By Eckhard M.S. Hitzer, University of Fukui, Prof. Ryosuke Nagaoka (University of the Air, Japan)
Yes, please have a look at the following pattern. So here we have the characteristic polynomial and we insert the matrix A. And the Hamilton-Cayley theorem says, that the result should be zero.
Now A is a square matrix, n by n square matrix, over the field of the complex numbers C. In this case we will always have n eigenvalues alpha one to alpha n, which are all element(s) of the field C. The characteristic polynomial for this matrix A is then given as the product of x minus the first eigenvalue, times x minus the second eigenvalue, and so on, until (we,) the final factor x minus the eigenvalue alpha n. Here now we replace each occurance of x by the matrix A itself here. Therefore we now have the characteristic polynomial of the matrix A, but we don't take if of x, (we take it of,) we replace x in this polynomial by the matrix A itself. Then we have this matrix product.
We know that if we have a square matrix, n dimensional square matrix over the field of complex numbers, we can always with a matrix P and its inverse transform the matrix into upper triangular form and on the main diagonal we have the list of eigenvalues: alpha one, alpha two until alpha n.
Now we can take the characteristic polynomial of the matrix A and also do this transformation: Apply the inverse of P from the left and P itself from the right. Then if we write it explicitly it will become this product of matrices. And now we can insert instead of the unit matrix here simply P times its inverse, P to the power of minus one, between each two factors. So a general factor will then be P to the power of minus one, times A minus alpha n E P. Now when we have the product P inverse times alpha k unit matrix E times P, this will be qual: P inverse times alpha k times P, and this is again the unit matrix alpha k E. So in the next step we can just simplify this whole product to become: P inverse times A times P minus alpha one E, P inverse A P minus alpha two E, etcetera.
Now we know, when we take for example the first matrix factor P inverse A P minus alpha one unit matrix E, that P inverse A P has in its main diagonal in the first place alpha one. And here we are exactly subtracting alpha one times unit matrix, so here we will have a zero.
The same occurs for the next factor in the product, just that we subtract alpha two times the unit matrix E. And this will produce a zero in the second position on the main diagonal.
And this continues until we reach the last position n, because we have minus alpha n times E. Now this will have the consequence, that we will have one zero here in the first position, one zero here in the second position, etcetera. If you continue to do this calculation explicitly, the result will be the zero matrix. This calculation actually is an exercise in the text book, which I strongly recommend you to perform.
So we have proved, that the inverse of P times the characteristic polynomial of the matrix A times P is zero. And it follows therefore, that also the characteristic polynomial itself of A must be zero. This is the zero matrix. And this here is already the proof of the Cayley-Hamilton theorem.
Soli Deo Gloria. Created by
Eckhard Hitzer (Fukui).
Last Modified 16 January 2004
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