From c4b9b32e16d94f99555c7d84eb14c6e6f7cdc57d Mon Sep 17 00:00:00 2001 From: Sungchan Yi Date: Fri, 2 Jan 2026 16:24:05 +0900 Subject: [PATCH] feat: matmul as tensors post --- .../algebra/2026-01-02-matmul-as-tensors.md | 385 ++++++++++++++++++ .../mathematics/algebra/matmul-tensor.png | Bin 0 -> 68231 bytes 2 files changed, 385 insertions(+) create mode 100644 _posts/Mathematics/algebra/2026-01-02-matmul-as-tensors.md create mode 100644 assets/img/posts/mathematics/algebra/matmul-tensor.png diff --git a/_posts/Mathematics/algebra/2026-01-02-matmul-as-tensors.md b/_posts/Mathematics/algebra/2026-01-02-matmul-as-tensors.md new file mode 100644 index 0000000..4218ce1 --- /dev/null +++ b/_posts/Mathematics/algebra/2026-01-02-matmul-as-tensors.md @@ -0,0 +1,385 @@ +--- +share: true +toc: true +math: true +categories: + - Mathematics + - Algebra +path: _ posts/mathematics +tags: + - math + - algebra +title: Matrix Multiplication as $(1, 1)$-Tensors +date: 2026-01-02 +github _ title: 2026-01-02-matmul-as-tensors +image: + path: /assets/img/posts/mathematics/algebra/matmul-tensor.png +--- + +## Introduction + +Matrices have various applications throughout many fields of science. However, when we first learn matrices and their operations, the most non-intuitive concept is the **multiplication** of matrices. + +Compared to addition and *scalar* multiplication, which are very intuitive, matrix multiplication has a rather strange definition. + +> **Definition.** (Matrix Multiplication) For $A = (a _ {ij}) _ {n\times n}$ and $B = (b _ {ij}) _ {n\times n}$, their product is defined as $C = AB = (c _ {ij}) _ {n\times n}$ where +> +> $$ +> c _ {ij} = \sum _ {k=1}^n a _ {ik}b _ {kj}. +> $$ + +This is often interpreted as "row times column", with the following diagram: + +$$ +\begin{aligned} +& \begin{bmatrix} & b _ {1j} & & & \\ +& b _ {2j} & & & \\ +& \vdots & & & \\ +& b _ {nj} & & & \\ +\end{bmatrix} \\ +\begin{bmatrix} \\ +a _ {i1} & a _ {i2} & \cdots & a _ {in} \\ +\\ \\ +\end{bmatrix} & \begin{bmatrix} +& & & & \\ +& c _ {ij} & & & \\ +& & & & \\ +& & & & +\end{bmatrix} +\end{aligned} +$$ + +But why is matrix multiplication defined this way, rather than multiplying element-wise just like addition? What is the meaning behind all this strange "row times column"? It is well-known that matrix multiplications can represent the **composition of linear transformations**, and we are pretty sure that one can derive the above formula so that the desired property holds. Still, why does it have this specific structure? + +In this article, we explore the deep reason behind the definition of matrix multiplication using **tensors**. + +We assume that all vector spaces are over a field $K$, and are *finite-dimensional*. + +## Linearity + +### Linear Maps + +> **Definition.** Let $(V, \oplus, \odot)$ and $(W, \boxplus, \boxdot)$ be vector spaces over $K$. A map $f : V \to W$ is **linear** if for all $v _ 1, v _ 2 \in V$ and $\lambda \in K$, +> +> $$ +>f \paren{(\lambda \odot v _ 1) \oplus v _ 2} = (\lambda \boxdot f(v _ 1)) \boxplus f(v _ 2). +> $$ + +We will drop the special notation for addition and scalar multiplication, and just write $+$ and $\cdot$. It should be clear from context and not cause any confusion. Then we can write the above definition in a familiar way. + +> For all $v _ 1, v _ 2 \in V$ and $\lambda \in K$, +> +> $$ +>f(\lambda v _ 1 + v _ 2) = \lambda f(v _ 1) + f(v _ 2). +> $$ + +Some definitions are in order. + +> **Definition.