Proving Rank(A) = k Through Im(A) = L(X)

  • Thread starter AndreTheGiant
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In summary, we can prove that the rank of the projection matrix A, given X with linearly independent columns, is equal to the number of columns of X by showing that the dimension of its image is equal to k. We can also show that the image of A is equal to the linear span of X by rewriting A as a product of linear transformations involving X. This proves that rank(A) = k.
  • #1
AndreTheGiant
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Homework Statement


Hi.

Given X E R^(n x k) and X has linearly independent columns, and A = X(X'X)^-1 X', I want to prove rank(A) = k. (A is the projection matrix basically).






Homework Equations





The Attempt at a Solution



I was told to show im(A) = L(X).

If i do that it would make sense since the Linear span of X is the basis of X with l.i columns, and dimL(X) = rank(X).

How do i show images though. This has to be a product of a linear transformation in order for it to be an image doesn't it? I'm really unsure at that point.
 
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  • #2



Hello!

To prove that rank(A) = k, we need to show that the dimension of the image of A (im(A)) is equal to k. Since A is a projection matrix, it projects any vector onto the subspace spanned by the columns of X. This means that every vector in im(A) can be written as a linear combination of the columns of X. Since X has k linearly independent columns, the dimension of im(A) is also k. Thus, rank(A) = k.

To show that im(A) = L(X), we need to show that every vector in im(A) can be written as a linear combination of the columns of X, and that every linear combination of the columns of X is in im(A). This can be done by showing that A can be written as a product of a linear transformation and X, as you mentioned.

To do this, we can rewrite A as A = X(X'X)^-1 X'. This can be seen as the composition of two linear transformations: first, the transformation X'X maps any vector onto the subspace spanned by the columns of X, and then the transformation (X'X)^-1 maps that subspace back onto the original vector space. Finally, the transformation X' maps the vector back onto the subspace spanned by the columns of X, which is the image of A.

Thus, every vector in im(A) can be written as a linear combination of the columns of X, and every linear combination of the columns of X is in im(A), proving that im(A) = L(X). Therefore, rank(A) = k.
 

What does it mean to prove that Rank(A) = k?

Proving that Rank(A) = k means showing that the rank, or the maximum number of linearly independent rows or columns, of a matrix A is equal to the value k.

How is the rank of a matrix related to its image?

The rank of a matrix A is equal to the dimension of its image, or the set of all possible outputs, written as Im(A). This means that the number of linearly independent rows or columns in A is equal to the number of dimensions in the image of A.

What is the significance of proving Rank(A) = k?

Proving Rank(A) = k is significant because it provides information about the linear independence of the rows and columns of a matrix A. It also helps determine the dimension of the image of A, which is useful in solving systems of linear equations and understanding the properties of a matrix.

What are the steps to prove Rank(A) = k through Im(A) = L(X)?

The steps to prove Rank(A) = k through Im(A) = L(X) are:1. Write out the matrix A.2. Find the dimension of the image of A by finding the number of linearly independent rows or columns.3. Use this dimension to determine the rank of A.4. Show that the rank of A is equal to the value k.

Can Rank(A) = k be proven through other methods?

Yes, there are other methods to prove that Rank(A) = k, such as using the determinant or using the Gaussian elimination method. However, proving it through Im(A) = L(X) is a commonly used method because it directly relates the rank of A to its image.

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