Question

Linear Algebra- kind of a long question but would really appreciate it if someone could help me explain why these statements are equivalent

Answer 1

If A is an nxn matrix then the following statements are equivalent
a. The reduced row echelon form of A is I_n
b. A is expressible as a product of elementary matrices
c A is invertible
d Ax=0 has only the trivial solution
e Ax=b is consistent for every vctor b in R^n
f Ax=b has exactly one solution for every vector b in R^n
g The column vectors of A are linearly independent
h The row vectors of A are linearly independent
I Det(A) does not equal 0

Answer 2

Here's my attempt:

a Okay so for the first one I assumed that since A has to be invertible reduced row form has to be an identity matrix
b. since it can be reduced to an identity matrix
c. invertible since determinant not 0 and it can be reduced to identity form?
Defgh I don’t understand
I the determinant can’t be 0 since it’s invertible

Answer 3

d. If A is an invertible matrix, then it makes sense that the only vector that you can multiply by to get the zero vector is the zero vector itself. This is a direct result from f, actually.
f. Since A is invertible, then each matrix b can be multiplied by A inverse to get a unique vector back out.
g./h. No column or row can be a linear combination of the other rows or columns. If this were not true, then there would not be a unique solution. Consider the matrix {{1, 2}, {2, 4}} and how setting it equal to a matrix b will yield an infinite number of solutions or no solution.

Answer 4

hm still don't really get g/h but that could partially be my fault

Answer 5

If you think about it, if you tried to row reduce the matrix and the column/row vectors were not linearly independent, then it would not be the same rank as the dimension of the vector space.

Answer 6

The point is, for it to be an invertible, or one to one and onto transformation it must have full rank.

Answer 7

If its linearly dependent, one row could be expressed as a linear combination of other rows. It would be eliminated.

Answer 8

could you do an example? like lets say in R^3?

Answer 9

so is g and h saying that everything has to be 0 below the leading number?

Answer 10

Ah, sorry. Think of it this way. If you had the following equations
\[x+2y=5\]\[2x+4y=6\]
This set of equations would be inconsistent and would yield no results. This is the same as setting {{1, 2}, {2, 4}}x={5, 6}

The row vectors are not linearly independent.
Equivalently, if the vectors are not linearly independent, then you'll have det=0

Answer 11

can you help me finish this ? when you get through

Answer 12

If a matrix is invertible you can row reduce it to the identity matrix. If there are linear dependencies you can eliminate one of the rows in th e process.

Answer 13

By eliminating a row you can no longer reduce it to the identity matrix. It basically demonstrates that the matrix doesnt have full rank.

Answer 14

By that I mean for an nxn matrix, the rank of the matrix is less than n.

Answer 15

Ohhh I think I understand, thanks everyone for their help!