Tensor #25. Review of Determinants텐서 #25. 행렬식 복습

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 25 of the playlist, Review of Determinants. We have built the exterior algebra and the exterior derivative up to the previous post, and the rules of the exterior algebra keep bringing in the determinant. So before going further we tidy up the determinant.

The determinant as a signed volume

At school we learned the determinant as a recipe of adding and subtracting products. This lecture starts from the meaning. In the plane take vectors $a = (a^1, a^2)$ and $b = (b^1, b^2)$ as two sides of a parallelogram. Let $\theta$ be the angle of $a$ with the $x$ axis and $\varphi$ the angle of $b$. The base is $|a|$ and the height is $|b| \sin(\varphi - \theta)$, so the area is as follows.

Insert relations like $|a| \cos\theta = a^1$ and the lengths all cancel, leaving only components. This value is the determinant of the $2 \times 2$ matrix whose rows are the two vectors. In three dimensions the volume of the parallelepiped spanned by three vectors is a $3 \times 3$ determinant, and in $n$ dimensions the volume spanned by $n$ vectors is an $n \times n$ determinant.

This reading agrees with intuition. Bring $b$ toward $a$ until the two are parallel and one dimension disappears, so the volume is 0. In the matrix this is the case where one row is a constant multiple of another, and that the determinant is 0 is something we already know. These vectors live in the tangent space at one point of spacetime, and their components are values written in the coordinate basis $\partial_i$. The calculation used lengths for a moment, but they cancel and no inner product remains in the final formula, so this quantity makes sense without an inner product.

Q. If the determinant is negative, is the area negative?
A. The determinant is not the area itself but a signed area that includes orientation. It is positive if $b$ lies counterclockwise from $a$ and negative if clockwise. The actual area is the absolute value. Swapping the two rows only flips the sign. (This explanation is added for understanding and was not covered in the lecture.)

Permutations and the Levi-Civita symbol

To write the determinant in $n$ dimensions we must fix the sign of every product, and that sign comes from permutations. A permutation $\sigma$ is a map that rearranges $1, 2, \ldots, n$. For example, $\sigma(1) = 2,\ \sigma(2) = 1,\ \sigma(3) = 4,\ \sigma(4) = 3$, with everything else unchanged. We also set $i_k = \sigma(k)$ and write the same permutation as the index list $(i_1, i_2, \ldots, i_n)$. The first notation puts subscripts on subscripts and is awkward to read, and the second grows long sideways, so we move between the two notations as needed.

Call swapping two entries an interchange. Any permutation can be built from some number of interchanges, and whether that number is even or odd is fixed for each permutation. The example above takes two, $1 \leftrightarrow 2$ and $3 \leftrightarrow 4$, so it is even, and the identity permutation that changes nothing takes 0 and is even. One interchange gives an odd permutation.

The Levi-Civita symbol writes this parity as a number.

The version with indices up, $\varepsilon^{i_1 \cdots i_n}$, has the same values. We define it this way so that, inside a sum, we only write indices and the sign attaches itself without separate bookkeeping.

(For example, for $n = 3$, $\varepsilon_{123} = \varepsilon_{231} = \varepsilon_{312} = +1$ and $\varepsilon_{132} = \varepsilon_{213} = \varepsilon_{321} = -1$, and $\varepsilon_{112}$ with a repeated number is 0. This example is added for understanding and was not covered in the lecture.)

This is a symbol, not a tensor. So its indices cannot be raised with the metric.

Q. If we raise the lower indices of the Levi-Civita symbol with the metric, do we get the upper-index version?
A. No. $g^{i_1 i_1'} \cdots g^{i_n i_n'} \varepsilon_{i_1' \cdots i_n'}$ is not equal to $\varepsilon^{i_1 \cdots i_n}$. The symbol is neither covariant nor contravariant, and that the up and down versions have the same values is only a convention. There is another convention that multiplies by $(-1)^t$, where $t$ is the number of negative eigenvalues of the metric, but we do not use it now. Exactly how the symbol transforms comes later.

The generalized Kronecker delta

The ordinary Kronecker delta $\delta^a{}_b$ is 1 if $a = b$ and 0 otherwise. Extending it to $n$ indices at a time gives the generalized Kronecker delta $\delta^{i_1 \cdots i_n}{}_{j_1 \cdots j_n}$. Consider the map $\sigma(i_k) = j_k$ taking the upper index list to the lower one. The value is $+1$ if that map is an even permutation, $-1$ if odd, and 0 if it is not a permutation. Not being a permutation means that a number repeats in the upper or lower list, or that the two lists are made of different numbers.

The same statement can be made with a product of two symbols.

The two factors each carry the sign of one list, so the result is the sign of the permutation between the two lists. This symbol is a notation that packs the product above into one letter.

The definition of the determinant

Write the linear transformation $\Lambda$ as a matrix $\Lambda^i{}_j$. It is a transformation from one basis to another, so it is a square matrix. We have avoided matrices so far, but the determinant is by nature about matrices. The definition is the following.

Sum over every $i_k$. Terms with a repeated number vanish since $\varepsilon = 0$, and only the $n!$ terms that are permutations remain. Each term picks one entry per column from distinct rows and multiplies them, and the sign is decided by whether the chosen row numbers form an even or odd permutation. There is also a form that sums over both index lists instead of fixing the column numbers to $1, \ldots, n$.

The factor $\frac{1}{n!}$ corrects for each term appearing $n!$ times. Books differ on where the indices go up or down, but if you match the index positions of $\varepsilon$ it is the same formula. Put it into $2 \times 2$ or $3 \times 3$ and the school formula comes out exactly.

