Rank and Nullity
Let A be a mxn Matrix
-The rowspace of a is the span of all the rows. It is a subspace of ℝ𝑛
𝑟𝑜𝑤 𝑠𝑝(𝐴) = 𝑠𝑝𝑎𝑛{𝑟⃗⃗⃗1 , ⃗⃗⃗
𝑟2 , … 𝑟⃗⃗⃗𝑛 , }
in echelon form 𝑠𝑝𝑎𝑛{𝑟⃗⃗⃗1 , ⃗⃗⃗
𝑟2 , … ⃗⃗⃗
𝑟𝑛 , } is a basis for 𝑟𝑜𝑤 𝑠𝑝(𝐴)
- the columns space iof A is the subspace of ℝ𝑚 , spanned by the columns.
𝑐𝑜𝑙𝑢𝑚𝑛 𝑠𝑝(𝐴) = 𝑠𝑝𝑎𝑛{𝑐⃗⃗⃗1 , ⃗⃗⃗
𝑐2 , … ⃗⃗⃗⃗
𝑐𝑛 , }
Thm
Let A be a mxn Matrix
Elementary row operation does not change the rowspace of A.
Thm
If A is a matrix in echelon form then the nonzero rows are linearly independent.
Therefore the rows of an echelon form matrix are a basis for the rowspace for that matrix (or of a
matrix that is this matrix but changed by elementary row ops, see above)
Defn
The Rank of a matrix is the dimension if of a basis for it’s rowspace.
The Nullity of a matrix is the dimension of a basic for it’s nullspace.
Finding Nullity
Null space of A is the solution (for x) to Ax=0,
Solve the system of linear equations to get the nullspace (x)
Find the null spaces basis, then get it’s nullity by the size of the basis.
Null space has dimension (nullity) equal to the number of free variables in the system of
linear equations. Ax=0
Thm (Rank/Nullity theorem)
if A is an mxn matrix then
𝑟𝑎𝑛𝑘 (𝐴) + 𝑁𝑢𝑙𝑙(𝐴) = 𝑛
n is number of columns of the matrix
rank(A) = dimention of the rowspace of A = number of basic (leading in echelon form) variables =
number of nonzero rows of A.
null (A)=nullity of A=number of free variables
Don’t confuse null(A) with nullspace of A.
Thus the theorem also means:
Number of Basic variable + number of free variables = number of variables.
Thm (rowspace column space)
rank(A) = Dimension row space= dimention column space for matrix.
Possible Questiuons:
1) what is the null space of A: answer: -Subspace
2) what is the nullity of A: answer: number
3) find a basis for the rowspace of A, answer: basis – set of vectors
4) find a basis for the column space of A, answer: basis – set of vectors
5) what is the rank of A, answer: number
6) find a subspace that is the basis for rowspace of A. answer - subspace
4) find a basis for the column space of A.
remember: elementary row operations change the column space.
So have to go back to original (non-echelon form) matrix.
and pick the number Rank(A) independent column vectors. Because if we have right number of
independent vectors of a set, it automatically will span the set.
Question 3.6.4.
Qi)suppose A is a 3x3 matrix whose nullspace is a line.
Give a geometric description of the rowspace.
ANS: a plane. Number of variables=3=Rank(A)+null(A)
3=Rank(A)+1⇛rank(A)=2 is a plane
Qii) is A is a 6x4 matrix s.t Ax=0 has a nontrivial solutions then what are possible values of rank(A)
ANS: Null (A)≥1, as nontrivial soltution exist to Ax=0
.’.Rank A=4-null(A)
Rank(A)=3 or 2 or 1 or 0
Theorem
if A is an nxn matrix then the following are equivalent (TFAE): (If one is true then all are true)
1) A is an invertible matrix (A has an inverse)
2) rank(A)=n (number of columns/variables)
- (if A is invertible, then in echelon form has no zero rows,)
3) null (A)=0
4) all rows of A are linearly independent
5) all columns of A are linearly independent