proof of properties of trace of a matrix
Proof of Properties:
-
1.
Let us check linearity. For sums we have
trace(A+B) = n∑i=1(ai,i+bi,i) (property of matrix addition) = n∑i=1ai,i+n∑i=1bi,i(property of sums) = trace(A)+trace(B). Similarly,
trace(cA) = n∑i=1c⋅ai,i(property of matrix scalar multiplication) = c⋅n∑i=1ai,i(property of sums) = c⋅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×m matrix and B is a m×n matrix. Then
traceAB = n∑i=1m∑j=1Ai,jBj,i = m∑j=1n∑i=1Bj,iAi,j(changing summation order) = traceBA. -
4.
The last property is a consequence of Property 3 and the fact that matrix multiplication
is associative;
trace(B-1AB) = trace((B-1A)B) = trace(B(B-1A)) = trace((BB-1)A) = 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 |