Worked examples — Diagonalization — conditions, procedure

You have seen the procedure. Now we drill it against every kind of matrix the topic can hand you — not just the friendly ones. The goal: after this page, no exam matrix should surprise you, because you will have seen its type already.

Before we start, one reminder of the vocabulary we will use non-stop (all built in the parent note and in Eigenvalues and Eigenvectors):

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The scenario matrix

Every diagonalization problem falls into exactly one of these cells. We will hit all of them.

#	Cell class	What is special	Diagonalizable?
A	$n$ distinct real eigenvalues	easiest case, free win	✅ always
B	Repeated eigenvalue, $g=a$	enough arrows despite repeat	✅
C	Repeated eigenvalue, $g<a$ (defective)	too few arrows	❌
D	Complex eigenvalues (real matrix)	rotation hidden inside	✅ over $\mathbb{C}$, ❌ over $\mathbb{R}$
E	Zero eigenvalue (singular $A$)	$\det A=0$, a squashed direction	depends ($g$ vs $a$)
F	Symmetric matrix	Spectral Theorem guarantee	✅ always, orthogonally
G	Real-world / word problem	population / [[Matrix Powers and Exponentials	$A^n$]]
H	Exam twist: given $D,P$, reconstruct $A$	run the machine backwards	—

Legend for signs we cover: eigenvalues that are positive, negative, zero, and complex ($\pm i$) all appear below, and both symmetric and non-symmetric matrices.

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Example A — distinct real eigenvalues (positive & negative)

Forecast: the two entries off the diagonal are equal, so this matrix is symmetric. Guess: two real eigenvalues, and (spoiler) perpendicular eigenvectors.

1. Characteristic equation. $\det (A-\lambda I)=(1-\lambda {)}^2-4={\lambda }^2-2\lambda -3=(\lambda -3)(\lambda +1).$ So ${\lambda }_1=3,{\lambda }_2=-1$. Why this step? Eigenvalues are the $\lambda$ that make $A-\lambda I$ singular; the roots of this polynomial are exactly those values. Notice one is positive, one negative — a valid, common case.

2. Eigenvector for $\lambda =3$. $A-3I=\left(\begin{array}{cc}-2& 2\\ 2& -2\end{array}\right)$, row $-2x_1+2x_2=0\Rightarrow x_1=x_2$. Take ${\mathbf{p}}_1=\left(\begin{array}{c}1\\ 1\end{array}\right)$. Why this step? We need the direction that is only stretched. The null space of $A-3I$ is that direction.

3. Eigenvector for $\lambda =-1$. $A+I=\left(\begin{array}{cc}2& 2\\ 2& 2\end{array}\right)\Rightarrow x_1=-x_2$. Take ${\mathbf{p}}_2=\left(\begin{array}{c}1\\ -1\end{array}\right)$. Why this step? Same reasoning for the second eigenvalue. Check: ${\mathbf{p}}_1\cdot {\mathbf{p}}_2=1-1=0$ — perpendicular, as forecast (symmetric matrices always give this, cell F).

4. Assemble (order must match!): $P=\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right),D=\left(\begin{array}{cc}3& 0\\ 0& -1\end{array}\right).$

Verify: $\det P=-2\ne 0$ (invertible ✓). Check column 1: $A{\mathbf{p}}_1=\left(\begin{array}{c}3\\ 3\end{array}\right)=3{\mathbf{p}}_1$ ✓; column 2: $A{\mathbf{p}}_2=\left(\begin{array}{c}-1\\ 1\end{array}\right)=-1\cdot {\mathbf{p}}_2$ ✓.

Look at the picture: the red and mint arrows are the eigenvector directions. The red one grows $3\times$; the mint one flips (that's the negative eigenvalue) and stays the same length. Nothing else in the plane keeps its direction.

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Example B — repeated eigenvalue, still diagonalizable ($g=a$)

Forecast: $\lambda =5$ appears twice. Repeated $\Rightarrow$ danger. But this is diagonal already, so surely it survives. Guess: yes, with $g(5)=2$.

