Row swap changing sign of determinant

Question

I was wondering if someone could help me clarify something regarding the effect of swapping two rows on the sign of the determinant. I know that if $A$ is an $n\times n$ matrix and $B$ is an $n\times n$ matrix obtained from $A$ by swapping two rows, then

$$\det(B)=-\det(A)$$

but I don't know how to prove this.

I have been looking for proofs at internet, and read in both in textbooks and lectures notes that are available that this result is very hard to prove and most approaches rely on induction and so was wondering if there is something wrong with using that $\det(AB)=\det(A)\det(B)$ and then writing $B=EA$ where $E$ is an elementary matrix swapping two rows and using this result to get $\det(B)=\det(E)\det(A)=-\det(A)$ (since showing that $\det(E)=-1$ in this case is not that hard).

Answer 1

The problem with your approach is that generally in order to prove that $\det(AB) = \det(A) \det(B)$, one uses the fact that swapping two rows of a matrix multiplies the determinant by $-1$ (see, for example, the second proof in http://www.proofwiki.org/wiki/Determinant_of_Matrix_Product).

So you are kind of stuck proving the desired result directly. There are several equivalent definitions of the determinant, and depending on which one you use, the proof looks a bit different. But I would agree with @Arthur that for at least some definitions, a direct proof is pretty straight-forward as long as you're comfortable with mathematical induction.

Answer 2

Yes, your method would work, and it is probably the most elegant possible.

We can without loss of generality assume that $E$ interchanges the first two rows. This means that we can write $E$ in block-diagonal form: $$ \left( \begin{array}[ccccc] 00 & 1 & 0 &\dots & 0 \\ 1 & 0 & 0 &\dots & 0 \\ 0 & 0 & 1 &\dots & 0 \\ ... & ... & ... & 1 & ... \\ 0 & 0 & 0 & ... & 1 \end{array}\right) $$

Now if you know how to calculate the determinant from the usual Laplace algorithm, starting at the bottom line, you see that the only nonzero terms are...

Also, why can we assume it interchanges the first two lines without loss of generality? (Think of what happens if we change a basis...)

Answer 3

If you have a matrix $\mathrm A \in \mathbb{R}^{n \times n}$ and you swap its $i$-th and $j$-th rows, you are left-multiplying $\mathrm A$ by the permutation matrix

$$\mathrm E := \mathrm I_n - \mathrm{e}_i \mathrm{e}_i^\top - \mathrm{e}_j \mathrm{e}_j^\top + \mathrm{e}_i \mathrm{e}_j^\top + \mathrm{e}_j \mathrm{e}_i^\top = \mathrm I_n - (\mathrm{e}_j - \mathrm{e}_i) (\mathrm{e}_j - \mathrm{e}_i)^\top$$

where $\mathrm{e}_k$ is a vector with $n-1$ zeros and with a one on the $k$-th entry. Thus, the determinant of the new matrix is

$$\det (\mathrm E \mathrm A) = \det(\mathrm E) \cdot \det (\mathrm A)$$

Using the Weinstein-Aronszajn determinant identity,

$$\begin{array}{rl} \det (\mathrm E) &= \det \left( \mathrm I_n - (\mathrm{e}_j - \mathrm{e}_i) (\mathrm{e}_j - \mathrm{e}_i)^\top \right)\\ &= \det \left( 1 - (\mathrm{e}_j - \mathrm{e}_i)^\top (\mathrm{e}_j - \mathrm{e}_i) \right)\\ &= 1 - \left( 1 + (-1)^2 \right) = \color{blue}{-1}\end{array}$$

Hence, swapping two rows of $\mathrm A$ does change the sign of the determinant.

Answer 4

It is a chicken and an egg kind of a problem if you think about it that way. All of the following ideas are connected to each other;

1- Swapping any 2 rows of a matrix, flips the sign of its determinant.

2- The determinant of product of 2 matrices is equal to the product of the determinants of the same 2 matrices.

3- The matrix determinant is invariant to elementary row operations.

4- Multiplying an entire row (or column) of a matrix by a constant, scales the determinant up by that constant.

If you assume any subset of these, the rest follow through.

I have used the elementary row operations and multiplying the entire row by a constant to show that the proof is quite straightforward.

Swapping 2 rows inverts the sign of the determinant.

For any square matrix you can generalize the proof of swapping two rows (or columns) being equivalent to swapping the sign of the determinant by using the axiom that the determinant is invariant under elementary row (or column) operations.

Consider a Matrix $\mathrm{A} \in \mathbb{F}^{n\times n} $ as shown below;

$$ \mathrm{A}= \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn}\\ \end{bmatrix} $$

Then the determinant of $\mathrm{A}$ is given by;

$$ \det{\mathrm{A}} = \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn}\\ \end{vmatrix} $$

Consider two separate rows that you want to swap and add one of them to the other. W.L.O.G here assume that the first row is added to the second row. This holds in general for any 2 rows (or columns).

$Row 2 := Row 2 + Row 1$

$$ \det{\mathrm{(A)}} = \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21}+a_{11} & a_{22}+a_{12} & \cdots & a_{2n}+a_{1n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn}\\ \end{vmatrix} $$

Next subtract the resulting row 2 from row 1.

$Row1 := Row1 - Row2$

$$ \det{\mathrm{(A)}} = \begin{vmatrix} -a_{21} & -a_{22} & \cdots & -a_{2n}\\ a_{21}+a_{11} & a_{22}+a_{12} & \cdots & a_{2n}+a_{1n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn}\\ \end{vmatrix} $$

Add the resulting Row 1 to Row 2.

