Diagonalization — conditions, procedure: Visual walkthrough

Before any symbol appears, let's agree on what the picture is.

Step 0 — What is a matrix, really?

Take a whole grid of arrows and feed every one through $A$ . Most arrows come out pointing a different way than they went in — they got turned. That turning is what makes $A^{100}$ (apply the machine 100 times) so painful: every application re-tangles everything.

Figure — Diagonalization — conditions, procedure

In the figure, the grey arrows are the inputs. The orange arrows are what $A$ did to them. Notice the two special arrows drawn thick: they came out pointing the exact same way they went in — only their length changed. Hold onto those. Everything below is the hunt for them.

Step 1 — The dream: a machine that only stretches

WHY we want it. Feed the arrow $(x, y)$ to $D$ : $D (x y) = (λ_{1} x λ_{2} y) .$

$λ_{1} x$ ::: the horizontal part $x$ , simply scaled by $λ_{1}$ — no mention of $y$ .
$λ_{2} y$ ::: the vertical part $y$ , simply scaled by $λ_{2}$ — no mention of $x$ .

PICTURE. A grid drawn along the axes stays a grid — cells only grow/shrink, never tilt. Applying it 100 times just raises each factor to the 100th power. That is the whole appeal.

Look at the teal (horizontal) and plum (vertical) directions: each simply scales. A diagonal machine is boring — and boring is exactly what we want to compute with.

Step 2 — Most matrices are NOT boring — so we change our graph paper

This is the idea of a Change of Basis: the transformation $A$ is fixed, but the numbers we write for it depend on which axes we chose. Choose the axes cleverly and the numbers become diagonal.

Left panel: standard graph paper (grey), where $A$ visibly skews the grid. Right panel: graph paper drawn along the two special arrows (teal & plum). In the right frame the same machine merely stretches along each new axis. Same machine, smarter paper.

Step 3 — Name the pieces: build $P$ and $D$

WHY columns? A matrix whose columns are your chosen axes is exactly the machine that translates "coordinates in the new paper" into "coordinates in the standard paper". Feed it the new-paper vector $(1, 0)$ and out comes $p_{1}$ — the first new axis, expressed in old coordinates. That is what $P$ is for.

The figure shows $P$ as two arrows placed as columns and $D$ as a table of stretch numbers sitting on the diagonal. We have named the pieces — now we prove the naming is forced.

Step 4 — Derive the eigenvalue equation (the heart)

Start from the goal and multiply both sides on the right by $P$ (so the messy $P^{- 1}$ cancels): $A = P D P^{- 1} ⟹ A P = P D .$

$A P$ ::: run the machine $A$ on each column (arrow) of $P$ .
$P D$ ::: scale each column of $P$ by the matching diagonal number of $D$ .
$P^{- 1} P = I$ ::: the inverse cancelled, which is why we multiplied by $P$ and not something else.

Now read this column by column. The $j$ -th column of the left side $A P$ is $A p_{j}$ . The $j$ -th column of the right side $P D$ is $λ_{j} p_{j}$ (because $D$ just scales column $j$ by $λ_{j}$ ). Equating them:

WHY this is the punchline. We never decided to use eigenvectors. We only asked "can $A$ look diagonal in some paper?" and algebra spat back: the paper's axes must be the arrows $A$ doesn't turn. See Eigenvalues and Eigenvectors.

The figure shows one column arrow $p_{1}$ (teal): input and output lie on the same line — output is just $λ_{1}$ times longer. A grey non-eigenvector is drawn beside it; its output swings off to a new direction. Only the on-line arrows survive as columns of $P$ .

Step 5 — Why $P$ must be invertible → the " $n$ independent arrows" condition

WHY. If two special arrows lay on the same line (or one were a blend of the others), $P$ would squash space flat — it would crush some directions to zero, and you can never un-crush that. No un-crushing = no $P^{- 1}$ = no diagonalization.

So the full condition is:

Left: two independent arrows span the plane — $P$ is honest, invertible, we win. Right: two arrows on the same line — the "grid" they make is a degenerate sliver, $det P = 0$ , no inverse, we lose.

Step 6 — The degenerate case: the shear that runs out of arrows

Take the shear $A = (2012)$ . Its only eigenvalue is $λ = 2$ (from $(2 - λ)^{2} = 0$ ). Solving $A x = 2 x$ forces the second coordinate to be $0$ , so every eigenvector lies on the horizontal line — there is only one independent special direction, not two.

Algebraic multiplicity $a (2) = 2$ ::: the eigenvalue $2$ is a double root of the characteristic equation (see Characteristic Polynomial).
Geometric multiplicity $g (2) = 1$ ::: but only one independent eigenvector actually exists.
$g < a$ ::: not enough arrows → cannot fill the plane → not diagonalizable (this is a defective matrix; its true simplest form lives in Jordan Normal Form).

The shear pushes the top of the square right while pinning the bottom. Every arrow except the horizontal one gets tilted. There is no second un-turned direction to serve as a second axis — the recipe stalls at "collect $n$ arrows".

Step 7 — All the cases on one map

A fourth, sneaky case: a real matrix with no real eigenvectors at all, like the $9 0^{\circ}$ rotation $(01 - 1 0)$ — it turns every real arrow, so no real special direction exists. It is diagonalizable, but only if we allow complex numbers ( $λ = \pm i$ ). Real matrix, complex answer.

The ordering warning from the parent is exactly the arrow-and-stretch pairing: column $j$ of $P$ must be the arrow whose stretch sits at $D_{j j}$ . Scramble them and $A p_{j} = λ_{j} p_{j}$ silently fails. See also Similar Matrices (why $P D P^{- 1}$ keeps the eigenvalues) and Matrix Powers and Exponentials (the payoff: $A^{100} = P D^{100} P^{- 1}$ ). For symmetric matrices the arrows are even guaranteed perpendicular — the Spectral Theorem.

The one-picture summary

One frame compresses the whole derivation: the twisting machine $A$ (grey skewed grid) becomes the pure-stretch machine $D$ (teal/plum axis grid) precisely because we redrew the paper along the arrows $A$ never turns. $P$ carries you between the two papers; $D$ is what the machine looks like once you're standing on eigenvector axes.

Recall Feynman: the whole walk in plain words

We wanted a machine so simple that repeating it a hundred times is easy — a machine that only stretches each axis and never twists. Real machines twist, so we can't change the machine. Instead we change our graph paper: we look for the few special arrows the machine never turns, only lengthens. We line our new axes up with those arrows. On that paper the machine finally looks like pure stretching — a diagonal table $D$ . The book of "which arrow is which axis" is the matrix $P$ ; its inverse $P^{- 1}$ translates back. Squeezing the equation $A = P D P^{- 1}$ proved the arrows have to be eigenvectors — we had no other choice. The catch: sometimes there aren't enough un-turned arrows to fill the whole space (a shear only has one). Then the new paper can't be built, and we call the machine "not diagonalizable."