Before we factor matrices, we need two numbers that summarize them: the determinant and the trace. The determinant measures how much a matrix scales volume (and whether it can be inverted); the trace sums its diagonal. Both are invariant under basis change, which is why they capture something intrinsic about a linear map — and both turn out to be built from the matrix’s eigenvalues.

A real-life example: does a mixture have a unique recipe?

You blend two base syrups to hit a target flavor with two properties (sweetness, acidity). Each syrup contributes a known amount of each. If the two syrups affect the properties in genuinely different ways, there’s a unique blend — but if one syrup is just a scaled version of the other, their “flavor vectors” are collinear and no unique recipe exists. The determinant of the $2\times2$ recipe matrix is exactly the test: nonzero → unique blend, zero → no unique solution.

The determinant

The determinant $\det(A)$ (also written $|A|$) is defined only for square matrices. Small cases have closed forms:

and for $3\times3$, Sarrus’ rule:

For a triangular matrix, the determinant is just the product of the diagonal, $\det(T) = \prod_i T_{ii}$ — which is why we compute large determinants by row-reducing to triangular form. Larger matrices use recursive Laplace expansion along a row or column.

The determinant is signed volume

The most useful picture: $\det(A)$ is the signed volume of the shape spanned by the columns of $A$. In 2-D that’s the area of the parallelogram formed by the two column vectors. Watch the area as one column rotates — it shrinks to zero exactly when the columns become collinear (a singular, non-invertible matrix), and the sign flips when their orientation flips:

That zero-area moment is the whole story of invertibility.

Determinant properties

• $\det(AB) = \det(A)\det(B)$ — determinant of a product is the product of determinants.
• $\det(A^\top) = \det(A)$ — transposition doesn’t change it.
• $\det(A^{-1}) = 1/\det(A)$ when $A$ is invertible.
• Similar matrices have the same determinant — so it’s invariant under basis change.
• Adding a multiple of one row to another leaves $\det$ unchanged; scaling a row by $\lambda$ scales $\det$ by $\lambda$; swapping two rows flips the sign.

Two headline theorems tie it to earlier chapters:

The trace

The trace is the sum of the diagonal entries:

It’s linear ($\text{tr}(A+B) = \text{tr}(A) + \text{tr}(B)$) and — crucially — invariant under cyclic permutation: $\text{tr}(AKL) = \text{tr}(KLA)$. In particular $\text{tr}(AB) = \text{tr}(BA)$ even when $AB \neq BA$, and $\text{tr}(S^{-1}AS) = \text{tr}(A)$, so the trace is also basis-independent.

The bridge to eigenvalues

The characteristic polynomial $p_A(\lambda) = \det(A - \lambda I)$ is where determinant and trace meet eigenvalues (next page). Two beautiful identities fall out:

The determinant is the product of the eigenvalues; the trace is their sum. Volume scaling = product of stretches; that’s the geometry made precise.

  flowchart LR
  A["Matrix A"] --> CP["char. polynomial
det(A − λI) = 0"]
  CP --> EV["eigenvalues λᵢ"]
  EV --> D["det(A) = Πλᵢ"]
  EV --> T["tr(A) = Σλᵢ"]

NumPy

import numpy as np
 
A = np.array([[4.0, 2.0],
              [1.0, 3.0]])
 
print("det:", np.linalg.det(A))          # 10.0
print("trace:", np.trace(A))             # 7.0
 
# invertible iff det != 0
print("invertible:", not np.isclose(np.linalg.det(A), 0))
 
# det = product of eigenvalues, trace = sum of eigenvalues
eig = np.linalg.eigvals(A)
print("eigenvalues:", np.round(eig, 4))
print("prod eig ≈ det :", np.isclose(np.prod(eig), np.linalg.det(A)))
print("sum  eig ≈ trace:", np.isclose(np.sum(eig), np.trace(A)))
 
# a singular matrix has det 0
S = np.array([[1.0, 2.0], [2.0, 4.0]])   # second column = 2 * first row-wise -> collinear
print("singular det:", round(np.linalg.det(S), 6))

import numpy as np
 
A = np.array([[4.0, 2.0],
              [1.0, 3.0]])
 
print("det:", np.linalg.det(A))          # 10.0
print("trace:", np.trace(A))             # 7.0
 
# invertible iff det != 0
print("invertible:", not np.isclose(np.linalg.det(A), 0))
 
# det = product of eigenvalues, trace = sum of eigenvalues
eig = np.linalg.eigvals(A)
print("eigenvalues:", np.round(eig, 4))
print("prod eig ≈ det :", np.isclose(np.prod(eig), np.linalg.det(A)))
print("sum  eig ≈ trace:", np.isclose(np.sum(eig), np.trace(A)))
 
# a singular matrix has det 0
S = np.array([[1.0, 2.0], [2.0, 4.0]])   # second column = 2 * first row-wise -> collinear
print("singular det:", round(np.linalg.det(S), 6))

det: 10.000000000000002
trace: 7.0
invertible: True
eigenvalues: [2. 5.]
prod eig ≈ det : True
sum  eig ≈ trace: True
singular det: 0.0

det: 10.000000000000002
trace: 7.0
invertible: True
eigenvalues: [2. 5.]
prod eig ≈ det : True
sum  eig ≈ trace: True
singular det: 0.0

Why this matters for ML

• Gaussian densities and normalizing flows need $\det$ of a covariance / Jacobian to normalize probabilities — the log-det term shows up in nearly every likelihood.
• Invertibility checks: a zero (or tiny) determinant flags singular / ill-conditioned matrices that break linear solves and blow up regression coefficients.
• The trace appears in loss functions (e.g. $\text{tr}(A^\top A) = \lVert A\rVert_F^2$), regularizers, and the cyclic trace trick that simplifies gradient derivations.

🧪 Try It Yourself

Exercise 1 – Determinant and invertibility

Exercise 2 – det = product of eigenvalues

Exercise 3 – Trace is the sum of eigenvalues

Recap

• The determinant is the signed volume a matrix scales space by; $\det = 0$ ⇔ singular ⇔ not full rank ⇔ no inverse.
• $\det(AB) = \det A \det B$, $\det(A^\top) = \det A$, and it’s basis-invariant.
• The trace sums the diagonal, is cyclic-invariant, and basis-independent.
• $\det(A) = \prod \lambda_i$ and $\text{tr}(A) = \sum \lambda_i$ — both are built from eigenvalues.

Next: those eigenvalues and the directions they belong to — Eigenvalues and Eigenvectors.