Square Matrices

These are matrices over some field $\mathbb{F}$ of size $n\times n$. There are exactly 2 types of square matrices: endomorphisms and bilinear forms.

Remark (Connecting Vector and Dual Spaces to Matrices)

The conceptual leap in courses like Math 100B often comes down to moving between coordinate-free, abstract definitions and coordinate-dependent, concrete computations. Whether you are puzzling over these proofs in the library or mulling them over at a local spot in La Jolla, the trick to bridging this gap is always the same: you must explicitly introduce a basis.

Abstract linear maps, such as endomorphisms ($U\to U$) or bilinear forms ($U\to U^{\ast }$), are entirely coordinate-free until we give them a coordinate system. Once we do, these abstract maps perfectly collapse into standard square matrices.

Here is the exact mechanical breakdown of how that happens.

1. Set Up the Bases

Let $U$ be an $n$-dimensional vector space. To get a matrix, we first choose a basis for $U$:

$$E=\{e_1,e_2,\dots ,e_n\}$$

Because we have a basis for $U$, we automatically get a strictly defined dual basis for the dual space $U^{\ast }$:

$$E^{\ast }=\{e^1,e^2,\dots ,e^n\}$$

The defining rule of this dual basis is how it acts on the original basis vectors. It outputs $1$ if the indices match, and $0$ if they do not (the Kronecker delta):

$$e^i(e_j)={\delta }_j^i$$

2. Map the Basis Vectors

We have two types of square matrices corresponding to our two maps. Let us see what they do to a single basis vector, $e_j$.

For an Endomorphism ($A:U\to U$): The output $A(e_j)$ is a vector in $U$. Because $E$ is a basis for $U$, $A(e_j)$ can be written as a linear combination of the basis vectors. Let us call the scalar coefficients $A_j^i$:

$$A(e_j)=\sum _{i=1}^n{A}_j^i{e}_i$$

For a Bilinear Form ($B:U\to U^{\ast }$): The output $B(e_j)$ is a covector in $U^{\ast }$. Because $E^{\ast }$ is a basis for $U^{\ast }$, $B(e_j)$ can be written as a linear combination of the dual basis vectors. Let us call the scalar coefficients $B_{ij}$:

$$B(e_j)=\sum _{i=1}^n{B}_{ij}{e}^i$$

By doing this for every basis vector $e_1$ through $e_n$, we generate an $n\times n$ grid of coefficients for both $A$ and $B$. That grid is your square matrix.

3. Evaluate the Map to Prove the Matrix

To see what those matrix entries actually mean, we evaluate the maps on a different basis vector, $e_i$.

For Bilinear Forms: We take our covector $B(e_j)$ and evaluate it on $e_i$. Substituting the linear combination from Step 2:

$$(B(e_j))(e_i)=\left(\sum _{k=1}^n{B}_{kj}{e}^k\right)(e_i)$$

Because the evaluation of covectors is linear, we pull the sum and scalars out to the front:

$$(B(e_j))(e_i)=\sum _{k=1}^n{B}_{kj}{e}^k(e_i)$$

Applying the rule of the dual basis ($e^k(e_i)={\delta }_i^k$), the sum collapses down to just one coefficient:

$$(B(e_j))(e_i)=B_{ij}$$

4. The Final Translation

For bilinear forms, evaluating the output of $B$ on a second vector is the definition of the bilinear form: $(B(e_j))(e_i)$ is identical to writing $B(e_i,e_j)$. Therefore:

$$B_{ij}=B(e_i,e_j)$$

This is the translation key. To build the square matrix $\mathbf{B}$ that represents $B:U\to U^{\ast }$, you evaluate the bilinear form on every possible pairing of your basis vectors. The scalar from $(e_1,e_2)$ goes into row 1, column 2, and so on.

If the tensor is symmetric ($B^{\ast }=B$), then $B(e_i,e_j)=B(e_j,e_i)$, which means $B_{ij}=B_{ji}$. The matrix equals its own transpose. This directly clarifies how the skew-symmetric case ($B^{\ast }=-B$) from Tensor Algebra alters the matrix structure: $B_{ij}=-B_{ji}$.

