Finding eigenspaces: Worked examples — Maths deep dive

This page is the drill floor for Finding eigenspaces. The parent note built the recipe; here we run it against every kind of matrix you can meet — clean roots, repeated roots, a singular matrix, a rotation with no real eigenvectors, a triangular matrix you read off by eye, a symmetric $3 \times 3$ , and a word problem. Each worked example is tagged with the cell of the scenario matrix it covers, so by the end you have hit every case.

Before we start, some notation we lean on, so nothing is used unearned:

Definition Size, dimension, rank, and the two multiplicities

We use these five words repeatedly below, so pin them down now in plain language:

$n$ = the size of the matrix: an $n \times n$ matrix acts on vectors with $n$ entries. For a $2 \times 2$ matrix $n = 2$ ; for the $3 \times 3$ example $n = 3$ .
$dim$ (dimension) of a space = the number of independent directions it contains. A line has $dim = 1$ ; a flat plane has $dim = 2$ ; a single point (just $0$ ) has $dim = 0$ .
rank of a matrix $M$ = the number of independent rows left after row-reduction — i.e. how many genuine equations $M v = 0$ actually imposes. A $2 \times 2$ matrix that reduces to one nonzero row has rank $1$ ; the zero matrix has rank $0$ .
algebraic multiplicity of $λ$ = how many times $(λ - that value)$ appears as a factor of the characteristic polynomial $det (A - λ I)$ (see Characteristic polynomial). A double root $(λ - 3)^{2}$ has algebraic multiplicity $2$ .
geometric multiplicity of $λ$ = the dimension of its eigenspace, $dim E_{λ}$ — how many independent eigenvectors it actually has. By the Rank-Nullity theorem these connect through $dim E_{λ} = n - rank (A - λ I),$ i.e. (number of variables) minus (number of independent equations) = (number of free directions).

The scenario matrix

Every eigenvalue problem falls into one of these boxes. Our job below is to fill all of them.

Cell	Scenario	What could go wrong / be surprising	Example
A	Two distinct real eigenvalues, both nonzero	routine, but watch sign of $λ$	Ex 1
B	An eigenvalue equal to zero ( $A$ singular)	"zero can't be an eigenvalue" trap	Ex 2
C	A negative eigenvalue (a flip)	eigenvector still a real line, just reversed	Ex 3
D	Repeated root, full (2D) eigenspace	geometric = algebraic multiplicity	Ex 4
E	Repeated root, deficient eigenspace	geometric < algebraic → not diagonalizable	Ex 5
F	No real eigenvalues (pure rotation)	roots are complex; no real line survives	Ex 6
G	Triangular matrix	eigenvalues read straight off the diagonal	Ex 7
H	Symmetric $3 \times 3$	always real, eigenspaces perpendicular	Ex 8
I	Word problem (population / limiting behaviour)	interpret dominant eigenvalue physically	Ex 9

Ex 1 — Cell A: two distinct nonzero real eigenvalues

Forecast: Guess before computing — will the two eigenvalues be far apart or close? (The row sums are $5$ and $5$ , a hint that $(1 1)$ might be special.)

Form $A - λ I$ . $A - λ I = (4 - λ 2 1 3 - λ) .$ Why this step? Eigenvectors exist only where this matrix is singular, so we must build it first.
Set the determinant to zero (see Determinants). $det = (4 - λ) (3 - λ) - (1) (2) = λ^{2} - 7 λ + 10 = 0.$ Why this step? A homogeneous system $M v = 0$ has a nonzero solution iff $det M = 0$ . No singularity, no eigenvector.
Solve the quadratic. $λ^{2} - 7 λ + 10 = (λ - 2) (λ - 5) = 0 \Rightarrow λ = 2, 5.$ Why this step? These are the two stretch factors; both positive, so both are genuine stretches (no flip).
Eigenspace for $λ = 5$ . $A - 5 I = (- 1 2 1 - 2) \to (10 - 1 0)$ , so $v_{1} = v_{2}$ . $E_{5} = span {(11)} .$ Why this step? Row-reducing turns the matrix equation into one plain equation $v_{1} - v_{2} = 0$ ; the free variable traces the whole line.
Eigenspace for $λ = 2$ . $A - 2 I = (2211) \to (2010)$ , so $2 v_{1} + v_{2} = 0$ , i.e. $v_{2} = - 2 v_{1}$ . $E_{2} = span {(1 - 2)} .$ Why this step? Same row-reduction idea for the second eigenvalue; the single surviving equation $2 v_{1} + v_{2} = 0$ defines the whole eigen-line.

