Characteristic Equation, Eigenvalue and Eigenvector

The key idea. Multiplying a vector by a matrix usually changes both its length and its direction. But for a few special directions, $A$ only stretches or shrinks the vector — the direction is preserved. Those directions are the eigenvectors, and the stretch factor is the eigenvalue.

Formally, a non-zero vector $\mathbf{x}$ is an eigenvector of $A$ with eigenvalue $\lambda$ if: $A\mathbf{x} = \lambda\mathbf{x}$

The right-hand side is the vector $\mathbf{x}$ simply scaled by the number $\lambda$.

Two facts to lock in:

• Eigenvectors are only defined up to a scalar: if $\mathbf{x}$ is an eigenvector, so is any non-zero multiple $k\mathbf{x}$. We usually quote the simplest integer vector.
• $\mathbf{x} = \mathbf{0}$ is never an eigenvector (it trivially satisfies $A\mathbf{0} = \lambda\mathbf{0}$ for every $\lambda$, so it carries no information).

In 9231 you only meet matrices with real, distinct eigenvalues — this keeps the algebra clean and guarantees the matrix can be diagonalised.

Cambridge tip. Always verify a found eigenvector by checking $A\mathbf{x} = \lambda\mathbf{x}$ directly. It takes seconds and catches arithmetic slips in the elimination.

Rearrange $A\mathbf{x} = \lambda\mathbf{x}$ as $A\mathbf{x} - \lambda\mathbf{x} = \mathbf{0}$, then factor out $\mathbf{x}$ using the identity matrix $I$: $(A - \lambda I)\mathbf{x} = \mathbf{0}$

For a non-zero $\mathbf{x}$ to exist, the matrix $(A - \lambda I)$ must be singular — otherwise the only solution is $\mathbf{x} = \mathbf{0}$. Singular means zero determinant: $\boxed{\;\det(A - \lambda I) = 0\;}$

This is the characteristic equation. For a $2\times2$ it is a quadratic in $\lambda$; for a $3\times3$, a cubic. Its roots are the eigenvalues.

$2\times2$ shortcut. For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$: $\det(A - \lambda I) = \lambda^2 - (\underbrace{a+d}_{\text{trace}})\lambda + (\underbrace{ad-bc}_{\det A}) = 0$

So the characteristic equation is $\lambda^2 - (\operatorname{tr}A)\lambda + \det A = 0$.

Worked sketch. For $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$: trace $= 4$, $\det = 3$, so $\lambda^2 - 4\lambda + 3 = 0 \;\Rightarrow\; (\lambda - 1)(\lambda - 3) = 0 \;\Rightarrow\; \lambda = 1 \text{ or } 3.$

Useful checks (catch sign errors before you go further):

• Sum of eigenvalues $=$ trace of $A$ (here $1 + 3 = 4$ ✓).
• Product of eigenvalues $= \det A$ (here $1 \times 3 = 3$ ✓).

Cambridge tip. Form $(A - \lambda I)$ by subtracting $\lambda$ from each diagonal entry only — never from off-diagonal entries. Then take the determinant carefully.

Finding the eigenvector for each eigenvalue $\lambda$: substitute $\lambda$ back into $(A - \lambda I)\mathbf{x} = \mathbf{0}$ and solve the (singular) system. Because the system is singular, you get a family of solutions — pick the simplest.

Example ($A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$, $\lambda = 3$): $(A - 3I)\mathbf{x} = \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{0} \;\Rightarrow\; -x + y = 0 \;\Rightarrow\; y = x.$ Take $\mathbf{x} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$. For $\lambda = 1$ the same method gives $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$.

Diagonalisation. Place the eigenvectors as the columns of $Q$, and the matching eigenvalues on the diagonal of $D$ (in the same order): $Q = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix},\qquad D = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}.$ Then $\boxed{\;M = QDQ^{-1}\;}$ holds whenever $M$ has a full set of independent eigenvectors (guaranteed here, since the eigenvalues are distinct).

Powers become easy. Since $D$ is diagonal, $D^n$ just raises each diagonal entry to the $n$: $M^n = QD^nQ^{-1},\qquad D^n = \begin{pmatrix} 3^n & 0 \\ 0 & 1^n \end{pmatrix}.$ This turns "multiply $M$ by itself $n$ times" into a single matrix triple-product — the central exam application.

Cambridge tip. The order MUST match: the eigenvector in column 1 of $Q$ pairs with the eigenvalue in position $(1,1)$ of $D$. Swapping one without the other gives a wrong $M$.

The Cayley–Hamilton theorem states: every square matrix satisfies its own characteristic equation. If the characteristic equation is $\lambda^2 - p\lambda + q = 0$, then replacing $\lambda$ by the matrix (and the constant by $qI$) gives: $A^2 - pA + qI = 0.$

This single identity is a powerful tool.

Use 1 — find the inverse. Rearrange to isolate $I$: $A^2 - pA + qI = 0 \;\Rightarrow\; qI = pA - A^2 \;\Rightarrow\; A^{-1} = \frac{1}{q}\left(pI - A\right)$ (after multiplying through by $A^{-1}$). For $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ with $\lambda^2 - 4\lambda + 3 = 0$: $A^{-1} = \frac{1}{3}(4I - A) = \frac{1}{3}\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}.$

Use 2 — reduce high powers. From $A^2 = pA - qI$ you can express $A^2$ (and hence $A^3, A^4, \ldots$) as a linear combination of $A$ and $I$. For example: $A^3 = A \cdot A^2 = A(pA - qI) = pA^2 - qA = p(pA - qI) - qA = (p^2 - q)A - pqI.$ So any power of $A$ collapses to "(number)$A$ + (number)$I$" — no repeated matrix multiplication needed.

Why it works. Each eigenvalue is a root of the characteristic polynomial, and the theorem promotes that scalar fact to the matrix itself. It is a guaranteed identity — you can always quote and use it.

Cambridge tip. When asked to find $A^{-1}$ or a high power "using the Cayley–Hamilton theorem", you must start from the characteristic equation and substitute the matrix — a direct inverse or repeated multiplication earns no method marks for that instruction.