Tensor #8. Linear Transformations텐서 #8. 선형 변환

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 8 of the playlist, Linear Transformations.

Inside one vector space

Let us forget the maps we used up to the previous post, and this time look only at the basic vector space $\V$. We will use neither the dual space nor tensor product spaces in this post. $\V$ is a vector space over the reals, and its elements are called vectors. What we watch is the basis. A basis $e_\mu$ is a set of vectors that generates all of $\V$, and there is no single fixed choice of basis.

A vector $w$ can be written $a^\mu e_\mu$ in the basis $e_\mu$. If we choose another basis $f_\nu$, it can be written $b^\nu f_\nu$.

This equality is the core. The vector $w$ is a single object inside the space, and we can write it in numbers only after choosing a basis. With a different basis the components differ, but the difference is made up by the difference of the bases. So the question to ask is how the two bases relate to each other, and this relation is the foundation of the topic called linear transformations.

Numbers that link two bases

The basis $f_\nu$ also consists of vectors of $\V$, so we can write them in the basis $e_\mu$. That is exactly what a basis is for. We write $f_0$ like this.

The coefficient $\Lambda$ has to tell us two things. The lower index says which of the new basis vectors we are dealing with, and the upper index says which of the old basis vectors the coefficient belongs to. The other new basis vectors have the same form, so with the Einstein summation convention they gather into one line.

The index $\mu$ is a dummy index that disappears once summed. It is used separately in each equation, or in each term of one equation, so reusing the name causes no mix-up.

This expression is a matrix product as we know it from linear algebra. Between the column of new basis vectors and the column of old basis vectors sits a square matrix $\Lambda$, and one row holds the coefficients of one new basis vector. In dimension 4 that is 16 numbers. Linear algebra books use a weaker convention that sums over a repeated letter without distinguishing up and down, but this series keeps up and down so that the summation convention works. And $\Lambda$ is not a tensor. It is only a table of numbers chosen to link two bases. The abstract bases $e$ and $f$ are merely letters, but $\Lambda$ requires fixing 16 numbers, and at that moment the relation between the two bases is fixed without ambiguity. (The matrix $\Lambda$ can also be read as the matrix of the linear transformation of $\V$ that sends $e_\nu$ to $f_\nu$. This reading is added here and was not covered in the lecture.)

Q. $\Lambda_\nu{}^\mu$ carries indices too, so isn’t it a tensor?
A. No. The components of a tensor product space pair with basis tensors, but $\Lambda$ is a matrix that collects the coefficients for combining several vectors. Having indices alone does not make something a tensor.

How the components change

Now back to the two expressions of $w$. Putting the expression above into $w$ written in the $f$ basis gives the form written in the old basis.

Compared with $w = a^\mu e_\mu$, the surviving index $\mu$ meets itself in the same place. The coefficients standing at each basis vector must be equal, so we get this.

This expression builds the components $a$ in the $e$ basis from the components $b$ in the $f$ basis. What we really want to know is the opposite direction. Since these are matrices, the opposite direction goes through the inverse matrix. The basis can be written backwards too.

Everything comes from the single collection of 16 numbers $\Lambda$. Once $\Lambda$ is fixed, the inverse is determined by computation. For a square matrix there is always a procedure to find the inverse. The two bases have the same number of elements (the dimension is the minimum number of vectors forming a basis, and it is a single number), so $\Lambda$ is square. And both bases are linearly independent, so $\Lambda$ has an inverse. (The explanation that linear independence gives the inverse is added here. The lecture only said that the inverse exists as long as the transformation makes sense.)

Let us handle it. Drag the points $f_0$ and $f_1$ to change $\Lambda$, and see how the two sets of components of the same vector $w$ build each other. (I reduced it to 2 dimensions.)

Q. Why do the components shrink when the basis is stretched?
A. Because the vector $w$ stays the same. In $w = b^\nu f_\nu$, if $f_0$ becomes twice as long, half of $b^0$ is enough to build the same $w$. The components change by $\Lambda^{-1}$ exactly as much as the basis vectors change by $\Lambda$, and the two cancel.

Putting it back together

We check that the transformations compensate each other. Starting from $w$ written in the $f$ basis, apply the inverse to the components and $\Lambda$ to the basis.

The two matrices in the middle are inverses of each other, so they become a delta and we return to the original $w$. The delta symbol is also used as an index, which is confusing, but the context separates them. The reason the transformations move in opposite ways is precisely to keep this equality from breaking.

Covariant and contravariant

The basis vectors and the components change by opposite matrices. The basis vector $f_\nu$ changes by the $\Lambda$ we fixed at the start, and the component $b^\nu$ changes by the inverse. There are words for this, and they are used heavily in general relativity. Whatever changes like the basis vectors is called covariant, and whatever changes the opposite way is called contravariant. Upper indices are contravariant and lower indices are covariant.

The basis vectors became the reference merely for convenience. When we think about a transformation we usually build the matrix that changes the basis vectors first, take that matrix as the reference, and call what goes the other way contra.

Q. Why do general relativity books talk about covariant and contravariant from the start?
A. Because in such books you hardly ever see the vector itself. They write only the component $b^\mu$ instead of $b^\mu e_\mu$, so the first thing you meet is the contravariant component. Without knowing that the hidden basis vector is changing covariantly, the word contravariant seems to come out of nowhere.

In the next post we look at how this transformation works in tensor product spaces.