2.2 Matrices

A matrix equation such as AX=B is solved by multiplying by A⁻¹ on the correct side when A is square and invertible: X=A⁻¹B. For systems, row reduction can avoid forming an inverse.

Check dimensions before multiplying, and interpret a singular matrix as a possible no-solution or many-solution case rather than forcing an inverse.

If A is 2×2 and AX=B, compute X=A⁻¹B and verify by multiplying AX back to B. If det A=0, use row reduction to test consistency instead.

Matrix division is not scalar division, and multiplying by A⁻¹ on the right generally solves a different equation.

For an invertible square matrix A, A⁻¹=adj(A)/det(A). The determinant must be non-zero; for a 2×2 matrix [[a,b],[c,d]], the inverse is (1/(ad−bc))[[d,−b],[−c,a]].

The adjugate is the transpose of the cofactor matrix. Use a determinant test first, then verify the inverse by multiplication or by checking the formula’s dimensions.

For [[2,1],[1,1]], det=1 and the inverse is [[1,−1],[−1,2]]. Multiplying the two matrices gives the identity.

Changing one sign in the 2×2 formula can invalidate every entry, and det=0 means the formula is undefined rather than an inverse with infinite values.

An eigenvector v of A is a non-zero vector satisfying Av=λv; λ is its eigenvalue. The vector’s direction is preserved, though it may be stretched, reversed or collapsed.

Eigenvectors are directions, not arbitrary points. The zero vector is excluded because it would satisfy every λ and carry no directional information.

A diagonal matrix [[3,0],[0,−1]] has eigenvectors along the coordinate axes: the x-direction is scaled by 3 and the y-direction is reversed by factor −1.

An eigenvector is not unique in length, and a repeated eigenvalue can have one or many independent eigendirections.

Eigenvalues satisfy det(A−λI)=0. For each root λ, solve (A−λI)v=0 to find the non-zero eigenvectors in its eigenspace.

The characteristic polynomial may have repeated roots. A repeated algebraic root does not automatically provide the same number of independent eigenvectors; solve the nullspace and count its dimension.

For a diagonal matrix with diagonal entries 3 and −1, det(A−λI)=(3−λ)(−1−λ), giving λ=3 and −1; the corresponding eigenspaces are the coordinate axes.

Do not solve det(A−λ)=0 without the identity matrix, and do not treat λ itself as an eigenvector.

If A has a basis of independent eigenvectors, place them as columns of P and their eigenvalues in matching diagonal positions of D. Then A=PDP⁻¹ and A^n=PD^nP⁻¹.

The order of columns in P must match the order of eigenvalues in D. Diagonal powers are easy, which is why diagonalisation helps with recurrences and repeated transformations.

If P=[v₁ v₂] and Av₁=3v₁, Av₂=−v₂, then D=diag(3,−1); A^n acts by multiplying the two eigen-components by 3^n and (−1)^n.

A matrix with eigenvalues is not automatically diagonalizable; it needs enough linearly independent eigenvectors.

For a square matrix, the sum of eigenvalues equals the trace and the product of eigenvalues equals the determinant, counting algebraic multiplicity.

These invariants quickly check a characteristic polynomial or numerical answer, but they do not determine eigenvectors or guarantee diagonalisation.

For a 2×2 matrix with trace 5 and determinant 6, possible eigenvalues 2 and 3 pass both checks; an alleged pair 1 and 5 fails the determinant check.

Matching trace and determinant is necessary, not sufficient, for a complete eigen-analysis; the eigenspaces still need to be solved.