# Thread: trace, jordan carnonical form

1. ## trace, jordan carnonical form

Can anyone help me showing that

a)Let A is a realnumber matrix nxn. Show that tr (A) =sum of all eigenvalues of A.
b)Let A is a realnumber matrix nxn. Show that det (exp (A)) = det (exp (J)) where J is
the Jordan canonical form of A.

2. ## Re: trace, jordan carnonical form

We know that $\mbox{Tr(A)}=\sum_{i=1}^{n}a_{ii}=a_{11}+a_{22}+.. .a_{nn}$

Now, have you any idea about the matrix of the linear map? (I would choose a base wherefore the matrix of the linear map is a diagonal matrix)

3. ## Re: trace, jordan carnonical form

I know TAT^-1 is D and D is the diagonal matrix whose diagonal entries are the eigenvalues of A. Hmm. I think I kind of getting it but I think I need a smoother answer please!

4. ## Re: trace, jordan carnonical form

Siron's answer is correct for diagonalizable matrices in $\mathbb{R}$ . In general:

Originally Posted by mathsohard
a)Let A is a realnumber matrix nxn. Show that tr (A) =sum of all eigenvalues of A.
Use that similar matrices have the same trace and $A$ is similar to its canonical form of Jordan (triangular matrix whose elements in the diagonal are exactly the eigenvalues of $A$ ) .

b)Let A is a realnumber matrix nxn. Show that det (exp (A)) = det (exp (J)) where J isthe Jordan canonical form of A.
Use that if $P^{-1}AP=J$ then, $e^A=Pe^JP^{-1}$ .

5. ## Re: trace, jordan carnonical form

Originally Posted by FernandoRevilla
Siron's answer is correct for diagonalizable matrices in $\mathbb{R}$ . In general:

Use that similar matrices have the same trace and $A$ is similar to its canonical form of Jordan (triangular matrix whose elements in the diagonal are exactly the eigenvalues of $A$ ) .

Use that if $P^{-1}AP=J$ then, $e^A=Pe^JP^{-1}$ .
Said well, support you ！

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