** (Isomorphism) A bijective linear map is a **isomorphism** of vector spaces. If vector spaces $V$ and $W$ are isomorphic, we write $V \approx W$. + +> **Definition.** The set of linear maps from $V$ to $W$ is denoted as $\rm{Hom}(V, W)$. + +> **Definition.** (Endomorphism) An **endomorphism** of $V$ is a linear map from $V$ to itself. We write $\rm{End}(V) = \rm{Hom}(V, V)$. + +### Bilinearity + +A map is *bilinear* if it is linear in its respective argument. Formally, + +> **Definition.** (Bilinearity) Let $V, W, Z$ be vector spaces over $K$. A map $f : V \times W \to Z$ is **bilinear** if +> 1. For any $w \in W$, $f(\lambda v _ 1 + v _ 2, w) = \lambda f(v _ 1, w) + f(v _ 2, w)$ for all $v _ 1, v _ 2 \in V$ and $\lambda \in K$. +> 2. For any $v \in V$, $f(v, \lambda w _ 1 + w _ 2) = \lambda f(v, w _ 1) + f(v, w _ 2)$ for all $w _ 1, w _ 2 \in W$ and $\lambda \in K$. + +In other words, + +1. The map $v \mapsto f(v, w)$ is linear for fixed $w \in W$. +2. The map $w \mapsto f(v, w)$ is linear for fixed $v \in V$. + +Note that this is *not* the same as linear maps in $V \times W \to Z$, since that would mean + +$$ +f\paren{\lambda(v _ 1, w _ 1) + (v _ 2, w _ 2)} = \lambda f(v _ 1, w _ 1) + f(v _ 2, w _ 2) +$$ + +for all $v _ 1, v _ 2 \in V$, $w _ 1, w _ 2 \in W$ and $\lambda \in K$. One can check that $(x, y) \mapsto x+ y$ is linear but not bilinear, while the map $(x, y) \mapsto xy$ is bilinear but not linear. + +**Example.** Famous examples of bilinear maps. + + - The standard inner product $\span{\cdot, \cdot} : \R^n \times \R^n \to \R$, where $\span{v, w} = v\trans w$. + - Determinants of $2 \times 2$ matrices $\det _ 2 : \R^2 \times \R^2 \to \R$, where + + $$\rm{det} _ 2 \paren{\begin{bmatrix} a \\ c\end{bmatrix}, \begin{bmatrix} b \\ d\end{bmatrix}} = \det \paren{\begin{bmatrix} a & b \\ c & d \end{bmatrix}} = ad - bc.$$ + +## Dual Spaces + +### Dual Vector Space + +To define tensors, we need *dual spaces*. + +> **Definition.** (Dual Space) The **dual vector space** of $V$ is defined as +> +> $$ +> V^\ast = \rm{Hom}(V, K) +> $$ +> +> where $K$ is considered as a vector space over itself. + +The dual vector space is the *vector space of linear maps from $V$ to the underlying field $K$*. Its elements are called **linear functionals**, **covectors**, or **one-forms** on $V$. + +We can also consider the *double dual* of a vector space. + +> **Definition.** The double dual vector space of $V$ is defined as +> +> $$ +>V^{\ast\ast} = \rm{Hom}(V^\ast, K). +> $$ + +We can continue on, but it is unnecessary due to the following result: $V \approx V^{\ast\ast}$. + +> **Theorem.** $V$ and $V^{\ast\ast}$ are *naturally isomorphic*. + +*Proof*. Define $\psi : V \ra V^{\ast\ast}$ as + +$$ +\psi(v)(f) = f(v) +$$ + +for $v \in V$ and $f \in V^\ast$. Then $\psi$ is a natural isomorphism, independent of the choice of basis on $V$. + +Furthermore, $V \approx V^\ast$ (finite-dimensional) but they are not naturally isomorphic since an isomorphism from $V \ra V^\ast$ requires a choice of basis. More about this in another article. + +### Dual Basis + +Since $V$, $V^\ast$ and $V^{\ast\ast}$ are all isomorphic, they all have the same dimension. + +> **Lemma.** $\dim V = \dim V^\ast = \dim V^{\ast\ast}$. + +Therefore, the bases of $V$ and $V^\ast$ have the same number of elements. We can construct a basis of $V^\ast$ from a basis of $V$. + +> **Definition.