11/4/2019 MATH 2243: Linear Algebra & Differential Equations

Discussion Instructor : Jodin Morey [email protected] Website : math.umn.edu/~moreyjc

4.5: Row and Column Spaces Row Space and Row Rank

a11 a12 3≥2 Row Vectors of A: Given A : a21 a22 , the row vectors are r 1 : Ωa11,a12 æ, r 2 : Ωa21,a22 æ,

a31 a32 2 and r 3 : Ωa31,a32 æ, which exist in Ü . 2 The subspace of Ü spanned by r 1, r 2, r 3 is called the row space Row ΩAæ of the matrix A. The dimension of the row space dim ΩRow ΩAææ is called the row rank of the matrix A.

Row Space of an Echelon Matrix (Theorem ): The non-zero row vectors of an echelon matrix are linearly independent and therefore form a basis of its row space.

Row Space of Equivalent Matrices (Theorem ): If two matrices A and B are (row) equivalent, then they have the same row space.

Basis for the Row Space (Algorithm ): To find a basis for the row space Row ΩAæ, use elementary row operations to reduce A to an echelon matrix E. Then the non-zero row vectors of E form a basis for Row ΩAæ. Column Space and Column Rank

a11 a12 m≥n 3≥2 Column Vectors of A : Given: A : a21 a22 , the column vectors of A are the vectors

a31 a32 3 3 c 1 : Ωa11,a21,a31 æ, and c 2 : Ωa12,a22,a32 æ existing in Ü . The subspace of Ü spanned by c 1, c 2 is called the column space Col ΩAæ of the matrix A. The dimension of the space dim ΩCol ΩAææ is called the column rank of the matrix A.

After transforming a matrix A into an echelon matrix E, the columns containing the leading entries are called the pivot columns of E.

Basis for the Column Space (Algorithm ): To find a basis for the column space of a matrix A, use elementary row operations to reduce A to an echelon matrix E. Then the column vectors of A (NOT E !!!) that correspond to the pivot columns of E form a basis for Col ΩAæ.

We can conclude from above that the column vectors in A that do not correspond to the pivot columns in E are linear combinations of those pivot columns. Rank and Dimension Equality of Row /Column Rank (Theorem ): The row rank and column rank of any matrix are equal.

So instead of the row rank or column rank of a matrix, we usually just refer to the rank of a matrix .

Null Space of A: The solution space of the homogeneous system Ax : 0 is called the null of A, denoted by Null ΩAæ.

For Am≥n, we have: rank ΩAæ + dim ΩNull ΩAææ : n. Non-homogeneous Linear Systems

If we can find a particular solution x 0 of the non-homogeneous system Ax : b, then we can solve the system by first solving the homogeneous system Ax : 0. How? Let’s say we find a basis x 1,q, x r for the solution space of the homogeneous system . Then the general solution to the original non-homogeneous system is then: x : x 0 + c1 x 1 +q+cr x r.

To make sense of this, let’s restrict ourselves to Ü3, and imagine our solution space of the homogeneous equation to be a subspace of Ü3, maybe a plane (intersecting the origin since we have a homogeneous equation). So when c1 :...cr : 0, we have x : 0, a solution to Ax : 0. However, for this plane to be situated correctly for our solution Ax 0 : b, we must move (translate) this plane so that when c1 :...cr : 0, we have Ax 0 : b. To ensure our subspace (plane) includes x 0, we can simply add x 0 to our homogeneous solution, as this will move the 0 solution to x 0. This has the effect of moving the plane in Ü3 away from the origin, and to the proper location intersecting x 0.

Problem 8: Find both a basis for the row space and also a basis for the column space of:

1 ?2 ?3 ?5 1 4 9 2 A : . 1 3 7 1 2 2 6 ?3

1 0 1 0 0 1 2 0 “ E : . 0 0 0 1 0 0 0 0

The row basis is the first three row vectors of E. Row ΩAæ : span ¡r1,r2,r3 ¬

The column basis is the first, second, and fourth column vectors of A. Col ΩAæ : span ¡c1,c2,c4 ¬

Problem 15: Let v 1 : Ω3,2,2,2æ, v 2 : Ω2,1,2,1æ, v 3 : Ω4,3,2,3æ, and v 4 : Ω1,2,3,4æ. Let 4 S : v 1, v 2, v 3, v 4 . Find a subset of S that forms a basis for the subspace of Ü spanned by S.

3 2 4 1 2 1 3 2 A : 2 2 2 3 2 1 3 4

Calculating the reduced echelon matrix, we get: 1 0 2 0 0 1 ?1 0 “ E : . 0 0 0 1 0 0 0 0

Linearly independent: v 1, v 2, and v 4.

n n Problem 18: Let S : v 1, v 2,q, v k be a basis for a subspace W of Ü . Then a basis T for Ü that contains S can be found by applying the method of Example 5 in the book to the vectors v 1,q, v k, e 1,q, e n. 3 Using this technique, find a basis T for Ü that contains the vectors v 1 : Ω3,2,?1æ and v 2 : Ω2,?2,1æ.

Calculating the reduced echelon matrix of A : v 1 q v k e 1 q e n , we get:

3 2 1 0 0 A : 2 ?2 0 1 0 ?1 1 0 0 1

1 2 1 0 5 0 ? 5 “ E : 1 3 . 0 1 5 0 5 0 0 0 1 2 The basis vectors are v 1, v 2, e 2.