Matrix Multiplication and Algebraic Properties of Matrix Operations

hkus10
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1) If A = [aij] is an n x n matrix, the trace of A, Tr(A), is defined as the sum of all elements on the main diagonal of A, Tr(A) = the sum of (aii) from i=1 to n. Show each of the following:
a) Tr(cA) = cTr(A), where c is a real number
b) Tr(A+B) = Tr(A) + Tr(B)
c) Tr(A(Transpose)) = Tr(A)

2) If r and s are real numbers and A and B are matrices of the appropriate sizes, then proves the following:
A(rB) = r(AB) =(rA)B

I have been thinking these for a long time with no directions to approach. Please help!
 
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hkus10 said:
1) If A = [aij] is an n x n matrix, the trace of A, Tr(A), is defined as the sum of all elements on the main diagonal of A, Tr(A) = the sum of (aii) from i=1 to n. Show each of the following:
a) Tr(cA) = cTr(A), where c is a real number
b) Tr(A+B) = Tr(A) + Tr(B)
c) Tr(A(Transpose)) = Tr(A)

2) If r and s are real numbers and A and B are matrices of the appropriate sizes, then proves the following:
A(rB) = r(AB) =(rA)B

I have been thinking these for a long time with no directions to approach. Please help!

For 1, you show the definition of Tr(A). What is Tr(cA)? Tr(A + B)? Tr(AT)?
For 2, look at a typical element of the matrices A(rB), r(AB), and (rA)B.
 
You don't do math problems by sitting and staring at a piece of paper so if you have "been thinking about these for a long time", you must have tried something. Show us what you have tried.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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