How to prove that
- the sum of the eigenvalues of a matrix equals to the trace of the matrix?
- the product of the eigenvalues equals to the determinant?
(Of course the matrix is a square matrix)
I would be very grateful if you could help me.
How to prove that
- the sum of the eigenvalues of a matrix equals to the trace of the matrix?
- the product of the eigenvalues equals to the determinant?
(Of course the matrix is a square matrix)
I would be very grateful if you could help me.
Consider then, is similar to its canonical form of Jordan .
Now use that similar matrices have the same trace and the same determinant.
Regards.
Fernando Revilla
Or you could notice that if the the characteristic polynomial is given by
Thus, recalling Vieta's formulas and noticing that the characteristic polynomial is monic we can see the the constant term of the above is . That said, we know that for any polynomial the constant term is merely . Thus, with equal validity we may conclude that the constant term of is
comparing these two gives as you required. A similar technique works for the trace.
Remark: Vieta's formula (for the above, not the trace proof) is a bit of an overkill. It's easy to see that what I claimed was the constant term is, in fact, the constant term.