Worked examples — Eigenvalues and eigenvectors — characteristic polynomial

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The scenario matrix

Every $2\times 2$ (and small $n$) matrix falls into exactly one of these case classes, decided by two things: the sign of the discriminant of the characteristic polynomial (how the two roots come out) and any degeneracy in the matrix.

#	Case class	What triggers it	Covered by
A	Two distinct real eigenvalues	$\Delta >0$	Ex 1
B	Both eigenvalues negative / sign check	$\Delta >0$, $\det >0$, $\mathrm{tr}<0$	Ex 2
C	Triangular shortcut	zero below diagonal	Ex 3
D	Repeated root, defective	$\Delta =0$, geo $<$ alg	Ex 4
E	Repeated root, NON-defective (scalar matrix)	$\Delta =0$, geo $=$ alg	Ex 5
F	Complex conjugate pair (rotation)	$\Delta <0$	Ex 6
G	Zero eigenvalue / singular matrix	$\det (A)=0$	Ex 7
H	$3\times 3$ + real-world word problem	degree-3 polynomial	Ex 8

Each row of that table is a trap the exam can set. We spring all of them below.

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Example 1 — Case A: two distinct real eigenvalues

Forecast (guess first): $\mathrm{tr}=7$, $\det =12-2=10$. Guess two positive reals whose sum is $7$ and product is $10$. Hold that thought.

1. Build the characteristic polynomial. $p(\lambda )=(4-\lambda )(3-\lambda )-(1)(2)={\lambda }^2-7\lambda +10.$ Why this step? This is ${\lambda }^2-\mathrm{tr}(A)\lambda +\det (A)$ — the direct route from the parent's $2\times 2$ formula. See Trace of a Matrix and Determinants for the two coefficients.

2. Compute the discriminant. $\Delta =7^2-4(10)=49-40=9>0.$ Why this step? $\Delta >0$ confirms we are in Case A — two different real roots.

3. Solve. $\lambda =\frac{7\pm \sqrt{9}}{2}=\frac{7\pm 3}{2}\Rightarrow {\lambda }_1=5,{\lambda }_2=2.$ Why this step? Quadratic formula; the square root of $9$ is clean, so the roots are exact.

4. Eigenvector for $\lambda =5$. Solve $(A-5I)\mathbf{x}=0$: $\left(\begin{array}{cc}-1& 1\\ 2& -2\end{array}\right)\mathbf{x}=0\Rightarrow -x_1+x_2=0\Rightarrow x_1=x_2\Rightarrow \mathbf{x}=\left(\begin{array}{c}1\\ 1\end{array}\right).$ Why this step? Eigenvectors live in the null space of $A-\lambda I$, never in a determinant.

5. Eigenvector for $\lambda =2$. $(A-2I)=\left(\begin{array}{cc}2& 1\\ 2& 1\end{array}\right)$, row gives $2x_1+x_2=0\Rightarrow x_2=-2x_1$, so $\mathbf{x}=\left(\begin{array}{c}1\\ -2\end{array}\right)$.

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Example 2 — Case B: sign-check with both roots negative

Forecast: $\mathrm{tr}=-8$ (sum of roots is negative), $\det =16-1=15>0$ (product positive). Two numbers whose sum is negative and product is positive must both be negative. Guess: two negative reals.

1. Polynomial. $p(\lambda )={\lambda }^2-(-8)\lambda +15={\lambda }^2+8\lambda +15$. Why this step? $-\mathrm{tr}=+8$; note the sign flips because trace is negative — a classic slip.

2. Discriminant. $\Delta =8^2-4(15)=64-60=4>0$ → Case A/B real distinct. Why this step? Confirms real roots before we claim signs.

3. Solve. $\lambda =\frac{-8\pm 2}{2}\Rightarrow {\lambda }_1=-3,{\lambda }_2=-5$. Both negative ✓ (matches forecast). Why this step? $\sqrt{4}=2$; the $-8$ up top drags both roots negative, exactly as the sign argument promised.

4. Eigenvector for $\lambda =-3$. $(A+3I)=\left(\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right)\Rightarrow x_1=x_2\Rightarrow \mathbf{x}=\left(\genfrac{}{}{0ex}{}{1}{1}\right)$.

