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it agrees with the (i, j) entry of the right side. First we make things clear by specifying that A has n columns and B has n rows (these dimensions have to agree, or the product is not defined). This gives us [AB]=b.j. (remember that we decided to use abbreviated notation, so we leave off the summation limits) so the (i, j) entry of the left-hand side is: [(AB)¹] =[AB]₁₁ Σassbus. At this point we can introduce A and B transpose because the j, k entry of any matrix is the k, j entry of its transpose: Σajabki = [4¹] [B]L.k. Σ{"]ky Since the terms of A and B are being expressed as a summation, they commute (i.e. order doesn't matter), which allows us to say (using our definition of matrix product): Ș[A¹] [B²].x = Σ [B"], [4"]µs = [B³A"],3 + Voilà, we have the (i, j) entry of the right-hand side, and the proof is complete. Exercise 11.5.2. Give a formula for (ABC)T, and prove your formula using summation notation.

it agrees with the (i, j) entry of the right side. First we make things clear by specifying that A has n columns and B has n rows (these dimensions have to agree, or the product is not defined). This gives us [AB]=b.j. (remember that we decided to use abbreviated notation, so we leave off the summation limits) so the (i, j) entry of the left-hand side is: [(AB)¹] =[AB]₁₁ Σassbus. At this point we can introduce A and B transpose because the j, k entry of any matrix is the k, j entry of its transpose: Σajabki = [4¹] [B]L.k. Σ{"]ky Since the terms of A and B are being expressed as a summation, they commute (i.e. order doesn't matter), which allows us to say (using our definition of matrix product): Ș[A¹] [B²].x = Σ [B"], [4"]µs = [B³A"],3 + Voilà, we have the (i, j) entry of the right-hand side, and the proof is complete. Exercise 11.5.2. Give a formula for (ABC)T, and prove your formula using summation notation.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please do Exercise 11.7.2 and Exercise 11.5.2 and please show step by step and explain

Proposition 11.5.1. If A and B are matrices such that the matrix product is defined, then (AB)™ = BT AT. PROOF. We'll prove this by expressing the (i,j) entry of the left-hand side in summation notation, doing some algebraic hocus-pocus, and showing that it agrees with the (i, j) entry of the right side. First we make things clear by specifying that A has n columns and B has n rows (these dimensions have to agree, or the product is not defined). This gives us [AB), = Σ with j (remember that we decided to use abbreviated notation, so we leave off the summation limits) so the (i, j) entry of the left-hand side is: [(AB)¹] =[AB] =Σajbi - At this point we can introduce A and B transpose because the j, k entry of any matrix is the k, j entry of its transpose: Σajbi-[A¹] [B¹] kk. Since the terms of A and B are being expressed as a summation, they commute (i.e. order doesn't matter), which allows us to say (using our definition of matrix product): Σ[A¹] kj [B¹k = Σ [B¹], [A¹]µ.5 = [B¹A¹]. Voilà, we have the (i.j) entry of the right-hand side, and the proof is complete. Exercise 11.5.2. Give a formula for (ABC)T, and prove your formula using summation notation. 0
Transcribed Image Text:Proposition 11.5.1. If A and B are matrices such that the matrix product is defined, then (AB)™ = BT AT. PROOF. We'll prove this by expressing the (i,j) entry of the left-hand side in summation notation, doing some algebraic hocus-pocus, and showing that it agrees with the (i, j) entry of the right side. First we make things clear by specifying that A has n columns and B has n rows (these dimensions have to agree, or the product is not defined). This gives us [AB), = Σ with j (remember that we decided to use abbreviated notation, so we leave off the summation limits) so the (i, j) entry of the left-hand side is: [(AB)¹] =[AB] =Σajbi - At this point we can introduce A and B transpose because the j, k entry of any matrix is the k, j entry of its transpose: Σajbi-[A¹] [B¹] kk. Since the terms of A and B are being expressed as a summation, they commute (i.e. order doesn't matter), which allows us to say (using our definition of matrix product): Σ[A¹] kj [B¹k = Σ [B¹], [A¹]µ.5 = [B¹A¹]. Voilà, we have the (i.j) entry of the right-hand side, and the proof is complete. Exercise 11.5.2. Give a formula for (ABC)T, and prove your formula using summation notation. 0
In the above exercise, we have considered the trace of the sum of two matrices. Now we consider the trace of the product of two matrices. To this end, let A and B be a n x n matrices. So first we have: - Σ(AB) = Σ(Σarabas) - Σ arbus ik All we've done here is take the matrix product formula, and set the second index of the second matrix entry equal to first index of the first matrix entry. Now to make things interesting, let's find the trace for the reverse order: Tr (BA) -- ΣΙΒΑ - Σ (Σκάκι) Let's play with this last equation a bit. As we mentioned before, we can change the sum over i, k to a sum over i, k without changing anything. Furthermore, since bi, and ai are numbers, they commute under multipli- cation: itaki-akibik- ik Tr (BA) = bak, akib.x. = Finally, we rename the indices by changing k to i and i to k. (Remember, it's the positions of the indices that are important, not the letters we call them by!) After renaming, we get: Tr (BA) = Σ @labwe which agrees with our original expression for Tr(AB). Exercise 11.7.2. In the above proof that Tr(AB) = Tr(BA), we assumed that both A and B were square matrices. Show that the formula is still true when A is a mx n matrix and B is a n x m matrix. (Notice that AB and BA are both square matrices, so that Tr (AB) and Tr (BA) are both well-defined.) 0
Transcribed Image Text:In the above exercise, we have considered the trace of the sum of two matrices. Now we consider the trace of the product of two matrices. To this end, let A and B be a n x n matrices. So first we have: - Σ(AB) = Σ(Σarabas) - Σ arbus ik All we've done here is take the matrix product formula, and set the second index of the second matrix entry equal to first index of the first matrix entry. Now to make things interesting, let's find the trace for the reverse order: Tr (BA) -- ΣΙΒΑ - Σ (Σκάκι) Let's play with this last equation a bit. As we mentioned before, we can change the sum over i, k to a sum over i, k without changing anything. Furthermore, since bi, and ai are numbers, they commute under multipli- cation: itaki-akibik- ik Tr (BA) = bak, akib.x. = Finally, we rename the indices by changing k to i and i to k. (Remember, it's the positions of the indices that are important, not the letters we call them by!) After renaming, we get: Tr (BA) = Σ @labwe which agrees with our original expression for Tr(AB). Exercise 11.7.2. In the above proof that Tr(AB) = Tr(BA), we assumed that both A and B were square matrices. Show that the formula is still true when A is a mx n matrix and B is a n x m matrix. (Notice that AB and BA are both square matrices, so that Tr (AB) and Tr (BA) are both well-defined.) 0
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