** (Dual Basis) Let $V$ be a $d$-dimensional vector space with basis $\mc{B} = \braces{e _ 1, \dots, e _ d}$. The **dual basis** is the unique basis +> +> $$ +>\mc{B}' = \braces{f^1, \dots, f^d} \subseteq V^\ast +> $$ +> +> such that +> +> $$ +>f^i (e _ j) = \delta^i _ j = \begin{cases} 1 & (i = j) \\ 0 & (i \neq j) \end{cases}. +> $$ + +Note that the superscripts are not exponents. + +## Tensor Spaces + +Tensors are not just high-dimensional arrays of numbers. We approach tensors from an algebraic perspective. + +### Tensors + +> **Definition.** (Tensor) Let $V$ be a vector space over $K$. A **$(p, q)$-tensor** $T$ on $V$ is a *multi-linear map* +> +> $$ +>T : \underbrace{V^\ast \times \cdots \times V^\ast} _ {p \text{ copies}} \times \underbrace{V \times \cdots \times V} _ {q \text{ copies}} \to K. +> $$ +> +> We write the set of $(p, q)$-tensors on $V$ as +> +> $$ +>T _ q^p V = \underbrace{V \otimes \cdots \otimes V} _ {p\text{ copies}} \otimes \underbrace{V^\ast \otimes \cdots \otimes V^\ast} _ {q\text{ copies}}. +> $$ + +Thus, **tensors** are **multi-linear maps**. + +**Remark.** The order of $V$ and $V^\ast$ are swapped in the definition of tensor and the set of tensors. Although this is just a matter of notation, it can be understood as follows. + +A $(p, q)$-tensor $T$ eats $p$ covectors and $q$ vectors to map them to a scalar. Notice that + +- An element of $V \approx V^{\ast\ast} = \rm{Hom}(V^\ast,K)$ eats a covector and maps it to a scalar. +- An element of $V^\ast = \rm{Hom}(V, K)$ eats a vector and maps it to a scalar. + +Thus, a $(p, q)$-tensor $T$ is sort of an element of $p$ copies of $V$ ($\approx V^{\ast\ast}$) and $q$ copies of $V^\ast$, which kind of justifies the notation in $T _ q^p V$. + +### Tensor Spaces + +The set $T _ q^pV$ can be equipped with a $K$-vector space structure by defining addition and multiplication component-wise, since tensors are linear maps. Now we have a **tensor space** as a $K$-vector space. + +**Example.** Some famous examples of tensor spaces. + +- $T _ 1^0 V = V^\ast$ is the set of linear maps $T : V \ra K$, which agrees with the definition of $V^\ast$. +- $T _ 1^1(V) = V \otimes V^\ast$ is the set of bilinear maps $T : V^\ast \times V \to K$. + +### Tensor Products + +> **Definition.** (Tensor Product) Let $T \in T _ q^pV$ and $S \in T _ s^r V$. The **tensor product** of $T$ and $S$ is the tensor $T \otimes S \in T _ {q+s}^{p+r} V$ defined as +> +> $$ +>\begin{aligned} +>&(T\otimes S)(w _ 1, \dots, w _ p, w _ {p+1}, \dots, w _ {p+r}, v _ 1, \dots, v _ q, v _ {q+1}, \dots, v _ {q+s}) \\ +>&\quad= T(w _ 1, \dots, w _ p, v _ 1, \dots, v _ q) \cdot S(w _ {p+1}, \dots, w _ {p+r}, v _ {q+1}, \dots, v _ {q+s}). +> \end{aligned} +> $$ +> +> for $w _ i \in V^\ast$ and $v _ i \in V$. + +The definition seems complicated but is actually very natural. $T \otimes S$ should eat $p+r$ covectors and $q+s$ vectors. Considering the arguments of $T$ and $S$, give $p$ covectors and $q$ vectors to $T$ and the remaining to $S$. Multiply the resulting scalar in $K$. + +Note that the $\otimes$ here is different from $\otimes$ used in the definition of $T _ q^p V$. + +### Tensors as Components + +With these tools, we can completely determine a tensor with its *components*, just like how linear maps are completely determined by its values on the basis vectors. + +> **Definition.