5. Eigenvector for $\lambda =-5$. $(A+5I)=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\Rightarrow x_1=-x_2\Rightarrow \mathbf{x}=\left(\genfrac{}{}{0ex}{}{1}{-1}\right)$.

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Example 3 — Case C: the triangular shortcut

Forecast: everything below the diagonal is $0$. The parent note said triangular matrices hand you the diagonal for free. Guess $\{6,-1,2\}$.

1. Write $\det (A-\lambda I)$. $\det \left(\begin{array}{ccc}6-\lambda & -3& 4\\ 0& -1-\lambda & 5\\ 0& 0& 2-\lambda \end{array}\right)=(6-\lambda )(-1-\lambda )(2-\lambda ).$ Why this step? The determinant of a triangular matrix is the product of its diagonal — the off-diagonal entries above never enter, because every term of the expansion needs a factor from a zero column.

2. Read off roots. Each factor $=0$ gives $\lambda =6,-1,2$. Why this step? A product is zero exactly when one factor is zero — no algebra needed.

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Example 4 — Case D: repeated root, DEFECTIVE

Forecast: triangular, so $\lambda =2$ twice. But how many directions? Guess before reading.

1. Polynomial. $p(\lambda )=(2-\lambda {)}^2$, so $\lambda =2$ with algebraic multiplicity 2. Why this step? The exponent on $(\lambda -2)$ counts how many times the root repeats.

2. Find eigenvectors. $(A-2I)=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$. The system forces $x_2=0$, $x_1$ free, so $\mathbf{x}=\left(\genfrac{}{}{0ex}{}{1}{0}\right)$ is the only direction. Why this step? Number of free variables = dimension of the null space = geometric multiplicity $=1$.

3. Diagnose. Geometric ($1$) $<$ algebraic ($2$) ⇒ defective, so $A$ is not diagonalizable (see Diagonalization).

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Example 5 — Case E: repeated root, NON-defective

Forecast: it's a scalar matrix; it scales everything by $2$. So maybe every vector is an eigenvector? Guess.

1. Polynomial. $p(\lambda )=(2-\lambda {)}^2$, $\lambda =2$, algebraic multiplicity $2$ — identical to Ex 4 so far. Why this step? Shows the polynomial alone cannot tell defective from non-defective — you must inspect the null space.

2. Eigenvectors. $(A-2I)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$, the zero matrix. Every vector satisfies $0\mathbf{x}=0$, so both $\left(\genfrac{}{}{0ex}{}{1}{0}\right)$ and $\left(\genfrac{}{}{0ex}{}{0}{1}\right)$ are independent eigenvectors. Why this step? The null space of the zero matrix is the whole plane — geometric multiplicity $=2$.

3. Diagnose. Geometric $=$ algebraic $=2$ ⇒ non-defective, fully diagonalizable (it is already diagonal).

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Example 6 — Case F: complex conjugate pair (a rotation)

Forecast: rotation turns every real vector to a new direction, so no real eigenvector can exist. Whatever roots appear must be non-real. Look at the figure.

The red vector and its image $A\mathbf{x}$ (chalk-blue) are perpendicular — no real vector stays on its own line, so no real $\lambda$.

1. Polynomial. $p(\lambda )=(0-\lambda )(0-\lambda )-(-1)(1)={\lambda }^2+1$. Why this step? Straight from ${\lambda }^2-\mathrm{tr}\lambda +\det$ with $\mathrm{tr}=0,\det =1$.

2. Discriminant. $\Delta =0^2-4(1)=-4<0$ → Case F, complex roots. Why this step? A negative under the root forces us into Complex Numbers.

3. Solve. ${\lambda }^2=-1\Rightarrow \lambda =\pm i$, where $i$ is the number with $i^2=-1$. Why this step? $\sqrt{-4}=2i$, so $\lambda =\frac{0\pm 2i}{2}=\pm i$. The two roots are conjugates, as they always are for a real matrix.

4. Read the geometry. The magnitude $|\lambda |=|\pm i|=1$ means the rotation preserves length (no stretch), and the roots being non-real means it genuinely turns vectors — matching the picture.