Verify: $A (1 1) = (5 5) = 5 (1 1)$ ✓ and $A (- 2 1) = (- 4 2) = 2 (- 2 1)$ ✓.

How to read this figure. The axes are the ordinary $x$ – $y$ plane (standard Cartesian coordinates, origin in the centre). The blue line is $E_{5}$ , the direction $(1, 1)$ ; the yellow line is $E_{2}$ , the direction $(1, - 2)$ . The grey arrow is a generic vector $v$ that lies on neither eigen-line, and the red arrow is its image $A v$ — notice it points in a different direction (it got rotated). The lesson to take away: only arrows sitting on the two coloured lines keep their direction under $A$ ; everything else swings to a new line.

Ex 2 — Cell B: an eigenvalue equal to zero

Forecast: The bottom row is half the top row. Predict: is this matrix invertible? If not, what does that force $det A$ — and hence one eigenvalue — to be?

Characteristic equation. $det (A - λ I) = (2 - λ)^{2} - 4 = λ^{2} - 4 λ = λ (λ - 4) = 0.$ Why this step? Standard: singularity condition. Notice the constant term is $det A = 4 - 4 = 0$ .
Read the roots. $λ = 0$ and $λ = 4$ . Why this step? The root $λ = 0$ appearing means $det A = 0$ : the matrix squashes some direction to nothing. That squashed direction is an eigenvector with eigenvalue $0$ .
Eigenspace $E_{0} = Null (A)$ . Solve $A v = 0$ : $(2142) \to (1020)$ , so $v_{1} = - 2 v_{2}$ . $E_{0} = span {(- 2 1)} .$ Why this step? $A - 0 \cdot I = A$ , so the eigenspace of $0$ is literally the null space of $A$ .
Eigenspace $E_{4}$ . $A - 4 I = (- 2 1 4 - 2) \to (10 - 2 0)$ , so $v_{1} = 2 v_{2}$ : $E_{4} = span {(1 2)}$ . Why this step? Same recipe for the nonzero eigenvalue; the surviving equation $v_{1} - 2 v_{2} = 0$ traces the eigen-line for $λ = 4$ .

Verify: $A (1 - 2) = (0 0) = 0 \cdot (1 - 2)$ ✓ and $A (1 2) = (4 8) = 4 (1 2)$ ✓.

Ex 3 — Cell C: a negative eigenvalue (a flip)

Forecast: This matrix swaps a vector's two entries: $(b a) \mapsto (a b)$ . Which arrow gets reversed by a swap? Which stays put?

Characteristic equation. $det (- λ 1 1 - λ) = λ^{2} - 1 = 0 \Rightarrow λ = \pm 1.$ Why this step? We expect a positive and a negative root because swapping keeps some direction and flips another.
$λ = 1$ . $A - I = (- 1 1 1 - 1) \to v_{1} = v_{2}$ : $E_{1} = span {(1 1)}$ . This is the "stays the same" direction (swap of $(1 1)$ is $(1 1)$ ). Why this step? Row-reducing gives the one equation $v_{1} = v_{2}$ , whose solution line is the fixed direction we predicted.
$λ = - 1$ . $A + I = (1111) \to v_{1} = - v_{2}$ : $E_{- 1} = span {(- 1 1)}$ . Why this step? $λ = - 1$ means the arrow is flipped through the origin — same line, opposite direction. Swapping $(- 1 1)$ gives $(1 - 1) = - (- 1 1)$ ✓.

Verify: $A (1 1) = (1 1) = 1 \cdot (1 1)$ ✓ and $A (- 1 1) = (1 - 1) = - 1 \cdot (- 1 1)$ ✓.

How to read this figure. Standard $x$ – $y$ axes again. The blue line is $E_{1}$ : the direction $(1, 1)$ that the swap leaves completely alone. The red line is $E_{- 1}$ : the direction $(1, - 1)$ . The red arrow on the lower-right is the eigenvector $(1, - 1)$ ; the red arrow on the upper-left is its image $A (1, - 1) = (- 1, 1)$ — the same line, but the arrowhead has flipped to the opposite end. The lesson: a negative eigenvalue keeps the vector on its own line but reverses which way it points.

