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By using induction, we can set our determinant to be n=2, then have have

[A]= \begin{vmatrix} a & b \\ a & b \end{vmatrix} = ab - ab=0

So we assume it works. Now we can work out for determinants of order n. A can be determinant of order n+1, with two equal rows, let us say i-th and j-th. If we expand the determinant by some k-th row, where k≠i, k≠j, therefore we can have a sum by having two part of a sum for n+1, where we have part for n and another for +1, and as we assumed it works for n, ergo the determinant will be 0.

But now what is my question, that I cannot fully prove the upper theorem by using the permutations.

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    The proof depends on your definition of determinant, so you need to say what that is. – 2017-02-22
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    I had this as an example. Arbitrary should be fine? – 2017-02-22
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    You can define the determinant a fair number of ways. If you define it as the product of the eigenvalues then you'll want to prove it differently than if you define it via minors/cofactors. – 2017-02-22
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    And if you define it as an alternating multilinear map on vectors that does the right thing to the standard basis, this property is part of the definition. – 2017-02-22
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    How would we prove it if we define it via minors/cofactors. I do not need anything too complicated. – 2017-02-22
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    Well, an easy way out is proving that if $A$ is transformed into $B$ by swapping two rows then $\det(A)=-\det(B)$. In the event that two rows are equal, by swapping two rows that are the same you get that $A=B$ and so $\det(A)=-\det(A)$, the only possible number that satisfies this is zero. – 2017-02-22
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    If you use the minor expansion along a column, it is easy to see that the determinant is $0$ if two rows are equal. – 2017-02-22
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    So all you want to say is to show when we have sign(p(p^-1))=sign(p)*sign(p^-1). By showing this, we assume A=B, therefore we will show for every odd permutation, we change the parity. And so the two will equal to 0. – 2017-02-22
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    The permutation formula for the determinant should be an easy way to prove this – 2018-01-02

3 Answers 3

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If the rows of a matrix are not linearly independent (can be expressed as linear combination of the other rows of the matrix) then the determinant is 0.

One interpretation of the determinant is how it dilates (or compresses) space after a transformation. If two rows are equal, then the principal components of space are being mapped onto a single line. 2D space is compressing onto 1D space, and the area of a line equals 0.

Or in a $3\times 3$ matrix 3D spaces is compressing onto a 2D plane and the volume is $0.$

Another attack, if you have two matrices, the determinant of the product equals the product of the determinants $det(AB) = det(A)det(B)$

If B is your target matrix, and A is a matrix that subtracts one row from another.

$A = \begin{bmatrix} 1&0&0\\0&1&-1\\0&0&1\end{bmatrix}$ would be one such matrix that subtracts the 3rd row of B from the second. If they are the same that would create a row of zero vectors. That would mean that $det(AB) = 0, det (A)$ is non-zero. So $det(B)$ must be $0.$

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    It is helpful. Thank you. – 2017-02-22
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Let [A]= \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}

if you have two (or more) rows which are linearly related, you can resolve some rows to zero. (Check out gaussian elimination to see how row reduction works.)

The determinant of a 3X3 matrix is given by $$ a_{11} \times a_{22} \times a_{33} \\ + a_{12} \times a_{23} \times a_{31} \\ + a_{21} \times a_{32} \times a_{13} \\ - a_{13} \times a_{22} \times a_{31} \\ - a_{12}\times a_{21} \times a_{33} \\ - a_{11} \times a_{23} \times a_{32} $$

Now, say row two and row three are the same value, we can subtract the element values of row two from row three, replacing row three instead with zeros. Look at the definition of the determinant of a 3x3 matrix again, replacing the $a_{3x}$ values with zero. Notice how you essentially multiply every row by a zero.

I hope this helps.

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    Good reference to gaussian elimination: http://mathworld.wolfram.com/GaussianElimination.html Whilst it does not include a not on determinants as discussed in my answer, it shows how to resolve rows to zero. If a column were to be filled with zeros, you would have a determinant equal to zero too. – 2017-02-22
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A simple explanation is by thinking of the determinant expansion as a sum of products. Say a row is repeated, call its members A1, A2, .... An.

In the determinant expansion, we can match the n! products in pairs:

I) the term (terms from first rows) x Ai x (terms from the rows in between) x Aj x (terms from remaining rows)

and,

II) the term (terms from first rows) x Aj x (terms from the rows in between) x Ai x (terms from remaining rows)

Observe that everything from I and II is matched, except "i" and "j" are switched. It takes 2*|i-j|-1 adjacent switches to interchange "i" and "j".

That's an odd number SO the corresponding permutations are of opposite parity (if one is odd the other is even and vice versa); so the sign in the expansion for I and II is opposite, but magnitude is the same.

THEN, those two terms add to ZERO. Ergo everything, having been split into pairs that add to zero, adds to ZERO. q.e.d.

(the difficulty for physics types, like me, is seeing the "source" of the property that the determinant switches sign when you swap columns/rows. why to swap two elements in a row you have to do an odd number of exchanges/neighbor swaps? You have to move BOTH of them over all the intervening elements - giving you an even number of neighbor swaps, and then ONE time over each other, making the total number of neighbor swaps odd.)