Exercise 12 from Hoffman and Kunze Linear Algebra

asmani · Mar 3, 2012

Hi all.

From Hoffman and Kunze's book Linear Algebra, page 27, exercise 12:

Printed exercise 12 showing an n by n matrix A with entries 1, 1/2, ..., 1/n in the first row, 1/2, 1/3, ..., 1/(n+1) in the second row, continuing down to 1/n, 1/(n+1), ..., 1/(2n-1) in the last row, followed by the question: prove A is invertible with A inverse having integer entries.

What's the most elementary way to solve this?

Any hint is appreciated.

Office_Shredder · Mar 3, 2012

Does example 16 calculate several small examples of these? The easiest way might be to use the examples to guess what the inverse is and then just prove through matrix multiplication that it works

AlephZero · Mar 3, 2012

Office_Shredder said:

Does example 16 calculate several small examples of these? The easiest way might be to use the examples to guess what the inverse is and then just prove through matrix multiplication that it works

If you can't guess, Google is your friend.

asmani · Mar 4, 2012

Thanks for the replies.

I've searched by Google (actually the book doesn't mention the name of the matrix), and all I found is some stronger results (like in Wikipedia: http://en.wikipedia.org/wiki/Hilbert_matrix) with not enough elementary proofs. I don't need to prove what the inverse matrix is, just need to prove it is invertible, and the inverse has integer entries.

Here is the example 16:

Gauss-Jordan row reduction of the 3x3 Hilbert matrix with rows (1, 1/2, 1/3), (1/2, 1/3, 1/4), (1/3, 1/4, 1/5), paired with the identity matrix, ending with inverse entries 9, -36, 30; -36, 192, -180; 30, -180, 180.

Exercise 12 from Hoffman and Kunze Linear Algebra

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