Ex 4 — Cell D: repeated root, full 2D eigenspace

Forecast: $λ = 3$ will be a double root. Guess: is the eigenspace a line or the whole plane?

Characteristic equation. $(3 - λ)^{2} = 0 \Rightarrow λ = 3$ , algebraic multiplicity $2$ . Why this step? Setting $det (A - λ I) = 0$ is always the first move; here it hands us a single repeated root, so we must next check how big its eigenspace actually is.
Build $A - 3 I = (0000)$ . Why this step? Whatever it row-reduces to tells us how many free variables (hence dimensions) the eigenspace has.
Every vector solves $0 \cdot v = 0$ . Both variables are free. $E_{3} = span {(10), (01)} = R^{2} .$ Why this step? With the zero matrix there are no equations to satisfy, so both coordinates are free — the null space, and hence the eigenspace, is the whole plane.

Here the size is $n = 2$ , and $rank (A - 3 I) = 0$ (the zero matrix imposes no equations), so $dim E_{3} = n - rank (A - 3 I) = 2 - 0 = 2$ — the geometric multiplicity (the dimension of the eigenspace) equals the algebraic multiplicity (the double root). So $A$ is diagonalizable (trivially, it's already diagonal). See Diagonalization.

Verify: $A (0 1) = 3 (0 1)$ and $A (1 0) = 3 (1 0)$ ✓.

Ex 5 — Cell E: repeated root, deficient eigenspace

Forecast: Same double root $λ = 3$ as Ex 4. Will the eigenspace again be the whole plane? Predict before step 3.

Characteristic equation. $(3 - λ)^{2} = 0 \Rightarrow λ = 3$ , algebraic multiplicity $2$ — identical to Ex 4. Why this step? We start with the singularity condition as always; the point of interest is that the polynomial is identical to Ex 4, yet the eigenspace will turn out different.
Build $A - 3 I = (0010)$ . Why this step? The off-diagonal $1$ survives, and it will cost us a dimension.
Solve. The single equation is $v_{2} = 0$ ; $v_{1}$ is free. $E_{3} = span {(10)} .$ Why this step? The surviving row forces $v_{2} = 0$ , leaving only one free variable — so the eigenspace is a single line, not the plane.

Now count the dimension. Here $n = 2$ and the matrix $(0010)$ has one independent nonzero row, so $rank (A - 3 I) = 1$ . Therefore $dim E_{3} = n - rank (A - 3 I) = 2 - 1 = 1$ (using the Rank-Nullity theorem relation from the definitions box). So the algebraic multiplicity is $2$ but the geometric multiplicity is $1$ , and since $1 < 2$ , this matrix is not diagonalizable.

Verify: $A (0 1) = (0 3) = 3 (0 1)$ ✓. And there is genuinely no second independent eigenvector for $λ = 3$ .

Recall Same root, different fate

Ex 4 and Ex 5 have the same characteristic polynomial $(λ - 3)^{2}$ . The repeat count only bounds the eigenspace dimension from above — the actual dimension comes from the rank of $A - λ I$ . Why does the off-diagonal 1 kill a dimension? ::: It raises $rank (A - 3 I)$ from $0$ to $1$ , so $dim E_{3} = 2 - 1 = 1$ .

Ex 6 — Cell F: no real eigenvalues (a rotation)

Forecast: A pure rotation turns every arrow. Predict: how many arrows can stay on their own line? What must that mean for real eigenvalues?

Characteristic equation. $det (- λ 1 - 1 - λ) = λ^{2} + 1 = 0.$ Why this step? Same recipe — but now watch the sign of the constant.
Solve. $λ^{2} = - 1$ . There is no real number whose square is $- 1$ ; the roots are $λ = \pm i$ (imaginary). Why this step? A real $λ$ would demand a real line held fixed. A $9 0^{\circ}$ rotation fixes no real direction, so the algebra must refuse a real answer — and it does.
Conclusion. Over the real numbers, $A$ has no eigenvalues and no eigenspaces. (Over the complex numbers it has $λ = \pm i$ , but no real invariant line.) Why this step? We report the real-vs-complex distinction because the whole point of this cell is a matrix with no real eigenvector to find.

