Diagonalizing a Matrix: Eigenvalues and Eigenvectors Explained

Once eigenvalues are found, substitute each back into (A - λI)v = 0 to find corresponding eigenvectors. For each eigenvalue λ_i, solve the homogeneous system to find non-trivial solutions for v. For the matrix [[0, 1], [-2, -3]] with λ = -1, substituting gives [[1, 1], [-2, -2]]v = 0, leading to v₁ = -v₂, so eigenvector v = k[1, -1] for any scalar k. Each eigenvalue corresponds to an eigenspace of solutions.

Eigenvalues (also called characteristic roots or latent roots) and eigenvectors (also called characteristic vectors or latent vectors) of a matrix are found by solving the characteristic equation det(A - λI) = 0, where λ represents the eigenvalues; once eigenvalues are determined, corresponding eigenvectors are found by solving the homogeneous system (A - λI)x = 0 for each eigenvalue, yielding infinitely many solutions that form the eigenspace associated with that eigenvalue.

To find eigenvectors and eigenspaces for a 3x3 matrix, first determine the eigenvalues by solving the characteristic equation det(λI - A) = 0. For each eigenvalue λ, compute the matrix (λI - A) and solve the homogeneous system (λI - A)v = 0 to find the eigenspace, which is the null space of this matrix. The eigenspace consists of all eigenvectors corresponding to that eigenvalue, forming a subspace of R³. For example, with eigenvalues λ = 3 and λ = -3, the eigenspace for λ = 3 is a plane spanned by vectors (1/2, 1, 0) and (1/2, 0, 1), while the eigenspace for λ = -3 is a line spanned by vector (-2, 1, 1).

To find eigenvectors associated with an eigenvalue λ, one solves the homogeneous system (A - λI)x = 0. The solution space (null space) of this system is the eigenspace corresponding to λ. The dimension of this eigenspace equals the geometric multiplicity of λ.

This comprehensive lesson covers the complete methodology for finding eigenvalues and eigenvectors of matrices. The process involves: (1) Forming the characteristic equation det(A - λI) = 0, (2) Calculating the determinant of (A - λI), (3) Solving the resulting characteristic equation (typically quadratic for 2x2 matrices) to find eigenvalues, (4) For each eigenvalue, solving the homogeneous system (A - λI)v = 0 to find corresponding eigenvectors through row reduction. The eigenvector represents a direction unchanged (except for scaling) under the linear transformation. The quadratic formula λ = [-b ± √(b² - 4ac)] / (2a) is used to solve the characteristic equation. The solution space for eigenvectors is parameterized, as any scalar multiple of an eigenvector is also an eigenvector.

Matrix multiplication requires columns of the first matrix to equal rows of the second. For A (m × n) and B (n × p), the product is m × p. Each element is computed by multiplying corresponding row-column elements and summing. Matrix multiplication is generally non-commutative (AB ≠ BA), though square matrices may commute. The inverse A^(-1) satisfies A × A^(-1) = I. For 2 × 2 matrices, A^(-1) = (1/det) × [d -b; -c a]. Inverse exists only for square matrices with non-zero determinant.

A square matrix A of order n is invertible if there exists matrix B such that AB = BA = I (identity matrix). A matrix is invertible if and only if its rank equals n, all rows/columns are linearly independent, and its determinant is non-zero. Key properties include: inverse of product equals product of inverses in reverse order, inverse of transpose equals transpose of inverse, and inverse of inverse returns original matrix. Matrix multiplication involves multiplying rows of the first matrix by columns of the second matrix element by element and summing the products.

To find the inverse of a matrix, use elementary row operations on the augmented matrix [A | I]. Transform the left side (A) into the identity matrix I through row operations. The right side will then become A⁻¹. For example, to find the inverse of [[2,1,1],[1,0,0],[0,1,1]], perform row operations to transform the left side to [[1,0,0],[0,1,0],[0,0,1]], and the right side becomes the inverse. Matrix multiplication involves multiplying rows by columns: for element (i,j), multiply row i of the first matrix by column j of the second matrix, then sum all products. For [[0,1,-1],[5,-4,0]] × [[5,-3,0],[1,1,1]], calculate each element by row-column multiplication.