Check it in the toy below. Pick the dimension and change the entries of the row vectors with the − + buttons, and the parallelogram or parallelepiped spanned by the rows is drawn. Below, one line per permutation shows the signed product, and their sum equals the determinant. You can also see the sign flip when two rows are swapped, and the value become 0 when one row is copied onto another.

Submatrices and minors

The determinant exists only for square matrices. But non-square matrices appear later. For example, in a vector space of dimension 9, the fifth exterior power gives a $9 \times 5$ array when the components of 5 vectors are stacked in columns. This array has no determinant, but if we cross out rows and columns it contains square submatrices. From $9 \times 5$ we can keep 5 of the 9 rows to make a $5 \times 5$, and crossing out more gives everything from $4 \times 4$ down to $1 \times 1$.

The determinant of a submatrix is called a minor. We name it by the kept row numbers $I = (i_1 < \cdots < i_k)$ and the kept column numbers $J = (j_1 < \cdots < j_k)$. In this post we write it as $m[I \mid J]$. For example, in a $3 \times 3$ matrix $M$, keeping rows 1, 3 and columns 2, 3 gives the following.

Complementary minors and cofactors

In a square $n \times n$ matrix $M$, every minor has a partner. The complementary minor crosses out the rows and columns the minor kept and keeps the ones it crossed out. In the example above, the complementary minor of $m[(1,3) \mid (2,3)]$ keeps only row 2 and column 1, the $1 \times 1$ value $d$. If the minor keeps $k$ rows, the complementary minor is $(n-k) \times (n-k)$.

The cofactor, at school, was the determinant that remains after crossing out one entry’s row and column, with the sign $(-1)^{i+j}$. The lecture generalizes it to any minor. Fix the first $k$ indices to the chosen rows and columns and sum only over the last $n-k$.

The product is the expansion of the complementary minor, and $\delta$ supplies the sign. The result equals the complementary minor times $(-1)^{(i_1 + \cdots + i_k) + (j_1 + \cdots + j_k)}$. For $k = 1$ the school cofactor comes out as it is. We define it this way to say, sign included and in one go, how much one minor contributes to the determinant.

(Let us check this on the example above. $\delta^{1 3 2}{}_{2 3 1}$ is the permutation $1 \to 2,\ 3 \to 3,\ 2 \to 1$, a single interchange, so it is $-1$, and $C[(1,3) \mid (2,3)] = -d$. This agrees with the sign formula $(-1)^{1+3+2+3} = -1$. This check is added for understanding and was not covered in the lecture.)

This pairing holds only for square matrices. The number $k$ of kept rows and columns and the number $n-k$ of crossed-out ones can both form squares only when the whole matrix is $n \times n$.

Laplace expansion and the relation between minors and the inverse

The formula that computes the determinant from cofactors is the Laplace expansion. Fix one column $j$, multiply each entry in that column by its cofactor, and add: that is the determinant.

The value is the same for whichever column we pick, so we leave $j$ as a letter. It is not efficient, but it speeds up if we first reduce the matrix to triangular form. The cofactor used here belongs to a single entry, so it is the school formula, an $(n-1) \times (n-1)$ determinant times $(-1)^{i+j}$. This expansion returns in later proofs.

There is one more relation we will use in proofs. For an invertible matrix $M$, a $k \times k$ minor can be written with a cofactor of the inverse.

The right side is the cofactor of the minor of $M^{-1}$ chosen with rows $J$ and columns $I$, so the roles of rows and columns are swapped, a transposed form. In the special case $k = 1$, it becomes the familiar fact that an entry of $M$ is given by a cofactor of the inverse (the adjugate). When a minor is hard to compute directly, this formula lets us rewrite it.

The transformation formula for the Levi-Civita symbol

There is one more formula that looks like the definition but whose lower indices are not fixed to $1, \ldots, n$.

If we regard $\Lambda$ as a coordinate transformation, this converts the Levi-Civita symbol of the $i$ coordinate system into the symbol of the $j$ system. The point is that the symbol does not transform like a tensor but picks up a factor $\det \Lambda$, and this formula is used when we build the Levi-Civita tensor.

(Here is a short reason it holds. The left side is antisymmetric, flipping sign when two of $j_1, \ldots, j_n$ are swapped, so it is a constant multiple of $\varepsilon_{j_1 \cdots j_n}$. Put $(j_1, \ldots, j_n) = (1, \ldots, n)$ and the left side is the definition of the determinant, so the constant is $\det \Lambda$. This explanation is added for understanding and was not covered in the lecture.)

The characteristic polynomial and a summary

A square matrix $\Lambda$ has a characteristic polynomial.

For an $n \times n$ matrix this is a polynomial of degree $n$. The coefficients $\mu_1, \ldots, \mu_n$ are invariants that do not change when the basis changes, so they are $n$ numbers belonging to the linear transformation itself. The first, $\mu_1$, is the trace and the last, $\mu_n$, is $\det \Lambda$. The middle coefficients are more complicated combinations than these two, and the ones to remember going forward are the trace and the determinant.

The lecture ends by retracing what we covered. The heart of the determinant is that it is a signed volume, and it becomes 0 when one vector falls onto another. The Levi-Civita symbol writes the sign of a permutation, it is not a tensor, and the generalized Kronecker delta abbreviates the product of two such symbols. With these two we can write the determinant in any dimension. A non-square matrix still has minors, and in a square matrix every minor is paired with a complementary minor and a cofactor. The Laplace expansion, the relation between minors and the inverse, and the transformation formula of the Levi-Civita symbol will be used as we go deeper into the exterior algebra. In the next post we recap p-vectors and p-forms.