Verify: Take any nonzero real $v = (b a)$ ; $A v = (a - b)$ . For this to equal $λ (b a)$ we would need $- b = λa$ and $a = λb$ , giving $a = - λ^{2} a$ , i.e. $λ^{2} = - 1$ — impossible for real $λ$ unless $a = b = 0$ . ✓ So no real eigenvector exists.

How to read this figure. Standard $x$ – $y$ axes, origin centred. The blue arrows are eight sample unit vectors $v$ pointing in all directions around the circle. Each green arrow is that vector's image $A v$ — the quarter-turn rotation. Compare any blue arrow with its green partner: the green one has swung $9 0^{\circ}$ counter-clockwise, so it lands on a different line every single time. The lesson: because no arrow ends up on its original line, there is no real eigenvector — the visual signature of complex eigenvalues.

Ex 7 — Cell G: triangular matrix (read off the diagonal)

Forecast: The lower-left corner is all zeros. Guess the eigenvalues without computing a determinant.

Use the triangular shortcut. For an (upper- or lower-) triangular matrix, $det (A - λ I)$ is the product of the diagonal entries of $A - λ I$ : $det (A - λ I) = (2 - λ) (5 - λ) (1 - λ) .$ Why this step? The determinant of a triangular matrix is just the product of its diagonal entries (all the off-diagonal cofactor terms vanish because of the zeros below the diagonal — see Determinants). So we get the whole characteristic polynomial already factored, with no expansion work.
Eigenvalues: $λ = 2, 5, 1.$ Why this step? Each factor $(d - λ)$ is zero exactly when $λ$ equals that diagonal entry $d$ , so the diagonal entries are the eigenvalues.
Eigenvector for $λ = 2$ . $A - 2 I = 000730 - 3 4 - 1$ . Rows 2 and 3 force $v_{3} = 0$ then $v_{2} = 0$ ; $v_{1}$ free. $E_{2} = span ⎩ ⎨ ⎧ 100 ⎭ ⎬ ⎫ .$ Why this step? We still solve the homogeneous system — the shortcut only found the $λ$ 's, not the vectors.

Verify: $A 100 = 200 = 2100$ ✓, and the product $(2 - λ) (5 - λ) (1 - λ)$ expands to the true characteristic polynomial (checked below).

Ex 8 — Cell H: symmetric $3 \times 3$ (perpendicular eigenspaces)

Forecast: $A$ equals its own transpose (mirror across the diagonal). By the Symmetric matrices and spectral theorem, predict two facts before solving: are the eigenvalues real? are the eigenvectors perpendicular?

Characteristic polynomial (expand along the first column since it isolates the block): $det (A - λ I) = (2 - λ) [(3 - λ)^{2} - 1] .$ Why this step? The top-left $2$ is decoupled (its row and column are zero elsewhere), so the problem splits into a $1 \times 1$ and a $2 \times 2$ piece — far easier than a full $3 \times 3$ expansion.
Solve. $(3 - λ)^{2} - 1 = 0 \Rightarrow 3 - λ = \pm 1 \Rightarrow λ = 2, 4$ . Together with the front factor $λ = 2$ : $λ = 2 (twice), 4.$ Why this step? Reading each factor's roots gives the eigenvalues; the two separate sources of $λ = 2$ warn us it is a double root (algebraic multiplicity $2$ ).
$E_{4}$ . $A - 4 I = - 2 00 0 - 1 1 01 - 1$ : forces $v_{1} = 0$ , $v_{2} = v_{3}$ . $E_{4} = span ⎩ ⎨ ⎧ 011 ⎭ ⎬ ⎫ .$ Why this step? Row-reducing yields $v_{1} = 0$ and $v_{2} - v_{3} = 0$ , leaving one free variable — a single eigen-line.
$E_{2}$ . $A - 2 I = 000011011$ : one equation $v_{2} + v_{3} = 0$ ; $v_{1}$ free, $v_{3}$ free. $E_{2} = span ⎩ ⎨ ⎧ 100, 01 - 1 ⎭ ⎬ ⎫, dim E_{2} = 2.$ Why this step? Here $n = 3$ and $A - 2 I$ has just one independent nonzero row, so $rank (A - 2 I) = 1$ and $dim E_{2} = 3 - 1 = 2$ . Symmetric matrices are always diagonalizable, so the double root $λ = 2$ must deliver a full 2-dimensional eigenspace — and the single surviving equation $v_{2} + v_{3} = 0$ leaves exactly two free variables, confirming it: geometric multiplicity $=$ algebraic multiplicity $= 2$ .

