the kernel and the row space, the cokernel is affected by changes in the column space, which may be altered under elementary row operations. Thus, a basis for the cokernel of the row echelon form does not necessarily provide a basis for the cokernel of the original matrix, as the transformation can affect the complement of the image in the codomain.

Row Echelon Form and Row Equivalence

A matrix in row echelon form is obtained from the original matrix by applying a sequence of elementary row operations, which are reversible. These operations establish that the row echelon form is row equivalent to the original matrix, meaning that key subspaces such as the row space (or coimage) and the null space (kernel) remain unchanged. However, elementary row operations can change the specific form of the columns, so while they preserve certain invariants (like rank), they do not necessarily preserve the column space directly.

Column Space (Image) and Pivot Columns

The column space (or image) of a matrix is the set of all possible linear combinations of its columns. While the basis for the column space of a row echelon form often consists of the pivot columns, selecting the corresponding columns from the original matrix yields a basis for the original matrix’s image. This result holds because the dependency relations among the columns are preserved under elementary row operations, meaning that the pivot columns of the original matrix remain linearly independent and span the image.

Transcript

00:02 For this exercise, suppose we have the matrix a, b, and n by m matrix, then we'll have image a defined as equal to a times v vector.

00:27 Then we have v vectors an element of r to the power of m space, and we have kernel a equal to vector v is an element of our space to the power of m, and then we have a times vector v equal to zero vector.

00:56 And according to that theorem, the image a and kernel a are subspaces, of r space to the power of m and r space to the power of n.

01:19 Now with that being said, for a, suppose we have vector w is an element of the image of a, then vector v is an element of three -dimensional space, such that a times vector v equals vector w, and zero vector is equal to a square times vector v, which is equal to a, and this is a times vector v, which equals a times vector w.

02:00 Thus, vector w is an element of the kernel of a, and the image of a is a subspace of the kernel of a.

02:27 Now for b, the dimensions here that we find for image a is that it's 1.

02:43 Let me write that a little more, dimensions of kernel a equal 2.

02:51 And then for c, an onto a vector that we'll be choosing is vector v .1 is an element of the image of a, where vector v .1 does not equal vector 0.

03:21 And we have vector v.

03:23 2 is an element of r space of power of 3, such that a times vector v.

03:34 2 equals vector v.

03:35 Sub 1.

03:38 And we have vector v...

Please give Ace some feedback