** Let $V$ be a finite-dimensional $K$-vector space with basis $\mc{B} = \braces{e _ 1, \dots, e _ d}$ and dual basis $\mc{B}' = \braces{f^1, \dots, f^{d}}$. +> +> The **components** of $T \in T _ q^p V$ are defined to be the numbers +> +> $$ +>T^{a _ 1 \dots a _ p} _ {b _ 1\dots b _ q} = T(f^{a _ 1}, \dots, f^{a _ p}, e _ {b _ 1}, \dots, e _ {b _ q}) \in K +> $$ +> +> where $1 \leq a _ i, b _ j \leq d = \dim V$. + +Notice that $a _ i, b _ j$ range from $1$ to $\dim V$. Since we have $p$ copies of $V^\ast$ and $q$ copies of $V$, the value of $T$ at *every possible combination of basis vectors* are the components. + +We can reconstruct a tensor from its components by the tensor product of vectors and covectors from the basis and the dual basis: + +$$ +T = \underbrace{\sum _ {a _ 1=1}^{\dim V} \cdots \sum _ {b _ q = 1}^{\dim V}} _ {p+q \text{ sums}} T^{a _ 1 \dots a _ p} _ {b _ 1\dots b _ q} e _ {a _ 1} \otimes \cdots \otimes e _ {a _ p} \otimes f^{b _ 1} \otimes \cdots \otimes f^{b _ q}. +$$ + +Here, $e _ {a _ i}$ are considered as elements of $T _ 0^1 V \approx V$ and $f^{b _ j}$ as elements of $T _ {1}^0 V \approx V^\ast$. + +## Vectors as Tensors + +Now that we have components for tensors in $T _ q^p V$ using a basis for $V$ and dual basis of $V^\ast$, we can write vectors and matrices as tensors. + +### Vectors + +It is well-known that any vector can be represented as a linear combination of the basis vectors. i.e., + +$$ +v = v^1 e _ 1 + \cdots + v^d e _ d \in V +$$ + +where $v^j$ are components of the vector $v$ with respect to each basis vector $e _ j$. We observe that vectors are indeed tensors in $T _ 0^1 V = V$. + +Conventionally, we often write vectors as *columns* of numbers. + +$$ +v = \sum v^i e _ i \quad \longleftrightarrow \quad v = \begin{bmatrix} v^1 \\ \vdots \\ v^d \end{bmatrix} +$$ + +### Covectors + +As for covectors $w \in V^\ast$, we can also represent them as a linear combination. i.e., + +$$ +w = w _ 1 f^1 + \cdots + w _ d f^d \in V^\ast +$$ + +where $w _ i$ are components of the covector $w$ with respect to each dual basis vector $f^i$. We also observe that covectors are tensors in $T _ 1^0 V = V^\ast$. + +Also, we often write covectors as *rows* of numbers. + +$$ +w = \sum w _ i f^i \quad \longleftrightarrow \quad w = \begin{bmatrix} w _ 1 & \dots & w _ d \end{bmatrix} +$$ + +## Matrices as Tensors + +As for matrices we need to prove a few things beforehand. We limit the discussion to square matrices. + +First, we know that **matrices are linear transformations**, so every matrix can be considered as a linear map $V \to V$, thus an element of $\rm{End}(V)$. + +> **Lemma.** $T _ 1^1 V = V \otimes V^\ast \approx \rm{End}(V^\ast)$. + +*Proof*. We must construct an invertible linear map between a tensor $T \in V \otimes V^\ast$ and a linear map $f : V^\ast \to V^\ast$. + +For $w \in V^\ast$, simply define $f(w) = T(\cdot, w)$, then $f(w)$ is a map from $V$ to $K$, where $f$ and $f(w)$ can be shown to be linear by the bilinearity of $T$. Thus $f(w) \in V^\ast$ and $f \in \rm{End}(V^\ast)$. + +This correspondence is invertible, since given a linear map $f : V^\ast \to V^\ast$, we can define $T$ as $T(v, w) = f(w)(v)$ for $v \in V$ and $w \in V^\ast$. Then $f(w) \in V^\ast$ and $f(w)(v) \in K$, so $T : V \times V^\ast \to K$. Linearity of $T$ follows directly from the linearity of $f$ and $f(w)$. + +> **Corollary.