For Endomorphisms and Matrix-Vector Multiplication: Endomorphisms are built using tensor products of the form $U^{\ast }\otimes U$. The endomorphism $A$ can be written as:

$$A=\sum _{i=1}^n\sum _{j=1}^n{A}_j^i(e^j\otimes e_i)$$

If you feed a column vector $\vec{v}={\sum }_k{v}^k{e}_k$ into this tensor, the covector $e^j$ “eats” the vector components $v^k$ (since $e^j(e_k)={\delta }_k^j$). The result collapses perfectly to:

$$A(\vec{v})=\sum _{i=1}^n\left(\sum _{j=1}^n{A}_j^i{v}^j\right)e_i$$

This precisely recovers the standard matrix-vector multiplication formula, where the $n\times n$ matrix $A_j^i$ multiplies the column vector $\vec{v}$ to produce a new column vector in $U$.

Theorem (Operations on Square Matrices)

You can view eigenvalues, trace, and the determinant. For the derivative of the determinant, see Jacobi’s Formula. Additionally, every square matrix is associated with its Adjugate Matrix, which relates the matrix to its inverse and cofactor expansion.

Definition (Endomorphisms)

Linear Maps from $V$ to itself are called endomorphisms.

$$\text{End}(V)=\left\{f:V\stackrel{\text{ linear}}{\to }V\right\}$$

Via index notation, we use $A_i^j$ to denote the matrix representation of an endomorphism.

Definition (Bilinear Forms)

A bilinear form on $V$ are linear maps $B:V\to V^{\ast }$. A bilinear form works like

$$\underset{\in V^{\ast }}{\underbrace{B(u)}}(v)$$

like a bilinear function on $V$. Via index notation, we use $B_{ij}$ to denote the matrix representation of a bilinear form.

Remark (Endomorphisms vs Bilinear Forms)

Operations on matrices such as taking trace, determinant, and eigenvalues; or properties of being identity, etc. makes sense only for endomorphisms.

Eigenvalues, trace, and determinants of $A_1{A}_2\dots A_k$ is invariant under cyclic permutations of the $A_i$‘s.

$$\text{tr}(A_1{A}_2\dots A_k)=\text{tr}(A_k{A}_1{A}_2\dots A_{k-1})=\cdots$$

Remark

Properties on matrices such as being symmetric (self-adjoint) makes sense only for bilinear forms.

$$B\in \left(V\stackrel{\text{ linear}}{\to }{V}^{\ast }\right)⟹B^{\top }\in \left(V^{\ast \ast }\stackrel{\text{ linear}}{\to }{V}^{\ast }\right)$$

where $V^{\ast \ast }$ is simply $V$. So, $B,B^{\top }$ are of the same type.

Definition (Symmetric Bilinear Forms)

A bilinear form $B$ is symmetric if $B=B^{\top }$. This is the same as $B(u)(v)=B(v)(u)$ or $B_{ij}=B_{ji}$.

Theorem (Symmetric Bilinear Forms and Quadratic Forms)

There is a one-to-one correspondence between symmetric bilinear forms and quadratic forms.

• From symmetric to quadratic: $q(u)=B(u)(u)$.
• From quadratic to symmetric: $q(u+v)=q(u)+q(v)+2B(u)(v)$

Definition (Positive Definite)

A symmetric bilinear form is positive definite if its quadratic form takes positive values $q(u)>0$ for all $u\ne 0$.

Definition (Inner Product Structure or Metric)

An inner product structure or metric on $V$ is a positive definite symmetric bilinear form. It is often denoted by $g_{ij}$ or ${♭}_{ij}$, the flat operator.

$$g(u,v)=⟨u,v⟩=⟨♭u|v⟩$$

There are infinitely many choices for a metric on $V$. That is, for a plain vector space, there is no canonical choice of metric. The metric (as any bilinear form) is a map from vectors to covectors, so the flat operator is known as index-lowering operator.

$$(♭u{)}_i={♭}_{ij}{u}^j$$

• The flat operator is invertible (due to positive-definiteness).
• The inverse of the flat operator is known as the index-raising operator or sharp operator $♯={♭}^{-1}$.

$${\left(♯\alpha \right)}^i={♯}^{ij}{\alpha }_j$$

Via index notation, the label is $g^{ij}=(g^{-1}{)}^{ij}$.

In particular, an inner product structure $♭:V\to V^{\ast }$ can turn vectors to covectors and vice versa with the inverse $♯:V^{\ast }\to V$. Geometrically, we can draw a quadratic surface with some vector $\vec{v}$. The covector is then the hyperplane ${\vec{v}}^{♭}$ where $♭(\vec{v})(\vec{v})=1$ denotes the unit sphere. The vectors are the tangent vectors to the surface. We construct the polar plane across the quadric surface.