proof of properties of trace of a matrix

Proof of Properties:

1. Let us check linearity. For sums we have

   $\mathrm{trace}(A+B)$ $=$ ${\displaystyle \sum _{i=1}^n}(a_{i,i}+b_{i,i})\mathit{\hspace{1em}\hspace{0.5em}\hspace{1em}}\text{(property of matrix addition)}$

   $=$ ${\displaystyle \sum _{i=1}^n}{a}_{i,i}+{\displaystyle \sum _{i=1}^n}{b}_{i,i}\text{(property of sums)}$

   $=$ $\mathrm{trace}(A)+\mathrm{trace}(B).$

   Similarly,

   $\mathrm{trace}(cA)$ $=$ ${\displaystyle \sum _{i=1}^n}c\cdot a_{i,i}\text{(property of matrix scalar multiplication)}$

   $=$ $c\cdot {\displaystyle \sum _{i=1}^n}{a}_{i,i}\text{(property of sums)}$

   $=$ $c\cdot \mathrm{trace}(A).$

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

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

   $\mathrm{trace}AB$ $=$ ${\displaystyle \sum _{i=1}^n}{\displaystyle \sum _{j=1}^m}{A}_{i,j}{B}_{j,i}$

   $=$ ${\displaystyle \sum _{j=1}^m}{\displaystyle \sum _{i=1}^n}{B}_{j,i}{A}_{i,j}\text{(changing summation order)}$

   $=$ $\mathrm{trace}BA.$

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

   $\mathrm{trace}(B^{-1}AB)$ $=$ $\mathrm{trace}\left((B^{-1}A)B\right)$

   $=$ $\mathrm{trace}\left(B(B^{-1}A)\right)$

   $=$ $\mathrm{trace}\left((B{B}^{-1})A\right)$

   $=$ $\mathrm{trace}(A).$

Title	proof of properties of trace of a matrix
Canonical name	ProofOfPropertiesOfTraceOfAMatrix
Date of creation	2013-03-22 13:42:54
Last modified on	2013-03-22 13:42:54
Owner	Daume (40)
Last modified by	Daume (40)
Numerical id	4
Author	Daume (40)
Entry type	Proof
Classification	msc 15A99