1. Eigenvalues. Being diagonal, they sit on the diagonal: $\lambda =5$ (twice), $\lambda =2$. So $a(5)=2,a(2)=1$. Why this step? For a diagonal (or triangular) matrix, $\det (A-\lambda I)=\prod ({\text{diag}}_i-\lambda )$, so roots = diagonal entries. No work needed.

2. Geometric multiplicity of $\lambda =5$. $A-5I=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& -3\end{array}\right)$. The null space is all vectors with third component $0$ — that's a 2-dimensional plane. So $g(5)=2$. Why this step? We must check $g=a$. Here $g(5)=2=a(5)$, so the repeat is "healthy."

3. Eigenvectors. For $\lambda =5$: ${\mathbf{e}}_1=(1,0,0),{\mathbf{e}}_2=(0,1,0)$. For $\lambda =2$: ${\mathbf{e}}_3=(0,0,1)$. Why this step? Three independent arrows for a $3\times 3$ matrix $\Rightarrow$ they span the space $\Rightarrow$ diagonalizable.

4. Assemble. $P=I$, $D=A$. (It was already diagonal.)

Verify: $g(5)+g(2)=2+1=3=n$ ✓. This is exactly the parent note's lesson — a repeated eigenvalue is fine when it hands you enough arrows.

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Example C — defective: repeated eigenvalue with $g<a$

Forecast: upper-triangular with a repeated $4$ and a nonzero corner — smells like a shear. Guess: only one eigenvector, so defective.

1. Eigenvalues. Triangular $\Rightarrow \lambda =4$ twice, $a(4)=2$. Why this step? Triangular matrices read eigenvalues off the diagonal instantly.

2. Eigenvectors. $A-4I=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$. Solve: second component $x_2=0$, $x_1$ free. Only one arrow $\mathbf{v}=(1,0)$, so $g(4)=1$. Why this step? We count independent solutions of $(A-4I)\mathbf{x}=0$. Only one $\Rightarrow$ trouble.

3. Compare. $g(4)=1<a(4)=2$. There are not two independent arrows, so no invertible $P$ can exist. Why this step? $P$ needs $n=2$ independent columns; we have $1$. Diagonalization is impossible — this matrix belongs in Jordan Normal Form territory instead.

Verify: $g<a$ confirmed. Sanity: a shear pushes points sideways; only the horizontal axis stays put, so geometrically there is genuinely one fixed direction — matches $g=1$.

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Example D — complex eigenvalues (a rotation)

Forecast: rotation turns every real direction, so there is no real eigenvector. Guess: eigenvalues are complex; diagonalizable only over $\mathbb{C}$.

1. Characteristic equation. $\det (A-\lambda I)={\lambda }^2+1=0\Rightarrow \lambda =\pm i$. Why this step? Where a real matrix has no real root, complex roots appear in a conjugate pair. Here $i=\sqrt{-1}$, the number whose square is $-1$.

2. Eigenvector for $\lambda =i$. $A-iI=\left(\begin{array}{cc}-i& -1\\ 1& -i\end{array}\right)$. Row 2: $x_1-i{x}_2=0\Rightarrow x_1=i{x}_2$. Take ${\mathbf{p}}_1=\left(\begin{array}{c}i\\ 1\end{array}\right)$. Why this step? The eigenvector lives in ${\mathbb{C}}^2$ now. That's allowed — we just can't draw it as a real arrow.

3. Eigenvector for $\lambda =-i$. Conjugate everything: ${\mathbf{p}}_2=\left(\begin{array}{c}-i\\ 1\end{array}\right)$. Why this step? For a real matrix, the eigenvector of $\bar{\lambda }$ is the conjugate of the eigenvector of $\lambda$ — free.

