# 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.

The second property follows since the transpose does not alter the entries on the main diagonal.

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