$Row2 := Row2 + Row1$

$$ \det{\mathrm{(A)}} = \begin{vmatrix} -a_{21} & -a_{22} & \cdots & -a_{2n}\\ a_{11} & a_{12} & \cdots & a_{1n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn}\\ \end{vmatrix} $$

Take $-1$ common from the first row, this falls out of the determinant.

$$ \det{\mathrm{A}} = (-1) \begin{vmatrix} a_{21} & a_{22} & \cdots & a_{2n}\\ a_{11} & a_{12} & \cdots & a_{1n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn}\\ \end{vmatrix} $$

Finally we have;

$$ \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn}\\ \end{vmatrix} = (-1) \begin{vmatrix} a_{21} & a_{22} & \cdots & a_{2n}\\ a_{11} & a_{12} & \cdots & a_{1n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn}\\ \end{vmatrix} \quad \blacksquare $$

The same argument holds for swapping any two rows (or columns).

Answer 5

Sorry for opening an old thread, but I have found that thinking about this abstractly is the most clarifying. Let the ground field be $\mathbb{R}$. For this, I need you to believe that the determinant of a matrix is built to measure the signed area of the parallelepiped that the image of the basis vectors span. In fact, some people define the determinant of a linear transformation in the following way: given $V$ a vector space and fix an ordered basis $e_1,e_2,...,e_n$, the determinant of a linear operator $T:V\to V$ is defined as the coefficient the induced map on the top exterior power: Namely $$T:\bigwedge^n V\to \bigwedge^nV $$ $$T:e_1\wedge...\wedge e_n\mapsto (\det T)e_1\wedge...\wedge e_n$$ If you haven't seen exterior powers, don't worry; it's essentially machinery you need to create to make the definition of determinant as signed area work out.

The thing to notice is that I needed to order my basis; this ordering of basis vectors gives an orientation on V. We want reflections of our vector space to reverse orientation. For example, think back to high school when you reflected a picture of a shape across the y-axis. For the sake of defining what a reflection is, let's put an inner product structure on our space. A reflection $R:V\to V$ is a linear map which take some $R:v\mapsto -v$ and fixes $v$'s orthogonal hyperplane. Suppose our basis vectors are orthogonal (no loss of generality because of Gram-Schmidt), then I claim the linear map which swaps two basis vectors, say $e_i$ with $e_j$ induces a reflection. Indeed, look at the 2-dimensional vector space spanned by $e_i,e_j$ and observe that $$R:e_i-e_j\mapsto e_j-e_i=-(e_i-e_j)$$ and its orthogonal hyperplane $$(e_i-e_j)^\perp=\text{Span}(e_i+e_j,e_1,e_2,...\hat{e}_i,...,\hat{e}_j...,e_n)$$ is fixed, where hat over $e_i,e_j$ denotes omit these two vectors. In high school language, this is what happens when you reflect about the line $y=x$, if we identify $\text{Span}(e_i)$ and $\text{Span}(e_j)$ with the x and y axes.

Under this definition of a determinant, several algebraic properties become geometrically transparent:

1. If the matrix isn't linearly independent, then the determinant is 0, because the (signed) area of a k-dimensional parallelepiped in an n-dimensional space is $0$ for $k<n$ (think of the volume of a square in $\mathbb{R}^3$).

2. $\det(T\circ S)=\det T\cdot \det S$. The intuition is in really in taking an orthonormal basis and considering diagonal matrices/linear transformation which only scale basis vectors. Then think about what happens with upper triangular matrices/linear transformation which preserve the flag, then use the QR decomposition of a linear transformation. For example, in one dimension, linear transformations are exactly multiplication by some fixed constant $\lambda$, so we can identify the scalars with one dimensional linear transformations. So for scalars $\lambda,\mu\in \mathbb{R}$, the for the linear transformation $\lambda\circ \mu$, the determinant is exactly $\lambda\cdot \mu$.

3. Swapping two columns (thus rows, as determinant is stable under transpose) can be though of as precompose $T$ with a swap which induces a reflection as we talked about. By 2, $$\det(T\circ R)=\det T\cdot\det R=-\det T$$

4. Scaling a linear transformation scales the determinant by the dimension of $V$. This is because if you scale all your basis vectors by $\lambda\in \mathbb{R}$, then the volume of the parallelepiped scales by $\lambda^n$.

Answer 6

The approach suggested by @NanoShaw is the easiest to understand. I cover this in a video explainer as well: https://www.youtube.com/watch?v=glxl6U3gqNQ

However, a much simpler approach is to consider the following:

Let original matrix be $M$, then $det(M) = x$ is its determinant.
Let, $M'$ be the matrix with columns, $a$ and $b$ swapped. Now $det(M') = y$.

Now construct $S = M + M'$, leading to $det(S) = det(M) + det(M') = x + y$

Since two columns $a$ and $b$ of $S$ are same with value $a+b$, $det(S) = 0$. Hence, $x + y = 0$, meaning $x = -y$. This proves $det(M) = -det(M')$.

Answer 7

If $~i~$ and $~j~$ are two rows of matrix $~A~$ that are interchanged to give matrix $~A^*~$.

Apply to $~A~$ and $~A^*~$ the general recursive formula twice (two-stage recursion) along the two interchanged rows $~i~$ and $~j~$.

Using the fact that $~A~$ and $~A^*~$ are identical when rows $~i~$ and $~j~$ are deleted (which is the case for the determinants of the resulting matrices after the two-stage recursion) and rearranging the co-factors in the sums, you will find that $~\det A = -\det A^*$.