Proving that a triangular matrix is invertible

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Homework Help Overview

The discussion revolves around proving the invertibility of a square triangular matrix, specifically focusing on the condition that the diagonal elements must be non-zero for the matrix to be invertible. Participants are exploring the relationship between the determinant of the matrix and its diagonal elements.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the determinant's dependence on the diagonal elements of a triangular matrix and the implications of having zero elements on the diagonal. There are attempts to connect the definition of the determinant with the properties of triangular matrices.

Discussion Status

The discussion is active, with various participants providing hints and insights into the proof. Some guidance has been offered regarding the determinant's formulation and its relationship to the diagonal elements, but there is no explicit consensus on the proof's structure yet.

Contextual Notes

Participants are navigating assumptions about linearity and the implications of having rows of zeros in the matrix. There is a focus on the definitions and properties of determinants in the context of triangular matrices.

Icheb
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I have a square triangular matrix with [tex]d_{ij} = 0[/tex] for all [tex]1 \le j < i \le n[/tex]. Now I have to prove that this matrix is only then invertible when [tex]d_{ii} \ne 0[/tex] for all [tex]1 \le i \le n[/tex].

From what I know a matrix is only then invertible when its determinant does not equal 0. I also think that the determinant of a triangular matrix is dependent on the product of the elements of the main diagonal and if that's true, I'd have the proof. However this is also where I'm stuck since I don't know how to prove that. Could someone help me there?
 
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Hint: Formulate the determinant using determinant expansion by minors on the first column.

You should get a very compact expression that only involves the diagonal elements.
 
Icheb said:
I also think that the determinant of a triangular matrix is dependent on the product of the elements of the main diagonal and if that's true, I'd have the proof. However this is also where I'm stuck since I don't know how to prove that. Could someone help me there?

It's a simple proof. Just use the definition of the determinant.
 
Ok, do I understand it correctly that if all elements of one line equal 0, the determinant equals 0? So I would just have to prove that the matrix is linear (Did I translate that word correctly?) and my proof would be complete? Or am I missing something?
 
You're missing something. It is straightfoward to show from the definition of determinant by expansion that the determinant of a triangular matrix is the product of the diagonal elements. A row of zeroes is neither here nor there.
 
But why wouldn't it suffice if I assume one element of the main diagonal to be 0, then I use Gaussian transformation to change the last line so it contains only 0s? Then if the last line contains only 0s the determinant would be 0 because of the linearity.
 
Yes, that works. However, you'd need a proof that you'll always get a row of zeros.

Besides, the proof that the determinant of a diagonal matrix is the product of the diagonal elements is similar enough.
 
Alright, thank you very much. :)
 

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