Study Vault Search notes, aliases and full text
Graph
ConceptLinear Algebra used inMSIAI-written · not yet proofread

Invertible Matrix

Also Inverse Matrix · Nonsingular Matrix · Regular Matrix · Singular Matrix · Invertierbare Matrix · Reguläre Matrix

A square matrix A is invertible if there is a matrix A⁻¹ with A A⁻¹ = A⁻¹ A = I; this holds iff A has full rank.

Know these firstMatrix Rank

Properties#

ProofWhy rank nn ⇔ invertible

By rank–nullity, rank⁡(A)=n  ⟺  dim⁡ker⁡(A)=0\operatorname{rank}(A) = n \iff \dim\ker(A) = 0, so the map x↦Axx \mapsto Ax is injective. Also dim⁡im⁡(A)=n\dim \operatorname{im}(A) = n, so im⁡(A)=Rn\operatorname{im}(A) = \mathbb{R}^n and the map is surjective. A bijective linear map has a linear inverse, and its matrix is A−1A^{-1}. Conversely, if rank⁡(A)<n\operatorname{rank}(A) < n, some x≠0x \neq 0 has Ax=0Ax = 0. Then A−1Ax=x≠0=A−10A^{-1}Ax = x \neq 0 = A^{-1}0 is a contradiction.

  • Products and transposes: (AB)−1=B−1A−1(AB)^{-1} = B^{-1} A^{-1} and (A⊤)−1=(A−1)⊤(A^\top)^{-1} = (A^{-1})^\top.
  • Positive definite ⇒ invertible: if A≻0A \succ 0 (positive definite), all eigenvalues are >0> 0, so 00 is not an eigenvalue. MSI: Φ⊤Φ≻0  ⟺  rank⁡(Φ)=d\Phi^\top \Phi \succ 0 \iff \operatorname{rank}(\Phi) = d.
  • 2×2 inverse: (abcd)−1=1ad−bc(d−b−ca)\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} for ad−bc≠0ad - bc \neq 0.

Example#

Sources#

  • MML p. 100 (Thm. 4.1: invertible ⇔ det⁡(A)≠0\det(A) \neq 0, 2×22 \times 2 inverse (4.2)), MML p. 103 (Thm. 4.3: det⁡(A)≠0\det(A) \neq 0 ⇔ full rank)
  • Use in MSI: MSI script p. 30 (Φ⊤Φ≻0  ⟺  rank⁡(Φ)=d\Phi^\top\Phi \succ 0 \iff \operatorname{rank}(\Phi) = d), MSI script p. 32 (ill-posed case: Φ⊤Φ\Phi^\top\Phi singular)
  • Exercise: MSI Sheet 00, task 1 (not published) (invertibility of a 3×33 \times 3 system matrix), task 2 (not published) (invertibility of a Jacobian)

Search

Type In
↑ ↓ move ↵ open ⇥ next section Ctrl↵ results page Prefix # to search text only