Proving Rank A=n Implies Basis of Rn for Matrix A

In summary, A is an m x n matrix with columns C1, C2, ..., Cn. If the rank of A is n, we need to show that {A^TC1, A^TC2, ..., A^TCn} is a basis of Rn. We know that the rank of A^T is also n, and the columns of A are the rows of A^T. We need to show that the columns of A^T multiplied by themselves maintain the rank of n. This means that the columns of A^T are linearly independent and form a basis for Rn. To determine the matrix (-x + ces) with the ATCi as its columns, we need to find a property
  • #1
stunner5000pt
1,465
4
Let A be an m x n matrix with columns C1, C2, ... Cn. If rank A = n show taht [tex] \{ A^{T}C_{1}, A^{T}C_{2}, ... , A^{T}C_{n} /} [/tex] is a basis of Rn.

ok [tex] \mbox{rank} A^{T} = n [/tex]
the columns of A are rows of A transpose
im not sure how to proceed though...
a column times itself with [tex] C_{1}^2 + C_{2} C_{1} + ... + C_{n}C_{1} [/tex] for the first term of [tex] A^{T} C_{1} [/tex] is the rank maintained through this multiplication? What justifies that?

help is greatly appreciated!

thank you!
 
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  • #2
What (product of) matrix (-x + ces) has the ATCi as its columns? And what property does an n x n matrix have to have for its rows to form a basis of R^n?
 
Last edited:
  • #3
what do u mean (-x + ces) ?
arent the rows of a square matrix A linearly independant if they form a basis for Rn?
 
  • #4
Sorry, it was supposed to be like "giraffe(s)," but that's not as easy when the word ends in an x. Anyway, right, they form a basis if the matrix containing them as columns has rank n.
 
  • #5
StatusX said:
Sorry, it was supposed to be like "giraffe(s)," but that's not as easy when the word ends in an x. Anyway, right, they form a basis if the matrix containing them as columns has rank n.
ok...
im still not sure to do with the -x + Ci part.
 

FAQ: Proving Rank A=n Implies Basis of Rn for Matrix A

What does "Proving Rank A=n Implies Basis of Rn for Matrix A" mean?

This statement means that if the rank of a matrix A is equal to its dimension n, then the columns of A form a basis for the vector space Rn. In other words, the columns of A are linearly independent and span the entire vector space.

How can I prove that the rank of a matrix A is equal to its dimension n?

To prove this, you can use the rank-nullity theorem, which states that the rank of a matrix A plus the nullity (dimension of the null space) is equal to the number of columns in A. If the rank is equal to the number of columns, then the nullity must be zero, meaning that the columns are linearly independent and form a basis for Rn.

What is the significance of the rank of a matrix?

The rank of a matrix is a measure of its linear independence and dimensionality. It tells us how many of the columns in the matrix are linearly independent, and therefore how many dimensions are needed to span the vector space. In other words, it tells us the maximum number of linearly independent columns in the matrix.

Can the rank of a matrix A be greater than its dimension n?

No, the rank of a matrix A can never be greater than its dimension n. This is because the rank is defined as the maximum number of linearly independent columns in the matrix, and the dimension represents the number of dimensions needed to span the vector space. If the rank were greater than the dimension, it would mean that there are more linearly independent columns than dimensions needed, which is not possible.

How does proving the rank of a matrix A=n imply the existence of a basis for Rn?

Proving that the rank of a matrix A is equal to its dimension n means that the columns of A are linearly independent and span the vector space Rn. This is because the rank represents the maximum number of linearly independent columns in the matrix, and the dimension represents the number of dimensions needed to span the vector space. Therefore, the columns of A form a basis for Rn, as they are both linearly independent and span the entire vector space.

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