# Thread: Matrix Help

1. ## 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!

2. Hello,

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

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

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

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

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

So, are they the same ?

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

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

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

therefore...

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

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

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

For the last one use fact that try to express $AA^T$ in terms of squares this will immediately imply the quantity is non-negative.

4. Thanks you guys! I was on the right track for most of them...just needed a second opinion!