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Does a lower triangular matrix have a determinant that is equal to the product of the elements in the diagonal similar to an upper triangular matrix.

Rodrigo de Azevedo
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J. Doe
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    Of course. A (square) matrix and its transpose have the same determinant. – hardmath Apr 20 '15 at 19:07
  • @hardmath elegant way to see that.Thanx – manifold Mar 18 '18 at 08:23
  • Related – Rodrigo de Azevedo May 24 '23 at 06:04
  • @hardmath is your comment referring to the fact that the only values that remain constant between a square matrix and its transpose are the diagonal elements, and therefore, the determinant remains unchanged? – CausalQuestions Nov 05 '24 at 22:51
  • @CausalQuestions: That is a fact, but I was making a different point. Any square matrix, not necessarily triangular, has the same determinant as its transpose. So if one knows the fact for upper triangular matrices, the fact for lower triangular matrices follows. – hardmath Nov 06 '24 at 02:06
  • @hardmath: Thank you for the clarification. I understand that cofactor expansion can be used to demonstrate this fact. Since the columns of matrix A are the rows of its transpose A^T, the cofactor expansion along row i of A is equivalent to the cofactor expansion along column i of A^T.

    Do you know of any accessible proof or resource that explicitly illustrates this relationship and does not involve such convoluted technical approaches?

     – CausalQuestions Nov 06 '24 at 21:05
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    @CausalQuestions: Have a look through these previous Math.SE posts on that topic. If I was trying to present it in a sophomore linear algebra class, I'd use the definition of determinant that sets a sign on a product of entries according to the parity of the permutation and argue the partity doesn't change under matrix transposition. – hardmath Nov 06 '24 at 22:02

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The matrix looks like this: $$\begin{bmatrix} a_{1,1}&0&\cdots&\cdots&0 \\b_{1,2}&a_{2,2}&\ddots&&\vdots \\ \vdots&\ddots&\ddots&\ddots&\vdots \\ \vdots&&\ddots&\ddots&0 \\ b_{1,n}&\cdots&\cdots&b_{n-1,n}&a_{n,n} \end{bmatrix}$$ The determinant can be written as the sum of the product of the elements in the top row with their associated minors: so this determinant would be: $$a_{1,1}*\begin{bmatrix} a_{2,2}&0&\cdots&0 \\b_{2,3}&a_{3,3}&\ddots&\vdots \\ \vdots&\ddots&\ddots&0 \\ b_{2,n}&\cdots&b_{n-1,n}&a_{n,n} \end{bmatrix}$$ This action eliminates the first column, $b_{1,x}$ and we are left with a similar matrix to find the determinant of. Reducing this one likewise we are left with $a_{1,1}*a_{2,2}*$(a matrix of remaining rows and columns) and so on until we have the product $a_{1,1}*a_{2,2}*\cdots*a_{n,n}$. The product of the diagonal.

Math Man
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