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Linear algebra, determinants, and transposes

  • Thread starter hgj
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hgj
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Okay, I need to prove that det(A^t) = det(A). I can see that it's true because I know columns and rows are interchangable (meaning you can use columns or rows when taking determinants), but I don't know how to prove this fact. Any help would be very appreciated.
 

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hgj said:
Okay, I need to prove that det(A^t) = det(A). I can see that it's true because I know columns and rows are interchangable (meaning you can use columns or rows when taking determinants), but I don't know how to prove this fact. Any help would be very appreciated.
Well, a determinant of a matrix is the sum of the products of its diagonals minus the products of its antidiagonals. How do these products change under a transpose?
 
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[tex]\det{A}=\sum_{i=1}^{m}\left(-1\right)^{i+j}a_{ij}\det{A_{ij}}[/tex]

Do you see what happens when you try to prove det(AT)=det(A) for 2x2 or 3x3 matrices? Use the definition.

Edit: To the above poster: Doesn't that definition only work for 3x3 matrices?
 
Last edited:
Hurkyl
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JoAuSc: that's only for 3x3 matrices.

apmcavoy: I don't believe that formula helps, at least not in a straightforward manner.

hgj: what are you using as the definition of a determinant? And have you yet proven that the determinant is multiplicative?
 
hgj
15
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If A is an nxn matrix, then
detA = a11det(A11) - a21*det(A21) + ... + (-1)^(n+1)*an1*det(An1)
(sorry, I don't know how to make things subscripts on this, so the 11, 21,...,n1 are supposed to be subscripts)

That's the definition we're using for a determinant.
 
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hgj said:
If A is an nxn matrix, then
detA = a11det(A11) - a21*det(A21) + ... + (-1)^(n+1)*an1*det(An1)
(sorry, I don't know how to make things subscripts on this, so the 11, 21,...,n1 are supposed to be subscripts)

That's the definition we're using for a determinant.
Ok, that's the same thing I posted above. I suppose you could write it out like you did for both A and AT, and then rearrange and show they are equal.
 

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