Let A and B be n times n matrices. Recall that the trace of A , written tr(A),equal sum_{i=1}^nA_{ii} Prove that tr(AB)=tr(BA) and tr(A)=tr(A^t)

Let A and B be $n×n$ matrices. Recall that the trace of A , written tr(A),equal
$\sum _{i=1}^{n}{A}_{ii}$
Prove that tr(AB)=tr(BA) and $tr\left(A\right)=tr\left({A}^{t}\right)$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

yagombyeR
Step 1
Consider A and B be $n×n$ matrices and trace of matrix is defined as,
$tr\left(A\right)=\sum _{i=1}^{n}{A}_{ii}$
step 2
To prove tr(AB)=tr(BA).
$tr\left(AB\right)=\sum _{i=1}^{n}\left(AB{\right)}_{ii}$
$=\sum _{i=1}^{n}\sum _{j=1}^{n}{a}_{ii}{b}_{jj}$
$=\sum _{j=1}^{n}\sum _{i=1}^{n}{b}_{jj}{a}_{ii}$
$=\sum _{i=1}^{n}\left(BA{\right)}_{ii}$
$=tr\left(BA\right)$
Hence, it is proved
Step 3
To prove $tr\left(A\right)=tr\left({A}^{t}\right)$
Let Then, ${b}_{ii}={a}_{ii}$
$tr\left(A\right)=\sum _{i=1}^{n}\left(A{\right)}_{ii}$
$=\sum _{i=1}^{n}{a}_{ii}$
$=\sum _{i=1}^{n}{b}_{ii}$
$=\sum _{i=1}^{n}\left({A}^{T}{\right)}_{ii}$
$=tr\left({A}^{T}\right)$
Hence, it is proved