To prove (AB)⁻¹ = B⁻¹A⁻¹, start with AB = I. Multiply both sides on the left by B⁻¹ and on the right by A⁻¹: B⁻¹(AB)A⁻¹ = B⁻¹IA⁻¹. This simplifies to (B⁻¹A)BA⁻¹ = B⁻¹A⁻¹, which further simplifies to I = B⁻¹A⁻¹. Thus, (AB)⁻¹ = B⁻¹A⁻¹. The algorithm to find the inverse of a square matrix A: (1) Write A = I × A, (2) Apply elementary row and column operations to convert A to the identity matrix I, (3) The same operations applied to I will transform it into A⁻¹. The step-by-step process: (1) Start with the element at position (1,1), convert it to 1 by dividing the entire row by that element, (2) Convert all other elements in that row to 0 using row operations, (3) Move to the next row and repeat the process, (4) Continue until the entire matrix is converted to identity matrix. The same operations applied to the identity matrix will yield the inverse.

Matrix multiplication is not commutative (AB ≠ BA in general). To multiply two matrices: (1) The number of columns in the first matrix must equal the number of rows in the second matrix, (2) Multiply corresponding elements of rows and columns, (3) Sum the products to get each element of the resulting matrix. For example, if A is 2x3 and B is 3x2, then AB is 2x2. The inverse of a matrix A, denoted A⁻¹, satisfies AA⁻¹ = A⁻¹A = I (identity matrix). For a 2x2 matrix [[a, b], [c, d]], A⁻¹ = (1/det(A)) × [[d, -b], [-c, a]].

A set of vectors forms a basis for a vector space if and only if: (1) The vectors are linearly independent; (2) The vectors span the space. To check if a set of vectors is linearly independent, form a matrix with the vectors as columns and check if the determinant is non-zero (for square matrices) or if the rank equals the number of vectors. If the determinant is non-zero, the vectors are linearly independent and can form a basis.

A set of vectors forms a basis for a vector space if it satisfies two properties: linear independence (the only solution to the equation c₁v₁ + c₂v₂ + ... + cpvp = 0 is the trivial solution where all coefficients are zero) and spanning the space (every vector in the space can be written as a linear combination of the basis vectors). To test linear independence, place the vectors in a matrix and reduce to echelon form to check for pivots in each column, or compute the determinant if the matrix is square (non-zero determinant indicates linear independence). To test spanning, check for pivots in each row of the echelon form matrix, or use the determinant test (non-zero determinant indicates spanning). If the number of vectors equals the number of components in each vector, the determinant can simultaneously verify both properties.

A set of vectors forms a basis for a vector space if they are linearly independent and span the space. For n vectors in an n-dimensional space, linear independence is sufficient to guarantee they form a basis. The condition for linear independence is that the only solution to c₁v₁ + c₂v₂ + ... + cₙvₙ = 0 is c₁ = c₂ = ... = cₙ = 0. This means no vector can be expressed as a linear combination of the others.

A set of n vectors in an n-dimensional vector space forms a basis if and only if they are linearly independent, which can be verified by checking that the determinant of the matrix formed by these vectors is non-zero. For example, in R³, the vectors [0,1,1], [1,0,1], and [1,1,0] are linearly independent (determinant = 2 ≠ 0) and thus form a basis for R³.

A basis of a vector space consists of n linearly independent vectors with n components. To determine if vectors form a basis, solve λ1*v1 + λ2*v2 = 0. If only the trivial solution (all λi = 0) exists, the vectors are linearly independent and form a basis. For example, E1 = (1,0) and E2 = (0,1) form a basis in 2D space. However, vectors A1 = (4,2) and A2 = (2,1) are linearly dependent because they yield non-trivial solutions (λ1 = 1, λ2 = -2), meaning they cannot form a basis.