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Example 7 — Case G: a zero eigenvalue (singular matrix)

Forecast: $\det =2\cdot 2-4\cdot 1=0$. Product of eigenvalues $=\det =0$, so at least one eigenvalue is $0$. Guess the roots.

1. Polynomial. $\mathrm{tr}=4,\det =0$, so $p(\lambda )={\lambda }^2-4\lambda +0=\lambda (\lambda -4)$. Why this step? With $\det =0$ the constant term vanishes, so $\lambda$ factors out immediately.

2. Roots. $\lambda =0$ and $\lambda =4$. Why this step? A zero eigenvalue is the algebraic fingerprint of a singular matrix — $A$ crushes a whole direction to the origin.

3. Eigenvector for $\lambda =0$. Solve $(A-0I)\mathbf{x}=A\mathbf{x}=0$: $2x_1+4x_2=0\Rightarrow x_1=-2x_2\Rightarrow \mathbf{x}=\left(\genfrac{}{}{0ex}{}{-2}{1}\right)$. This is exactly the null space of $A$. Why this step? The $\lambda =0$ eigenvectors ARE the null space — because $A\mathbf{x}=0\mathbf{x}=0$.

4. Eigenvector for $\lambda =4$. $(A-4I)=\left(\begin{array}{cc}-2& 4\\ 1& -2\end{array}\right)\Rightarrow -2x_1+4x_2=0\Rightarrow x_1=2x_2\Rightarrow \mathbf{x}=\left(\genfrac{}{}{0ex}{}{2}{1}\right)$.

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Example 8 — Case H: a $3\times 3$ from a real-world story

Forecast: each column sums to $1$ (a stochastic matrix), so a "do-nothing" eigenvalue $\lambda =1$ should exist (steady state). The bottom row is decoupled, giving another $\lambda =1$. Guess the third.

1. Exploit block structure. The third row/column is separate, so $\lambda =1$ comes from it directly, and the top-left $2\times 2$ block $B=\left(\begin{array}{cc}0.7& 0.2\\ 0.3& 0.8\end{array}\right)$ handles the rest. Why this step? A block-diagonal (here block-triangular) matrix's characteristic polynomial factors into the blocks' — no need to expand a full $3\times 3$.

2. Block polynomial. $\mathrm{tr}(B)=1.5,\det (B)=0.56-0.06=0.5$, so $p_B(\lambda )={\lambda }^2-1.5\lambda +0.5=(\lambda -1)(\lambda -0.5).$ Why this step? $\det (B)=0.7\cdot 0.8-0.2\cdot 0.3=0.56-0.06=0.5$; Vieta gives the clean factoring.

3. All eigenvalues. From the block: $\lambda =1,0.5$; from the decoupled entry: $\lambda =1$. So the full spectrum is $\{1,1,0.5\}$. Why this step? Collect roots from every block.

4. Interpret. The mode with $\lambda =1$ is the steady state (amounts settle, never fade). The mode with $\lambda =0.5$ halves each day ($0.5^k\to 0$), so any imbalance dies out — the ponds equalise. Its eigenvector for $\lambda =1$: $(B-I)=\left(\begin{array}{cc}-0.3& 0.2\\ 0.3& -0.2\end{array}\right)\Rightarrow 3x_1=2x_2\Rightarrow \left(\genfrac{}{}{0ex}{}{2}{3}\right)$, the equilibrium split.

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Recall

Recall Which sign of

$\Delta$ gives which roots? $\Delta >0$ real distinct ::: $\Delta =0$ repeated real ::: $\Delta <0$ complex conjugate pair.

Recall Same repeated

$\lambda$, why can Ex 4 be defective but Ex 5 not? Defectiveness is decided by the rank of $A-\lambda I$ (geometric multiplicity), not by the polynomial. ::: Ex 4 has rank 1 (one eigenvector); Ex 5 has rank 0 (two eigenvectors).

Recall What does a

$\lambda =0$ eigenvalue tell you instantly? The matrix is singular ($\det =0$) and its $\lambda =0$ eigenvectors are its null space.

See also the Cayley–Hamilton Theorem: every one of these matrices satisfies its own characteristic polynomial.