Verify: First check the eigen-equations: $A 011 = 044 = 4011$ ✓, $A 100 = 200 = 2100$ ✓, and $A 01 - 1 = 02 - 2 = 2 01 - 1$ ✓. Now confirm the spectral-theorem promise that eigenvectors from different eigenspaces are perpendicular (dot product $= 0$ ): $011 \cdot 100 = 0, 011 \cdot 01 - 1 = 0, 100 \cdot 01 - 1 = 0.$ All three dot products vanish ✓ — exactly as the Symmetric matrices and spectral theorem guarantees: a real symmetric matrix always has real eigenvalues and a full set of mutually perpendicular eigenvectors.

Ex 9 — Cell I: word problem (limiting behaviour)

Forecast: In the long run the split stops changing. What eigenvalue corresponds to "stops changing"? Guess it before step 2.

Steady state = eigenvector for $λ = 1$ . A distribution that no longer moves satisfies $A v = v = 1 \cdot v$ . Why this step? "Unchanged from year to year" is precisely the eigenvalue- $1$ condition — the steady state is the eigenvector we want.
Confirm $λ = 1$ is an eigenvalue. Column sums of $A$ are $1$ (it's a stochastic matrix), which forces $λ = 1$ . Check: $det (A - I) = det (- 0.2 0.2 0.3 - 0.3) = (- 0.2) (- 0.3) - (0.3) (0.2) = 0.06 - 0.06 = 0$ ✓ (singular, so $1$ is indeed an eigenvalue). Why this step? We must verify $λ = 1$ really is an eigenvalue before solving for its eigenvector; the zero determinant confirms $A - I$ is singular.
Solve $E_{1}$ . $A - I = (- 0.2 0.2 0.3 - 0.3)$ : equation $- 0.2 v_{1} + 0.3 v_{2} = 0 \Rightarrow v_{1} = \frac{3}{2} v_{2}$ . $E_{1} = span {(32)} .$ Why this step? Row-reducing $A - I$ leaves the single relation $v_{1} = \frac{3}{2} v_{2}$ , whose direction $(3, 2)$ is the unchanging split.
Interpret as percentages. Scale so entries sum to $1$ : split $= (\frac{3}{5}, \frac{2}{5}) = (60%, 40%)$ . Why this step? Populations are proportions, so we normalise the eigenvector to sum to $1$ to read it as percentages.

Verify: $A (2 3) = (0.2 \cdot 3 + 0.7 \cdot 2 0.8 \cdot 3 + 0.3 \cdot 2) = (0.6 + 1.4 2.4 + 0.6) = (2 3)$ ✓ — the split $60%/40%$ reproduces itself forever. (The other eigenvalue is $λ = 0.5$ , which decays away, so any starting split funnels toward $60/40$ .)

Recall

Which cell has NO real eigenspace, and why? ::: Cell F (rotation): $λ^{2} + 1 = 0$ has no real roots because no real line is fixed by a rotation. How do you get eigenvalues of a triangular matrix instantly? ::: Read them off the diagonal — $det (A - λ I)$ is the product of diagonal entries. In a stochastic (column-sum-1) matrix, which eigenvalue always appears? ::: $λ = 1$ ; its eigenvector (normalised) is the steady-state distribution. Two matrices share polynomial $(λ - 3)^{2}$ — why can their eigenspaces differ in dimension? ::: $dim E_{3} = n - rank (A - 3 I)$ ; the rank can be $0$ (full 2D space) or $1$ (deficient line). What is the difference between algebraic and geometric multiplicity? ::: Algebraic = how many times $λ$ is a root of the characteristic polynomial; geometric = $dim E_{λ}$ . Always $1 \leq$ geometric $\leq$ algebraic.

Connections

Finding eigenspaces — the parent recipe these examples drill.
Characteristic polynomial — every "characteristic equation" step above comes from it.
Null space and solving homogeneous systems — each $E_{λ}$ is a null space we row-reduce.
Determinants — the $det (A - λ I) = 0$ singularity test.
Rank-Nullity theorem — gives $dim E_{λ} = n - rank (A - λ I)$ (Ex 5).
Diagonalization — Ex 4 vs Ex 5 is exactly the diagonalizable / not test.
Symmetric matrices and spectral theorem — Ex 8's real, perpendicular eigenspaces.