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Transpose of a matrix product

Taking the transpose of the product of two matrices. Created by Sal Khan.

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• Forgive my ignorance, but what is the practical application of transposing a matrix product? When will we actually use this type of math? Thanks, T.S.
• From the top of my head I can think of two applications of the transpose:
1) Within maths, but potentially useful due to extensions: We know we can take the transpose of a determinant without changing it. So by performing a transpose we might be able to transform a determinant into a more familiar form, making it easier to solve for any applications which require a determinant.

2) In quantum physics, we use matrices and other linear algebra objects and their compositions quite regularly. We also take their transpose fairly frequently (specifically, we take the Hermitian conjugate, which combines transpose and complex conjugate). So that is a notable application.
• At Sal says that Cij is equivalent to Dji, but both elements were apparently selected at random so there is no guarantee that the i in C is the same value as the i in D. Wouldn't one have to specify a connection between Cij and Dji when selecting Dji in order to assert that they are equal? How would such a restriction be stated mathmatically?
• Maybe Sal should have been clearer. While the i,j in Cij are arbitrary, the i,j in Dji are not. Instead, they are automatically fixed by the i, j in Cij. For example, if we chose (at random) i,j to be 2,3, then element C23 would be selected (at random), BUT the corresponding element in D would be fixed to be D32. No random selection allowed here.
The idea is this: Every element in C has one associated element in D equal to its value. Its (row, col) location turns out to be a swap of the (row_, col_) location of the element in C. Thus, if the element in C was Cij (row_ = i, col_ = j), then the "partner" element in D is Dji (row = j, col = i).
• At , Sal concludes that the transpose of the product of two matrices is simply the matrices themselves transposed and then multiplied in reverse order. Would this technique work for products of more than two matrices? For example, if we have (ABC)^T, would that equal (C^T) (B^T) (A^T)?
• Great question,

Let us use the fact that matrix multiplication is associative, that is (AB)C=A(BC). Then we can write (ABC)^T=((AB)C)^T. AB is just a matrix so we can use the rule we developed for the transpose of the product to two matrices to get ((AB)C)^T=(C^T)(AB)^T=(C^T)(B^T)(A^T).

That is the beauty of having properties like associative. It might be hard to believe at times but math really does try to make things easy when it can.
• Ok, is it necessary for A to be m*n and B to be n*m. Can't B be n*k? Will this still work?
• Well yes, it won't work. Because if your matrix A is an n x m matrix, and B is an m x k matrix, the product of AB will be defined, but the product of A^T and B^T will not be defined. So this works only for matrices of the same type.
(1 vote)
• There is a serious problem. If matrix A is a m x n matrix, B is a n x k matrix , the product will be mxk. If I take the transpose of the matrix, it will be the transpose of B times the transpose of A, which is not possible, a n x k matrix can not mutiply a m x n matrix! What's wrong?
(1 vote)
• If you take the transpose of B, you will get a k x n matrix, and if you take the transpose of A, you will get an n x m matrix. A k x n matrix can be multiplied by an n x m matrix.
• At he says "we've seen this before." I watched all five videos in the matrix multiplication section and I don't remember anything about finding a specific entry in a matrix product. What video was it in?
(1 vote)
• at what happens when there are 3 matrices? can we apply the same on them as well?