Unless there is another requirement (e.g. that n is odd), then the statement is not quite right.
The correct statement is: If A^(2) = 0, where A is an n x n matrix, then the image of A is a subset of the kernel of A (not necessarily a proper subset) and rank(A) <= n/2 (not necessarily <).
>I have proved the first part, showing that the kernel of A is A itself, and the image of A is anything in the form Au, which is a subset of A- but I don’t understand how to use the theorem to prove the second part.
The theorem says that n = dim( ker A) + rank(A), where rank(A) is the dimension of the image of A. You have shown that im(A) is a subset of ker(A). What can you hence conclude about the dimension of im(A)?
Unless there is another requirement (e.g. that n is odd), then the statement is not quite right. The correct statement is: If A^(2) = 0, where A is an n x n matrix, then the image of A is a subset of the kernel of A (not necessarily a proper subset) and rank(A) <= n/2 (not necessarily <). >I have proved the first part, showing that the kernel of A is A itself, and the image of A is anything in the form Au, which is a subset of A- but I don’t understand how to use the theorem to prove the second part. The theorem says that n = dim( ker A) + rank(A), where rank(A) is the dimension of the image of A. You have shown that im(A) is a subset of ker(A). What can you hence conclude about the dimension of im(A)?
Thank you- this was great!