Eigenvalues and eigenvectors — characteristic polynomial

WHAT are we defining?

WHY does $\det(A-\lambda I)=0$ find the eigenvalues? (Derivation from scratch)

We don't memorise — we build it.

Step 1 — Rewrite the defining equation. $A\mathbf{x} = \lambda\mathbf{x} \quad\Longleftrightarrow\quad A\mathbf{x}-\lambda\mathbf{x}=\mathbf{0}.$ Why this step? We want everything on one side so we can factor out $\mathbf{x}$.

Step 2 — Factor out $\mathbf{x}$ using the identity. We can't write $(A-\lambda)\mathbf{x}$ because $A$ is a matrix and $\lambda$ is a scalar — they live in different worlds. Insert the identity: $\lambda\mathbf{x}=\lambda I\mathbf{x}$. Then $(A-\lambda I)\mathbf{x}=\mathbf{0}.$ Why this step? Now $A-\lambda I$ is a genuine matrix, so this is an ordinary homogeneous linear system.

Step 3 — Demand a nonzero solution. A homogeneous system $M\mathbf{x}=\mathbf{0}$ has a nonzero solution iff $M$ is singular (not invertible). If $M=(A-\lambda I)$ were invertible, multiplying by $M^{-1}$ would force $\mathbf{x}=\mathbf{0}$, which we forbade. Why this step? The whole point is $\mathbf{x}\neq\mathbf0$, and singularity is the gateway condition.

Step 4 — Singular $\Leftrightarrow$ zero determinant. $\boxed{\det(A-\lambda I)=0.}$ Why this step? "Not invertible" and "determinant $=0$" are the same statement. This is a polynomial equation in $\lambda$ — solvable!

Two structural facts (derived, not memorised)

For a $2\times2$ matrix $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$: $p(\lambda)=\det\begin{pmatrix}a-\lambda&b\\c&d-\lambda\end{pmatrix}=(a-\lambda)(d-\lambda)-bc.$ Expand: $p(\lambda)=\lambda^2-(a+d)\lambda+(ad-bc)=\lambda^2-\operatorname{tr}(A)\,\lambda+\det(A).$

Worked Example 1 — a clean $2\times2$

Let $A=\begin{pmatrix}2&1\\1&2\end{pmatrix}$.

Characteristic polynomial: $p(\lambda)=(2-\lambda)^2-1.$ Why this step? Direct $(a-\lambda)(d-\lambda)-bc$ with $a=d=2,\ bc=1$.

$p(\lambda)=\lambda^2-4\lambda+3=(\lambda-1)(\lambda-3).$ So $\lambda_1=1,\ \lambda_2=3$.

Check: $\operatorname{tr}=4=1+3$ ✓, $\det=3=1\cdot3$ ✓.

Eigenvector for $\lambda=3$: solve $(A-3I)\mathbf x=0$: $\begin{pmatrix}-1&1\\1&-1\end{pmatrix}\mathbf x=0\ \Rightarrow\ x_1=x_2\ \Rightarrow\ \mathbf x=\begin{pmatrix}1\\1\end{pmatrix}.$ Why this step? Both rows give $-x_1+x_2=0$; one free direction remains.

Eigenvector for $\lambda=1$: $\begin{pmatrix}1&1\\1&1\end{pmatrix}\mathbf x=0\Rightarrow x_1=-x_2\Rightarrow \mathbf x=\begin{pmatrix}1\\-1\end{pmatrix}$.

Worked Example 2 — a triangular matrix (the shortcut)

Let $A=\begin{pmatrix}5&7\\0&-2\end{pmatrix}$.

$p(\lambda)=\det\begin{pmatrix}5-\lambda&7\\0&-2-\lambda\end{pmatrix}=(5-\lambda)(-2-\lambda)-0.$ Why this step? The lower-left entry is $0$, so the determinant of a triangular matrix is just the product of diagonal entries.

Roots: $\lambda=5,\ \lambda=-2$ — exactly the diagonal entries.

Worked Example 3 — repeated root (algebraic vs geometric)

Let $A=\begin{pmatrix}3&1\\0&3\end{pmatrix}$. $p(\lambda)=(3-\lambda)^2 \Rightarrow \lambda=3 \text{ (twice)}.$ The root $3$ has algebraic multiplicity 2. Eigenvectors: $(A-3I)=\begin{pmatrix}0&1\\0&0\end{pmatrix}$, so $x_2=0$, giving only $\mathbf x=\binom{1}{0}$ — a single independent eigenvector (geometric multiplicity 1).

Common mistakes (Steel-manned)

Active recall

Flashcards

Connections

• Determinants — the singularity test that powers the whole method.
• Null Space and Rank — eigenvectors live in $\ker(A-\lambda I)$.
• Diagonalization — when independent eigenvectors fill space, $A=PDP^{-1}$.
• Trace of a Matrix — sum-of-eigenvalues sanity check.
• Complex Numbers — roots may be complex even for real $A$.
• Cayley–Hamilton Theorem — $A$ satisfies its own characteristic polynomial.

Concept Map

Hinglish (regional understanding)