Finding eigenspaces: Visual walkthrough — Maths deep dive

We will use one running machine the whole way: $A = (2112) .$ Nothing here assumes you know what a determinant "means" yet — we build it visually where it's needed.

Step 1 — What a matrix does to an arrow

WHAT. An arrow (a vector) is just a point-with-direction, drawn from the origin. A matrix $A$ is a machine: feed it an arrow $v$ , it hands you back a new arrow $A v$ .

Here $v$ means "the input arrow" and $A v$ means "what the machine returns."

WHY. Before we hunt for special arrows, we must see the typical thing that happens: most arrows come out pointing somewhere new. That turning is the enemy we are looking to avoid.

PICTURE. Two ordinary input arrows (black) and where $A$ sends them (the red arrows have clearly rotated to a new direction).

Step 2 — The special arrow that refuses to turn

WHAT. Some input directions come out exactly parallel to how they went in. For our $A$ , the arrow $(1 1)$ is such a one: $A (11) = (2 \cdot 1 + 1 \cdot 1 1 \cdot 1 + 2 \cdot 1) = (33) = 3 (11) .$

Read the equation left to right: the machine $A$ ate $(1 1)$ and returned $(3 3)$ , which is just the same arrow scaled by $3$ . That scale factor $3$ is the eigenvalue, written $λ$ (Greek "lambda", our name for "the stretch amount").

WHY. This is the entire phenomenon. If we can characterise these no-turn arrows algebraically, we can go find them for any machine.

PICTURE. The red arrow $(1 1)$ and its image $(3 3)$ lie on the same dashed line — pure stretch, zero rotation.

Step 3 — Writing "same direction" as an equation

WHAT. "Output stays on the input's line" means output $=$ some number times input: $A v = λ v .$

$A v$ — the machine's output arrow.
$λ$ — the single number the arrow got scaled by.
$v$ — the input arrow, which must be nonzero (the zero arrow has no direction to preserve, so it tells us nothing).

WHY. We've traded a geometric wish ("don't rotate") for an algebraic equation we can manipulate. That is the whole move of linear algebra: turn a picture into symbols, solve, then read the picture back.

PICTURE. The same arrow with the equation annotated on it — input in black, scaled output in red, the " $= λ \times$ " sitting between them.

Step 4 — Collapse both sides onto one machine

WHAT. Move everything to one side: $A v - λ v = 0 .$ We cannot write $(A - λ) v$ — $A$ is a grid of numbers (a matrix), $λ$ is a single number; subtracting them is like subtracting an apple from an orchard. The repair: replace $λ$ by $λ I$ , where $I = (1001)$ is the identity machine (the one that returns every arrow unchanged, so $I v = v$ ). Now both terms are machines acting on $v$ : $(A - λ I) v = 0 .$

$A - λ I$ — a new single machine: take $A$ and subtract $λ$ from each diagonal entry only.
$0$ — the zero arrow (a point at the origin), not the number zero.

For our $A$ : $A - λ I = (2 - λ 1 1 2 - λ) .$

WHY. One machine $(A - λ I)$ times one arrow equalling zero is a shape we know how to solve — a homogeneous system.

PICTURE. Build-up of the new machine: $A$ 's grid, minus $λ$ slid down the diagonal (red), giving $A - λ I$ .

Step 5 — Why the machine must flatten a direction

WHAT. We need $(A - λ I) v = 0$ for some nonzero $v$ . That says: the machine $M := A - λ I$ takes a nonzero arrow and crushes it to the origin. A machine that squashes a whole direction to a point is called singular (not invertible) — it has lost information, so you cannot undo it.

WHY. If $M$ weren't singular, it would be reversible: the only arrow it sends to $0$ would be $0$ itself — no eigenvector. So we require $M$ to be singular. That is not a trick; it is the precise condition for a non-turning arrow to exist.

PICTURE. Left: an invertible machine spreads the unit square into a tilted parallelogram with real area — nothing is crushed. Right: a singular machine collapses the square flat onto a red line — area zero, one direction annihilated.

Step 6 — The determinant is the "how flat?" meter

WHAT. The determinant $det (M)$ measures the area the machine $M$ produces from the unit square (signed, but ignore the sign for now). " $M$ crushes a direction flat" is the same statement as "that area is $0$ ": $det (A - λ I) = 0.$ For our $2 \times 2$ machine: $det (2 - λ 1 1 2 - λ) = (2 - λ) (2 - λ) - (1) (1) = (2 - λ)^{2} - 1.$

$(2 - λ) (2 - λ)$ — product of the diagonal entries.
$- (1) (1)$ — minus the product of the off-diagonal entries.

This is the characteristic equation — an equation whose unknown is $λ$ .

WHY. We have converted "find the stretch amounts that make the machine collapse" into a plain equation in one variable $λ$ . Solve it and the eigenvalues fall out.

Set it to zero: $(2 - λ)^{2} = 1 \Rightarrow 2 - λ = \pm 1 \Rightarrow λ = 1 or λ = 3.$

PICTURE. A dial: as $λ$ slides, the area (determinant) rises and falls; it touches zero exactly at $λ = 1$ and $λ = 3$ — the red zero-crossings.

