Matrix Help

• Sep 7th 2008, 07:53 PM
Brokescholar
Matrix Help
If A = [aij] is an n x n matrix then the trace of A, Tr(A) is defined as the sum of all elements on the main diagonal A, Tr(A) = n SUM i = 1 aij. Prove

Tr(AB) = Tr(BA)

Tr(AT) = Tr(A)

Tr(ATA) > 0

I've got some ideas down about this but I'm just not totally sure how to prove these. Thanks!
• Sep 8th 2008, 10:45 AM
Moo
Hello,

1/ Let $\displaystyle A=\left[a_{ij}\right]$ and $\displaystyle B=\left[b_{ij}\right]$, with $\displaystyle 1 \le i,j \le n$

The trace is $\displaystyle \sum_{i=1}^n a_{ii}$, not $\displaystyle a_{ij}$ !

The matrix product $\displaystyle C=AB$ is defined as being $\displaystyle C=\left[\sum_{k=1}^n a_{ik} b_{kj}\right]=\left[c_{ij}\right]$

The same way, $\displaystyle D=BA=\left[\sum_{k=1}^n b_{ik} a_{kj}\right]=\left[d_{ij}\right]$

$\displaystyle \left. \begin{array}{lll}\text{Tr}(AB)=\sum_{i=1}^n c_{ii}=\sum_{i=1}^n \sum_{k=1}^n a_{ik} b_{ki} \\ \\ \text{Tr}(BA)=\sum_{i=1}^n d_{ii}=\sum_{i=1}^n \sum_{k=1}^n b_{ik} a_{ki} \end{array} \right\} \text{Just by substituting j by i}$

So, are they the same ?

2/ $\displaystyle A^T=\left[a_{ji}\right]=\left[a'_{ij}\right]$ if $\displaystyle A=\left[a_{ij}\right]$
Does that change the trace ?

3/ $\displaystyle A^TA=e_{ij}$ where $\displaystyle e_{ij}=\sum_{k=1}^n a_{ik} a'_{kj}=\sum_{k=1}^n a_{ik} a_{jk}$

$\displaystyle \text{Tr}(A^TA)=\sum_{i=1}^n e_{ii}=\sum_{i=1}^n \sum_{k=1}^n a_{ik} a_{ik}$

therefore... :p
• Sep 8th 2008, 10:45 AM
ThePerfectHacker
Quote:

Originally Posted by Brokescholar
If A = [aij] is an n x n matrix then the trace of A, Tr(A) is defined as the sum of all elements on the main diagonal A, Tr(A) = n SUM i = 1 aij. Prove

Tr(AB) = Tr(BA)

Tr(AT) = Tr(A)

Tr(ATA) > 0

I've got some ideas down about this but I'm just not totally sure how to prove these. Thanks!

Let $\displaystyle A = (a_{ij})$ and $\displaystyle B = (b_{ij})$ then $\displaystyle AB= (c_{ij})$ where $\displaystyle c_{ij}=(\Sigma_{k=1}^n a_{ik}b_{kj})$. This means $\displaystyle \text{Tr}(AB) = c_{11}+...+c_{nn} = \Sigma_{k=1}^n a_{1k}b_{k1}+ ... + \Sigma_{k=1}^n a_{nk}b_{kn} = \Sigma_{s=1}^n \Sigma_{r=1}^n a_{rs}b_{sr}$.

Now do a similar computation with $\displaystyle \text{Tr}(BA)$.

To show that $\displaystyle \mbox{Tr}(A^T) = \mbox{Tr}(A)$ just use the definition and if $\displaystyle A = (a_{ij}) \implies A^T = (a_{ji})$.

For the last one use fact that try to express $\displaystyle AA^T$ in terms of squares this will immediately imply the quantity is non-negative.
• Sep 8th 2008, 02:35 PM
Brokescholar
Thanks you guys! I was on the right track for most of them...just needed a second opinion!