# proof of properties of trace of a matrix

Proof of Properties:

1. 1.

Let us check linearity. For sums we have

 $\displaystyle\operatorname{trace}(A+B)$ $\displaystyle=$ $\displaystyle\sum\limits_{i=1}^{n}(a_{i,i}+b_{i,i})\,\,\,\,\,\,\,\,\,\,\,\mbox% {(property of matrix addition)}$ $\displaystyle=$ $\displaystyle\sum\limits_{i=1}^{n}a_{i,i}+\sum\limits_{i=1}^{n}b_{i,i}\,\,\,\,% \mbox{(property of sums)}$ $\displaystyle=$ $\displaystyle\operatorname{trace}(A)+\operatorname{trace}(B).$

Similarly,

 $\displaystyle\operatorname{trace}(cA)$ $\displaystyle=$ $\displaystyle\sum\limits_{i=1}^{n}c\cdot a_{i,i}\,\,\,\,\,\mbox{(property of % matrix scalar multiplication)}$ $\displaystyle=$ $\displaystyle c\cdot\sum\limits_{i=1}^{n}a_{i,i}\,\,\,\,\,\mbox{(property of % sums)}$ $\displaystyle=$ $\displaystyle c\cdot\operatorname{trace}(A).$
2. 2.
3. 3.

The proof of the third property follows by exchanging the summation order. Suppose $A$ is a $n\times m$ matrix and $B$ is a $m\times n$ matrix. Then

 $\displaystyle\operatorname{trace}AB$ $\displaystyle=$ $\displaystyle\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}A_{i,j}B_{j,i}$ $\displaystyle=$ $\displaystyle\sum\limits_{j=1}^{m}\sum\limits_{i=1}^{n}B_{j,i}A_{i,j}\,\,\,\,% \mbox{(changing summation order)}$ $\displaystyle=$ $\displaystyle\operatorname{trace}BA.$
4. 4.

The last property is a consequence of Property 3 and the fact that matrix multiplication  is associative;

 $\displaystyle\operatorname{trace}(B^{-1}AB)$ $\displaystyle=$ $\displaystyle\operatorname{trace}\big{(}(B^{-1}A)B\big{)}$ $\displaystyle=$ $\displaystyle\operatorname{trace}\big{(}B(B^{-1}A)\big{)}$ $\displaystyle=$ $\displaystyle\operatorname{trace}\big{(}(BB^{-1})A\big{)}$ $\displaystyle=$ $\displaystyle\operatorname{trace}(A).$
Title proof of properties of trace of a matrix ProofOfPropertiesOfTraceOfAMatrix 2013-03-22 13:42:54 2013-03-22 13:42:54 Daume (40) Daume (40) 4 Daume (40) Proof msc 15A99