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Proving that a triangular matrix is invertible

  • Thread starter Icheb
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  • #1
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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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  • #2
D H
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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.
 
  • #3
radou
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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.
 
  • #4
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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?
 
  • #5
matt grime
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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.
 
  • #6
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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.
 
  • #7
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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.
 
  • #8
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Alright, thank you very much. :)
 

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