Proving that a triangular matrix is invertible

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A square triangular matrix is invertible if and only if all diagonal elements are non-zero. This conclusion is based on the property that the determinant of a triangular matrix equals the product of its diagonal elements. The proof involves using the definition of the determinant and expansion by minors, demonstrating that if any diagonal element is zero, the determinant becomes zero, confirming non-invertibility. Gaussian transformations can also illustrate that a row of zeros leads to a determinant of zero.

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

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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