4. Assemble over $\mathbb{C}$. $P=\left(\begin{array}{cc}i& -i\\ 1& 1\end{array}\right),D=\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right).$

Verify: $A{\mathbf{p}}_1=\left(\begin{array}{c}-1\\ i\end{array}\right)=i\left(\begin{array}{c}i\\ 1\end{array}\right)$ ✓ (since $i\cdot i=-1$). Over $\mathbb{R}$: no real eigenvector exists, so not diagonalizable over $\mathbb{R}$ — the parent note's second "steel-man" mistake made concrete.

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Example E — zero eigenvalue (singular matrix)

Forecast: this flattens the plane onto the $x$-axis. A flattened direction gets multiplied by $0$. Guess: eigenvalues $1$ and $0$, both real, diagonalizable.

1. Eigenvalues. Diagonal $\Rightarrow \lambda =1,\lambda =0$. Why this step? $\lambda =0$ is a perfectly legal eigenvalue; it means $A$ is singular ($\det A=0$) and has a nonzero null space.

2. Eigenvector for $\lambda =1$. $A-I=\left(\begin{array}{cc}0& 0\\ 0& -1\end{array}\right)\Rightarrow x_2=0$. ${\mathbf{p}}_1=(1,0)$. Why this step? The $x$-axis is untouched (kept at full length), so its eigenvalue is $1$.

3. Eigenvector for $\lambda =0$. $A-0I=A\Rightarrow x_1=0$. ${\mathbf{p}}_2=(0,1)$. Why this step? The $y$-direction is crushed to zero — that IS the eigenvector for $\lambda =0$, and it's exactly $\ker A$.

4. Assemble. $P=I,D=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$. Two independent arrows $\Rightarrow$ diagonalizable.

Verify: $g(1)=1,g(0)=1$, sum $=2=n$ ✓. A zero eigenvalue does not block diagonalization; it only means one axis is squashed flat.

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Example F — symmetric matrix (Spectral Theorem, orthogonal $P$)

Forecast: symmetric $\Rightarrow$ Spectral Theorem promises real eigenvalues and perpendicular eigenvectors we can turn into a rotation matrix.

1. Eigenvalues. $\det (A-\lambda I)=(2-\lambda {)}^2-1={\lambda }^2-4\lambda +3=(\lambda -3)(\lambda -1)$. So $\lambda =3,1$ — both real, as guaranteed. Why this step? Symmetric matrices never produce complex eigenvalues; this is why cell F is always ✅.

2. Eigenvectors. $\lambda =3$: $x_1=x_2\Rightarrow (1,1)$. $\lambda =1$: $x_1=-x_2\Rightarrow (1,-1)$. Dot product $=0$ ✓ perpendicular. Why this step? We want an orthonormal basis; distinct-eigenvalue eigenvectors of a symmetric matrix are automatically orthogonal.

3. Normalize (divide each by its length $\sqrt{2}$) to get an orthogonal $P$: $P=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right),D=\left(\begin{array}{cc}3& 0\\ 0& 1\end{array}\right).$ Why this step? With unit, perpendicular columns, $P^{\top }P=I$, so $P^{-1}=P^{\top }$ — no messy inverse needed.

Verify: $P^{\top }P=I$ (check: rows are unit and orthogonal) ✓, and $\det P=-1$ (an orthogonal matrix, $|\det |=1$) ✓.

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Example G — word problem: population model & $A^n$

Forecast: column sums $=1$ always forces an eigenvalue $\lambda =1$ (steady state). Guess: the other eigenvalue is between $0$ and $1$, so its contribution dies out under repeated powers.

1. Eigenvalues. $\det (A-\lambda I)=(0.9-\lambda )(0.8-\lambda )-0.02={\lambda }^2-1.7\lambda +0.7=(\lambda -1)(\lambda -0.7)$. So $\lambda =1,0.7$. Why this step? $\lambda =1$ confirms a steady state; $\lambda =0.7<1$ is the "transient" that shrinks.

