# trace, jordan carnonical form

• Dec 12th 2011, 08:54 AM
mathsohard
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.
• Dec 12th 2011, 09:36 AM
Siron
Re: trace, jordan carnonical form
We know that $\displaystyle \mbox{Tr(A)}=\sum_{i=1}^{n}a_{ii}=a_{11}+a_{22}+.. .a_{nn}$

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)
• Dec 12th 2011, 09:47 AM
mathsohard
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!
• Dec 12th 2011, 01:59 PM
FernandoRevilla
Re: trace, jordan carnonical form
Siron's answer is correct for diagonalizable matrices in $\displaystyle \mathbb{R}$ . In general:

Quote:

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 $\displaystyle A$ is similar to its canonical form of Jordan (triangular matrix whose elements in the diagonal are exactly the eigenvalues of $\displaystyle A$ ) .

Quote:

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 $\displaystyle P^{-1}AP=J$ then, $\displaystyle e^A=Pe^JP^{-1}$ .
• Dec 13th 2011, 09:02 PM
mdiendd123456
Re: trace, jordan carnonical form
Quote:

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

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

Use that if $\displaystyle P^{-1}AP=J$ then, $\displaystyle e^A=Pe^JP^{-1}$ .

Said well, support you ！

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