Identity, Inverse, and the Matrix 'Undo' Button

The Multiplicative Identity

For scalars: $a \cdot 1 = a$ . The number 1 is the identity. For matrices: $A I = I A = A$ . The identity matrix $I$ — diagonal of 1s, zeros elsewhere — plays the same role.

This isn't trivia. The identity matrix is the anchor every "undo" definition leans on.

The Inverse — Matrix Division, Sort Of

For scalars: $a \cdot a^{- 1} = 1$ (with $a^{- 1} = 1/ a$ ). For matrices: $A \cdot A^{- 1} = A^{- 1} \cdot A = I$ . The inverse "undoes" the original transformation. Rotated 30° clockwise? $A^{- 1}$ rotates 30° counter-clockwise. Doubled in size? $A^{- 1}$ halves it.

Not Every Matrix Has an Inverse

Two requirements:

Square. Rectangular matrices change dimensionality, and you can't fully undo "throwing away a dimension."
Non-zero determinant. If the determinant is 0, the matrix collapses space — multiple inputs map to the same output, so there's no way to undo it. Singular = un-invertible.

For singular or rectangular matrices, the pseudoinverse ( $A^{+}$ ) gives a best-effort generalization. Used everywhere in regression and least-squares.

Practically: Don't Compute the Inverse

To solve $A X = B$ , the textbook says $X = A^{- 1} B$ . The textbook is teaching you the concept, not the algorithm. In practice you call np.linalg.solve(A, B), which uses LU decomposition — faster and more numerically stable than inverting $A$ and multiplying.

The inverse is a concept; solve is the implementation. Use the inverse when you're reasoning. Use solve when you're computing.

Code

Concept vs implementation·python

import numpy as np

A = np.array([[2, 1],
              [1, 3]], dtype=float)

# Identity check
I = np.eye(2)
print(np.allclose(A @ I, A))      # True

# Inverse — the concept
A_inv = np.linalg.inv(A)
print(np.allclose(A @ A_inv, I))  # True (modulo float)

# Solving — the implementation
B = np.array([5, 6])
X_via_inverse = A_inv @ B         # textbook, slower
X_via_solve   = np.linalg.solve(A, B)  # what you should actually use
print(np.allclose(X_via_inverse, X_via_solve))   # True

Exercise

Construct a 2×2 matrix that doubles its input. Find its inverse. Verify that the inverse halves vectors. Then construct a singular 2×2 matrix (e.g. one with two identical rows) and try np.linalg.inv — what error do you get?

Hint

Doubling matrix = 2*np.eye(2). Inverse = 0.5*np.eye(2). Singular = [[1, 1], [2, 2]]. The inversion attempt raises LinAlgError: Singular matrix.

Identity, Inverse, and the Matrix 'Undo' Button

The Multiplicative Identity

The Inverse — Matrix Division, Sort Of

Not Every Matrix Has an Inverse

Practically: Don't Compute the Inverse

Code

External links

Exercise

Progress

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