Linear Algebra

Eigenvalues and determinants of matrices

For a square matrix A of order n, a scalar \lambda is an eigenvalue if Ax=\lambda x for some nonzero vector x. Equivalently, \det(\lambda I-A)=0, where \det(\lambda I-A) is the characteristic polynomial of degree n. Two identities connect eigenvalues with simple matrix quantities: the sum of the eigenvalues equals the trace, \operatorname{tr}(A)=\sum_i \lambda_i, and the product of the eigenvalues equals the determinant, \det A=\prod_i \lambda_i. For a 2\times 2 matrix the characteristic equation is \lambda^2-\operatorname{tr}(A)\,\lambda+\det A=0. So a matrix with zero trace and determinant -1 has eigenvalues 1 and -1, while zero trace and determinant +1 gives \pm j. Eigenvalues are not simply the diagonal entries or products of entries; they come from the whole characteristic polynomial.

A square matrix is invertible if and only if its determinant is nonzero, which is the same as saying that 0 is not an eigenvalue. If two rows are proportional, the determinant is zero and the matrix is singular, whereas a nonzero determinant from cofactor expansion certifies invertibility.

• Shift rule — if Ax=\lambda x, then (A+cI)x=(\lambda+c)x, so every eigenvalue of A+cI is \lambda+c. Example: a matrix with eigenvalues 2 and -3 has A+I with eigenvalues 3 and -2.

If A is nilpotent, meaning A^k=0 for some k, then every eigenvalue satisfies \lambda^k=0, so all eigenvalues are zero. Then all eigenvalues of A+I equal 1, and \det(A+I), the product of these eigenvalues, equals 1.

If Av=\lambda v, then A^k v=\lambda^k v for every k. Applying this term by term to a power series shows that e^{A}v=e^{\lambda}v, so the eigenvector is preserved and the eigenvalue is mapped by the same function.

Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic polynomial. When A is invertible, this lets A^{-1} be written as a polynomial in A by multiplying the identity through by A^{-1}, and comparing coefficients of the minimal polynomial then fixes unknown constants.

If two matrices A and B share the same nonzero eigenvector x with eigenvalue 2, then (A-2I)x=0 and (B-2I)x=0, so both A-2I and B-2I are singular. Adding gives (A+B)x=4x, so A+B-4I is also singular. However, (A+B-2I)x=2x\neq 0, so its singularity is not implied.

Since the eigenvalues of A^2 are the squares \lambda_i^2, the sum of the eigenvalues of A^2 equals \operatorname{tr}(A^2). For instance, a matrix with eigenvalues 1 and 3 has A^2 with eigenvalues 1 and 9, whose sum 10 equals \operatorname{tr}(A^2).

Matrix symmetry types

A square matrix A is symmetric if A^T=A, that is a_{ij}=a_{ji} for all i,j. It is skew-symmetric if A^T=-A, that is a_{ij}=-a_{ji}. Diagonal entries of a skew-symmetric matrix must be zero, because a_{ii}=-a_{ii} forces a_{ii}=0. A quick test for the type of an entry rule is to swap the indices: if the entry formula f(i,j) becomes its negative, the matrix is skew-symmetric; if it is unchanged, the matrix is symmetric. For example, a_{ij}=i+j is symmetric, while a_{ij}=(i-j) is skew-symmetric. An odd function of i-j, such as (i-j)^3, also gives a skew-symmetric matrix.

For a real skew-symmetric matrix A and a real vector x, the scalar x^TAx equals its own transpose, x^TAx=x^TA^Tx=-x^TAx. Hence x^TAx=0 for every real x, regardless of the norm of x.

• Real symmetric matrix — all eigenvalues are real, and eigenvectors belonging to distinct eigenvalues are orthogonal. If Av_1=\lambda_1 v_1 and Av_2=\lambda_2 v_2 with \lambda_1\neq\lambda_2, then \lambda_1 v_1^Tv_2=\lambda_2 v_1^Tv_2, so v_1^Tv_2=0. Two such eigenvectors are never scalar multiples of each other, because they are linearly independent.

Consistency of linear systems

A linear system Ax=b is consistent, meaning it has at least one solution, if and only if \operatorname{rank}(A)=\operatorname{rank}([A\mid b]). If the system is consistent with \operatorname{rank}(A)=r and n unknowns, the solution set has n-r free parameters. It is unique when r=n and has infinitely many solutions when r<n. A quick check is to see whether one equation is a scalar multiple of another, including the right-hand side. If it is, the augmented rank equals the coefficient rank and the system is consistent, but a free variable remains if the rank is less than n.

For a square system Ax=b with a unique solution, the coefficient matrix must be nonsingular, so A^{-1} exists and \operatorname{rank}(A)=n. A product of invertible matrices is invertible, so each power A^m is nonsingular and A^mx=b also has a unique solution with \operatorname{rank}(A^m)=n=\operatorname{rank}(A). Since the system is consistent, \operatorname{rank}([A\mid b])=\operatorname{rank}(A), so the strict inequality \operatorname{rank}(A)<\operatorname{rank}([A\mid b]) cannot hold.