In Exercises 1–4, assume that the matrix A is row equivalent...

In Exercises 1–4, assume that the matrix A is row equivalent to B. Without calculations, list rank A and dim Nul A. Then find bases for Col A, Row A, and Nul A. $$ A=\left[\begin{array}{rrrrr}{1} & {-3} & {4} & {-1} & {9} \\ {-2} & {6} & {-6} & {-1} & {-10} \\ {-3} & {9} & {-6} & {-6} & {-3} \\ {3} & {-9} & {4} & {9} & {0}\end{array}\right] $$ $$ B=\left[\begin{array}{rrrrr}{1} & {-3} & {0} & {5} & {-7} \\ {0} & {0} & {2} & {-3} & {8} \\ {0} & {0} & {0} & {0} & {5} \\ {0} & {0} & {0} & {0} & {0}\end{array}\right] $$

In Exercises 1–4, assume that the matrix A is row equivalent to B. Without calculations, list rank A and dim Nul A. Then find bases for Col A, Row A, and Nul A.
$$
A=\left[\begin{array}{rrrrr}{1} & {-3} & {4} & {-1} & {9} \\ {-2} & {6} & {-6} & {-1} & {-10} \\ {-3} & {9} & {-6} & {-6} & {-3} \\ {3} & {-9} & {4} & {9} & {0}\end{array}\right]
$$
$$
B=\left[\begin{array}{rrrrr}{1} & {-3} & {0} & {5} & {-7} \\ {0} & {0} & {2} & {-3} & {8} \\ {0} & {0} & {0} & {0} & {5} \\ {0} & {0} & {0} & {0} & {0}\end{array}\right]
$$

Key Concepts

Matrix Rank

Matrix rank is defined as the maximum number of linearly independent rows or columns in a matrix. It indicates the dimension of the row space or column space and is invariant under elementary row operations. Determining the rank can help understand the solutions to a system of linear equations and is directly related to the degrees of freedom within the matrix.

Null Space and Nullity

The null space (kernel) of a matrix consists of all the solutions to Ax = 0. Its dimension, called the nullity, is determined by the difference between the number of columns of the matrix and its rank. The null space is important because it reveals whether a system of linear equations has nontrivial solutions, and how many free variables are present.

Column Space

The column space of a matrix is the set of all possible linear combinations of its column vectors. It represents the range of the linear transformation associated with the matrix. Finding a basis for the column space involves identifying a set of linearly independent columns that span this space, and this subspace's dimension is equal to the matrix's rank.

Row Space

The row space is the set of all linear combinations of the row vectors of a matrix. Like the column space, its dimension equals the rank of the matrix. Since elementary row operations do not change the row space, converting a matrix to its row-echelon or reduced row-echelon form simplifies the process of finding a basis for the row space.

Basis for a Subspace

A basis of a subspace is a set of linearly independent vectors that span the subspace. In the context of matrix theory, one often finds bases for the column space, row space, and null space in order to succinctly describe these vector spaces. Selecting appropriate basis vectors reveals the structure and dimension of the subspaces and is essential for understanding the solutions to systems of linear equations.

Transcript

00:01 Alright, so in this problem we're given the matrix a and row equivalent matrix b.

00:07 So the first thing we need to do is to, without calculation, to list the rank a and dimension of noa.

00:15 So the rank a is just, we just need to count the non -zero rows, in the row reduced matrix b.

00:25 So that is actually 3, non -zero rows.

00:29 So the next thing, the dimension of now a is to count the number of three variables.

00:39 So in this case, x5 is always zero.

00:44 If we consider bx equals zero, the equation, linear equation, then x5 is always zero.

00:52 And x3 will be determined by x4 and x5, but x5 is zero.

00:58 So x3 will only be determined by x4.

01:03 So x4 is a free variable.

01:08 Now, x1 will be determined by x2, x3, x4, and x5.

01:13 But only x2 can be free variable because x3 will be determined by x4 and x5 is 0.

01:21 So x2 is a free variable.

01:25 So the dimension turns out to be 2.

01:30 Okay, so the next thing is to find the basis for column a, for column a, the basis, the only thing we need to do is to find out the people columns and write down the people columns in terms of entries of a.

01:52 So by observing the matrix b, we find that people of columns are column 1, column 3 column.

02:02 So we just write down the first column, third column, and the fifth column of a.

02:09 So that is 1, negative 2, negative 3, and 3, and 4.

02:28 We have negative 6, negative 6, and 4.

02:35 The 5th column is 9, negative 10, and negative 3, and 0.

02:47 So we're done.

02:50 Okay, next thing is to find row a.

02:58 So in order to find the row a, we still need to observe the reduced row for, reduce the row matrix.

03:10 To observe that, we just write down the rows that are not all zeros...