Answer the following questions related to the Rank Theorem a

Answer the following questions related to the Rank Theorem and the Rank and Nullity Theorem:

a) Suppose A is a 6×8 matrixIf dim(row(A)) = 4, then dim(col(A)) = 0

b) Suppose A is a 3×4 matrixIf dim(row(A)) = 2, then dim(null(A)) = 0

c) Suppose A is a 5×6 matrixIf A has rank 3, then dim(null(A)) = 0

d) Suppose A is a 6×8 matrixIf A has rank 4, then rank(A^T) = 0

e) Suppose A is a 5×7 matrixThe smallest value dim(null(A)) could possibly have is 0

Solution

first find the some definition about dimension of rowspave dimension of column space and dimension of null space as below :

If A is a mxn matrix with m rows and n columns then,

dim(row(A) = rank of matrix = r

dim(col(A)) = rank of matrix = r = dim(row(A))

dim(null(A)) = n - r

a)

Suppose A is a 6×8 matrix If dim(row(A)) = 4, then dim(col(A)) = 0

so we have m = 6 , n = 8

we know that dim(row(A)) = rank of matrix = r = 4 = dim(col(A))

so given statement is wrong

if f dim(row(A)) = 4, then dim(col(A)) = 4

b)

Suppose A is a 3×4 matrixIf dim(row(A)) = 2, then dim(null(A)) = 0

we have m = 3 and n = 4

dim(row(A)) = rank of matrix = r = 2

so dim(null(A)) = n - r = 4 - 2 = 2

so given statement is wrong,

If dim(row(A)) = 2, then dim(null(A)) = 2

c)

Suppose A is a 5×6 matrix If A has rank 3, then dim(null(A)) = 0

we have m = 5 and n = 6

rank of matrix = r = 3 = dim(row(A))

so dim(null(A)) = n - r = 6 - 3 = 3

If A has a rank = 3, then dim(null(A)) = 3

d)

Suppose A is a 6×8 matrix If A has rank 4, then rank(AT) = 0

we know that rank of matrix A is equal to the rank of matrix A^T so given statement is wrong.

if rank of A = 4 then rank of A^T = 4

e)

Suppose A is a 5×7 matrixThe smallest value dim(null(A)) could possibly have is 0

The given statement is wrong.

we have m = 5 and n - 7

we know that if A is full rank matrix then maximum rank of matrix A = number of non zero rows of A = 5

we know that dim(null(A)) = n - r = rank of the matrix

if A is full rank matrix then dim(null(A)) = n - r = 7 - 5 = 2

so minimum possible dimension of null is dim(null(A)) = 2