Step 7 — Solve the crushed system to get the eigenspace

WHAT. For each eigenvalue, ask which arrows get crushed. Take $λ = 3$ : $A - 3 I = (2 - 3 1 1 2 - 3) = (- 1 1 1 - 1) .$ Row-reduce to $(10 - 1 0)$ , which reads $v_{1} - v_{2} = 0$ , i.e. $v_{1} = v_{2}$ . Let the free variable $v_{2} = t$ : $v = t (11) .$ Every scaling $t$ of $(1 1)$ works — that whole line is the eigenspace: $E_{3} = Null (A - 3 I) = span {(11)} .$ Doing the same for $λ = 1$ gives $v_{1} = - v_{2}$ , so $E_{1} = span {(- 1 1)}$ .

WHY. The null space of $A - λ I$ is the eigenspace — it collects all arrows the crushed machine sends to $0$ , which are exactly the no-turn arrows for that $λ$ .

PICTURE. Two red lines through the origin: $E_{3}$ (the $(1 1)$ direction) and $E_{1}$ (the $(- 1 1)$ direction) — the machine's two non-turning axes.

Step 8 — Degenerate cases you must never be surprised by

WHAT. The recipe never breaks, but the shape of the answer varies. Three things can happen when an eigenvalue repeats (algebraic multiplicity $2$ for a $2 \times 2$ ):

Machine	$A - λ I$ crushes…	Eigenspace $E_{λ}$	$dim$
$(5005)$ , $λ = 5$	nothing left to keep — it is the zero machine, crushes all of $R^{2}$ into $0$	whole plane $R^{2}$	$2$
$(5015)$ , $λ = 5$	only a single line	one line, $span {(0 1)}$	$1$
$λ = 0$ case (any singular $A$ )	$A$ itself is already singular	$E_{0} = Null (A)$	$\geq 1$

The middle case is the trap: the eigenvalue repeats twice yet the eigenspace is only a line. The repeat count (algebraic multiplicity) is only an upper bound on the eigenspace dimension (geometric multiplicity). See Diagonalization for why the gap matters.

WHY. A reader who only saw the clean $λ = 1, 3$ case would wrongly expect "repeat count = eigenspace dimension." Showing all three closes every escape route.

PICTURE. Three panels side by side: (left) whole plane shaded red — full eigenspace; (middle) a single red line — deficient eigenspace; (right) a red line through the origin — the $λ = 0$ / singular case.

The one-picture summary

One frame ties it together: input arrows (black) hit the machine $A$ ; almost all rotate; on the two red eigen-lines they only stretch (by $λ = 3$ and $λ = 1$ ). The stretch factors are the $λ$ where the machine goes flat, i.e. where $det (A - λ I) = 0$ ; each such $λ$ 's crushed direction is the eigenspace.

Recall Feynman retelling — the whole walkthrough in plain words

Think of the matrix as a wind that blows every arrow to a new spot. Most arrows swing to a new angle. But a couple of directions are like a flagpole standing straight into the wind — they just grow or shrink, never swing. To find those directions, I subtract off a candidate "growth amount" $λ$ from the machine's diagonal, making a new machine $A - λ I$ . If, for the right $λ$ , this new machine flattens some arrow down to nothing, that arrow was a no-swing direction. "Flattens a direction to nothing" is the same as "makes zero area," and the area meter is the determinant — so I hunt for the $λ$ that make $det (A - λ I) = 0$ . For each such $λ$ I collect every arrow that gets flattened; that collection is a whole line (or plane) called the eigenspace. Sometimes a $λ$ shows up twice but only flattens one line — then the eigenspace is smaller than the repeat count warned, and that's the whole warning of the last picture.

Active recall

Why does an eigenvector's direction stay fixed?

Because

A v = λ v

lands the output on the same line as the input — only the length/sign changes.

In the picture, what does

det (A - λ I) = 0

look like?

The machine

A - λ I

collapses the unit square to zero area — it flattens a whole direction onto the origin.

Why insert the identity

I

A - λ I

So we subtract a matrix from a matrix;

A - λ

is meaningless since

A

is a grid and

λ

a single number.

For

A = (2112)

, name the eigen-lines.

E_{3} = span {(1 1)}

and

E_{1} = span {(- 1 1)}

Can a doubly-repeated eigenvalue give only a 1D eigenspace?

Yes — e.g.

(5015)

gives

dim E_{5} = 1

; geometric

\leq

algebraic.

Connections

Finding eigenspaces — the parent recipe this page draws out step by step.
Determinants — the "how flat" meter driving Step 6.
Null space and solving homogeneous systems — Step 7's engine.
Characteristic polynomial — the equation the determinant produces.
Rank-Nullity theorem — gives $dim E_{λ} = n - rank (A - λ I)$ .
Diagonalization — why the deficient case of Step 8 matters.
Symmetric matrices and spectral theorem — when repeats always fill the space.