2. Steady-state eigenvector ($\lambda =1$). $A-I=\left(\begin{array}{cc}-0.1& 0.2\\ 0.1& -0.2\end{array}\right)\Rightarrow x_1=2x_2$. So ${\mathbf{p}}_1=(2,1)$; as fractions of the population, $(\frac{2}{3},\frac{1}{3})$. Why this step? This is the distribution the town converges to, regardless of the start.

3. Transient eigenvector ($\lambda =0.7$). $A-0.7I=\left(\begin{array}{cc}0.2& 0.2\\ 0.1& 0.1\end{array}\right)\Rightarrow x_1=-x_2$, so ${\mathbf{p}}_2=(1,-1)$. Why this step? Under $A^n$, this direction is multiplied by $0.7^n\to 0$, so it vanishes — the town forgets its initial imbalance.

4. Long run. $A^n=P{D}^n{P}^{-1}$ and $D^n=\left(\begin{array}{cc}1& 0\\ 0& 0.7^n\end{array}\right)\to \left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$. So every start funnels to the $(2:1)$ split, i.e. $\approx 66.7\%$ city, $33.3\%$ suburb.

Verify: column sums of $A$ are $1$ ✓, so $\lambda =1$ must be an eigenvalue. Eigenvector fractions $\frac{2}{3}+\frac{1}{3}=1$ ✓. Units: people/people (a fraction) — dimensionless ✓.

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Example H — exam twist: rebuild $A$ from $D$ and $P$

Forecast: running diagonalization backwards — you already have the eigen-data, just multiply out. This is where Similar Matrices and Change of Basis meet.

1. Compute $P^{-1}$. For $2\times 2$: $P^{-1}=\frac{1}{\det P}\left(\begin{array}{cc}-1& -1\\ -1& 1\end{array}\right)$ with $\det P=-2$, giving $P^{-1}=\left(\begin{array}{cc}1/2& 1/2\\ 1/2& -1/2\end{array}\right)$. Why this step? $A=PD{P}^{-1}$ needs the inverse; the $2\times 2$ inverse formula (swap diagonal, negate off-diagonal, divide by det) does it fast.

2. Multiply $PD$. $\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)\left(\begin{array}{cc}2& 0\\ 0& 4\end{array}\right)=\left(\begin{array}{cc}2& 4\\ 2& -4\end{array}\right)$ (scale column 1 by $2$, column 2 by $4$). Why this step? Multiplying by a diagonal matrix on the right just scales columns — cheap.

3. Multiply by $P^{-1}$. $\left(\begin{array}{cc}2& 4\\ 2& -4\end{array}\right)\left(\begin{array}{cc}1/2& 1/2\\ 1/2& -1/2\end{array}\right)=\left(\begin{array}{cc}3& -1\\ -1& 3\end{array}\right)$. Why this step? This finishes $PD{P}^{-1}$, giving the original transformation back in standard coordinates.

Verify: $A=\left(\begin{array}{cc}3& -1\\ -1& 3\end{array}\right)$. Check trace: $3+3=6=2+4$ (sum of eigenvalues) ✓. Check determinant: $9-1=8=2\cdot 4$ (product of eigenvalues) ✓. Symmetric, as forced by orthogonal-ish columns of $P$.

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Recall Which cell, which verdict?

Distinct real eigenvalues — diagonalizable? ::: Yes, always (Cell A). Repeated eigenvalue with $g=a$ — diagonalizable? ::: Yes (Cell B). Repeated eigenvalue with $g<a$ (defective) — diagonalizable? ::: No (Cell C). Real rotation with eigenvalues $\pm i$ — diagonalizable over $\mathbb{R}$? ::: No; only over $\mathbb{C}$ (Cell D). A zero eigenvalue ($\det A=0$) — does it block diagonalization? ::: No; it just marks a squashed direction (Cell E). Symmetric matrix — diagonalizable? ::: Always, and orthogonally, by the Spectral Theorem (Cell F). Quickest sanity checks on a final $A$? ::: Trace $=\sum {\lambda }_i$ and $\det =\prod {\lambda }_i$.

Back to the parent topic.