How to prove the product of upper triangular matrices is upper triangular?

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To prove that the product of upper triangular matrices is upper triangular, one effective method involves expressing the entries of the product matrix as dot products of rows from the first matrix and columns from the second matrix. For upper triangular matrix A, the elements satisfy the condition that a_{ij} = 0 when i > j. When multiplying two upper triangular matrices A and B to get C, it is necessary to demonstrate that c_{ij} = 0 for i > j. This approach simplifies the proof by focusing on the structure of the matrices rather than listing all entries. Ultimately, this method provides a clearer and more efficient way to establish the desired result.
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This seems easy but when I tried to do this, the best way I came up with is to list all entries and then do the multiplication work. Is there any better ,clearer and more simple way to do the proof?
 
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Brucezhou said:
This seems easy but when I tried to do this, the best way I came up with is to list all entries and then do the multiplication work. Is there any better ,clearer and more simple way to do the proof?

Did you try to express the entries in the product matrix in terms of the dot products of the row of one matrix with the corresponding column in the second matrix?
 
Suppose A is an upper triangular matrix with elements ##a_{ij}##. Then you know that ##a_{ij}=0## if ##i>j##. If C=AB where B is also upper triangular, you want to show that ##c_{ij}=0## if ##i>j##.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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