** For finite dimensional $K$-vector space $V$, $T _ 1^1 V \approx \rm{End}(V)$. + +This directly follows from the fact that $V\approx V^\ast$. + +### Matrices as $(1, 1)$-Tensors + +Finally, since matrices are elements of $\rm{End}(V)$, we can conclude that **square matrices are $(1, 1)$-tensors**. Therefore, we can write a tensor $\phi \in T _ 1^1 V$ with respect to the chosen basis as + +$$ +\phi = \sum _ {i=1}^{\dim V} \sum _ {j=1}^{\dim V} \phi^i _ j \; e _ i \otimes f^j, +$$ + +where $\phi^i _ j = \phi(f^i, e _ j)$. Now it is *very tempting* to think of $\phi^i _ j$ as a **square array of numbers**. + +$$ +\phi = \sum _ {i=1}^{\dim V} \sum _ {j=1}^{\dim V} \phi^i _ j \; e _ i \otimes f^j \quad \longleftrightarrow \quad \phi = +\begin{bmatrix} +\phi _ 1^1 & \phi _ 2^1 & \cdots & \phi _ d^1 \\ +\phi _ 1^2 & \phi _ 2^2 & \cdots & \phi _ d^2 \\ +\vdots & \vdots & \ddots & \vdots \\ +\phi _ 1^d & \phi _ 2^d & \cdots & \phi _ d^d +\end{bmatrix} +$$ + +The convention is to consider the top index $i$ as a *row* index and bottom index $j$ as a *column* index. + +### Matrix Multiplication + +However, the above *arrangement* is a pure convention. + +Consider $\phi \in \rm{End}(V)$ also as a tensor $\phi \in T _ 1^1 V$. Abusing the notation, we can write + +$$ +\phi(w, v) = w(\phi(v)) +$$ + +for $v \in V$ and $w \in V^\ast$. For clarity, $\phi : V^\ast \times V \to K$ on the left hand side. As for the right hand side, $\phi : V \to V$. $w$ eats a vector $\phi(v) \in V$ and maps it to $K$ as it is an element of $V^\ast$. + +Thus, the components of $\phi \in \rm{End}(V)$ are $\phi _ j^i = \phi(f^i, e _ j) = f^i(\phi(e _ j))$. + +Since **matrix multiplication represents the composition of linear transformation**, for $\phi, \psi \in \rm{End}(V)$, let us consider the components of $\phi \circ \psi$ as a $(1,1)$-tensor. We have + +$$ +\begin{aligned} +(\phi \circ \psi) _ j^i &= (\phi \circ \psi)(f^i, e _ j) \\ +&= f^i\big( (\phi\circ\psi)(e _ j) \big) \\ +&= f^i\big( \phi(\psi (e _ j)) \big) \\ +&\overset{(\ast)}{=} f^i\paren{ \phi\paren{\sum _ k \psi^k _ j e _ k} } \\ +&\overset{(\star)}{=} \sum _ k \psi _ j^k f^i \big( \phi(e _ k) \big) \\ +&= \sum _ k \psi _ j^k \phi _ k^i. +\end{aligned} +$$ + +- $(\ast)$ follows by $\psi(e _ j) = \sum _ k \psi _ j^k e _ k$. This holds since for $w \in V^\ast$, + + $$ + \begin{aligned} + \psi(w, e _ j) &= \sum _ {k, l} \psi _ l^k \; (e _ k \otimes f^l)(w, e _ j) \\ + &= \sum _ {k, l} \psi _ l^k \; e _ k(w) \cdot f^l(e _ j) \\ + &= \sum _ {k} \psi _ j^k e _ k (w) + \end{aligned} + $$ + + because $f^l (e _ j) = \delta _ {j}^l$, leaving only the terms on $l=j$. +- $(\star)$ follows by the linearity of $f^i$ and $\phi$. + +Doesn't this expression look familiar? Since $K$ is a field, + +$$ +\sum _ {k} \psi _ j^k \phi _ k^i = \sum _ k \phi^i _ k \psi _ j^k, +$$ + +which implies that the $(i, j)$-th component of $(1, 1)$-tensor $\phi \circ \psi$ exactly matches the $(i, j)$-th entry of the matrix product $\phi \psi$. + +Now we know why matrix multiplication is defined as "row times column". Tensor spaces, tensor products and dual spaces were behind all this. + +## Notes + +- The above argument should be generalizable to rectangular matrices, although I haven't checked yet. +- Now we know why matrix multiplication is defined in a strange way, but the question is now why tensors are defined in such a strange way, involving vector spaces and their duals. +- (Not sure) Matrix transpose should be interpretable with tensors? + +## References + +- [Geometrical Anatomy of Theoretical Physics](https://youtube.com/playlist?list=PLPH7f _ 7ZlzxTi6kS4vCmv4ZKm9u8g5yic&si=dJj6nmoAc944YOgl), Lecture 8 diff --git a/assets/img/posts/mathematics/algebra/matmul-tensor.png b/assets/img/posts/mathematics/algebra/matmul-tensor.png new file mode 100644 index 0000000000000000000000000000000000000000..62f4d1c81b6fe49132b3e36ca97ea6cf523255d6 GIT binary patch literal 68231 zcmeFZbyQSq8#ayzGAQaGQc?pbAvuB!0t2IfC}~kr0@B?`I)c(5A|aikQqo9E%!mjG 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