Calculating Eigenvalues and Eigenspaces

Definition Let $A$ be a real $n \times n$ matrix. Its characteristic polynomial is

where $t$ is a scalar variable. Expanding the determinant gives a polynomial in $t$ of degree $n$ with leading coefficient $1$. A real number $\lambda$ is an eigenvalue of $A$ exactly when $\chi_A(\lambda) = 0$. Indeed, $A\mathbf{v} = \lambda\mathbf{v}$ is the same as $(A - \lambda I)\mathbf{v} = \mathbf{0}$, which has a nonzero solution exactly when the square matrix $A - \lambda I$ is singular, that is, when $\det(A - \lambda I) = 0$. Since $\lambda I - A = -(A - \lambda I)$, the two determinants differ only by the factor $(-1)^n$ and vanish together. For a $2 \times 2$ matrix with rows $(a, b)$ and $(c, d)$,

Only square matrices have a characteristic polynomial and eigenvalues.

Definition The characteristic equation of a real $n \times n$ matrix $A$ is $\chi_A(t) = 0$, and its real solutions are exactly the real eigenvalues of $A$. For a $2 \times 2$ matrix it is a quadratic equation $t^2 + pt + q = 0$. When the left side factors as $(t - r)(t - s)$, the roots are $r$ and $s$, because a product of real numbers is zero only when a factor is zero. Whether or not a factorization is easy to see, the number $p^2 - 4q$, the discriminant, decides how many real roots there are. If it is positive there are two distinct real roots, if it is zero there is one repeated real root, and if it is negative there is no real root, so the matrix has no real eigenvalue. When $p^2 - 4q \ge 0$, the quadratic formula gives the real roots

Definition Let $\lambda$ be an eigenvalue of a real $n \times n$ matrix $A$. The eigenspace of $A$ for $\lambda$ is

As a null space it is a subspace of $\mathbb{R}^n$. It contains $\mathbf{0}$, its nonzero vectors are exactly the eigenvectors of $A$ for $\lambda$, and it is not $\{\mathbf{0}\}$ because $\lambda$ is an eigenvalue. The equation $\chi_A(\lambda) = 0$ says only that a nonzero solution exists. The homogeneous system $(A - \lambda I)\mathbf{v} = \mathbf{0}$ still has to be solved to find which vectors they are.

Definition To find a basis of $E_\lambda$, row reduce $A - \lambda I$ to echelon form, solve for the pivot variables in terms of the free variables, and write the general solution as a linear combination with one vector for each free variable. Those vectors are linearly independent and span $E_\lambda$, so they form a basis, and

the number of free variables. A basis is not unique. Any nonzero multiple of a basis vector of a line spans the same line. A request to find the eigenspace asks for all solutions, usually given as a basis, and one eigenvector is not enough when the eigenspace has dimension greater than one.

Definition The trace $\operatorname{tr} A$ of a square matrix is the sum of its diagonal entries. For a real $2 \times 2$ matrix,

If $\chi_A(t) = (t - \lambda_1)(t - \lambda_2)$, comparing coefficients gives

The same holds for a real $n \times n$ matrix whose characteristic polynomial factors into linear factors over the scalars in use, $\chi_A(t) = (t - \lambda_1)\cdots(t - \lambda_n)$, with each eigenvalue listed as many times as its factor appears. The coefficient of $t^{n-1}$ in $\chi_A$ is $-\operatorname{tr} A$ and the constant term is $\chi_A(0) = \det(-A) = (-1)^n \det A$, so the eigenvalues add up to $\operatorname{tr} A$ and multiply to $\det A$. If $\chi_A$ has nonreal roots, the factorization needs complex scalars and the list $\lambda_1, \dots, \lambda_n$ includes those roots, so the real eigenvalues alone cannot complete the check. The checks concern eigenvalues only. They say nothing about the eigenvectors, which are checked by computing $A\mathbf{v}$.