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The determinant of a square matrix is the same as the determinant of its transpose. The transpose of a matrix can be obtained by reflecting the matrix across its main diagonal.

I was wondering how the determinant might change if we transform the matrix in other ways.

Setup

Suppose we have an $NxN$ matrix

$$ a_{11}\ a_{12}\ ...\ a_{1N}\\ a_{21}\ a_{22}\ ...\ a_{2N}\\ .\\ .\\ .\\ a_{N1}\ a_{N2}\ ...\ a_{NN} $$

whose determinant is $D$.

Rotation

If we rotate $M$ $90$ degrees clockwise to get $M_{rot}$

$$ a_{N1}\ a_{N-1,1}\ ...\ a_{21}\ a_{11}\\ a_{N2}\ a_{N-1,2}\ ...\ a_{22}\ a_{12}\\ .\\ .\\ .\\ a_{NN}\ a_{N-1,N}\ ...\ a_{2N}\ a_{1N} $$

how are the determinant of $M$ and $M_{rot}$ related?

Flip/reflection across horizontal axis

If we flip $M$ across a horizontal axis to get $M_{horizflip}$

$$ a_{N1}\ a_{N2}\ ...\ a_{NN}\\ .\\ .\\ .\\ a_{21}\ a_{22}\ ...\ a_{2N}\\ a_{11}\ a_{12}\ ...\ a_{1N} $$

how are the determinant of $M$ and $M_{horizflip}$ related?

Flip/reflection across vertical axis

If we flip $M$ across a vertical axis to get $M_{vertflip}$

$$ a_{1N}\ a_{1,N-1}\ ...\ a_{12}\ a_{11}\\ a_{2N}\ a_{2,N-1}\ ...\ a_{22}\ a_{21}\\ .\\ .\\ .\\ a_{NN}\ a_{N,N-1}\ ...\ a_{N2}\ a_{N1} $$

how are the determinant of $M$ and $M_{vertflip}$ related?

joseville
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    A vertical flip of the columns of a matrix $A$ is equivalent to a post-multiplication $AJ$ with matrix $J$ whose entries are equal to $1$ along the secondary diagonal, and $0$ elsewhere ($J_{i,j}=\delta(i+j-(n+1)$). – Jean Marie Sep 02 '22 at 21:53

2 Answers2

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If you flip across vertical or horizontal axis, the only thing you are doing is transposing rows or columns several times. Transposing two rows or two columns will multiply the determinant by -1. So if the size of the matrix is $n \times n$, the determinant will be multiplied by $(-1)^{\lfloor\frac{n}{2}\rfloor}$ where $\lfloor\frac{n}{2}\rfloor$ is the total number of transposition necessary to do the flip operation.

For the rotated case, we can see that rotating by $90$ degrees clockwise is equivalent to first transpose the matrix (which does not change the determinant), and then flip vertically. Therefore it also multiplies the determinant by $(-1)^{\lfloor\frac{n}{2}\rfloor}$.

Lelouch
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  • Thanks! "Transposing" here means swapping, right? Not related to matrix transpose, right? – joseville Sep 05 '22 at 02:30
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    In the first paragraph transposing two rows/columns means swapping or permuting. In the second paragraph I truly mean "matrix transposition". – Lelouch Sep 05 '22 at 12:42
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I can answer your question for $M_{horizflip}$ and $M_{vertflip}$. The key principle is that of elementary row (or column) operations, particularly that of interchanging rows (or columns). This operation has the effect of changing the sign of $\det(M)$. There are a lot of proofs of this online, or you can try to prove it yourself using the definition of determinant.

Suppose you had a $3\times 3$ matrix $M$. Then $M_{vertflip}$ is obtained by interchanging the first and third column of $M$. Therefore $\det(M_{vertflip}) = -\det(M)$.

But in the $4\times 4$ case, we would perform two operations: interchange columns 1 and 4, and then 2 and 3. Since we do two interchanges, we find $\det(M_{vertflip}) = \det(M)$.

Since we see there are only two options, we naturally ask when the determinant changes sign and when it stays the same, and we see this is dependent on the size of $M$. In particular, we are curious about the parity of the number of pairs needed to interchange. One way to describe the behavior is by examining $N \mod4$.

If $N \mod 4 \in \{0,1\}$, then $\det(M) = det(M_{vertflip})$.

If $N \mod 4 \in \{2,3\}$, then $\det(M) = -det(M_{vertflip})$.

Of course, my choice of vertflip and columns was arbitrary. As you note in your first sentence, you could do all of this in the transpose to find the same result about horizflip and rows. I'm sorry I can't help with $M_{rot}$; I'm sure some smarter person will come along and help us both with that.

Britt Q
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