5.4 Diagonalization and Similarity

The chapter so far has been a census: real eigenlines that a matrix stretches, and complex pairs that make it spiral. What the census does not yet supply is a method. That arrives by combining eigenvectors with the idea of Chapter 2 — that a basis is a coordinate system — because in the right coordinate system the matrix from section 5.1 stops being a grid of nine numbers and becomes a list of three.

That matrix was

$A = (\begin{matrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & - 1 \end{matrix})$

with characteristic polynomial $- (λ - 3) (λ - 1) (λ + 1)$ , so its eigenvalues are $3$ , $1$ and $minus 1$ . The eigenvectors are $open parenthesis 1, 1, 0 close parenthesis$ , $open parenthesis 1, minus 1, 0 close parenthesis$ and $open parenthesis 0, 0, 1 close parenthesis$ — check the first: $A (1, 1, 0) = (2 + 1, 1 + 2, 0) = (3, 3, 0) = 3 (1, 1, 0)$ , as an eigenvector must satisfy.

Three eigenvectors, and they are independent, so they are a basis of $ℝ^{3}$ . Coordinates in that basis are what make $A$ simple: the matrix does nothing to the frame itself, it just scales along each axis of it.

Looking down at the horizontal plane. A faint dashed gray circle of radius one sits at the origin, and two dashed lines cross there at right angles running diagonally: the red one along (1, 1, 0) and the blue one along (1, -1, 0). A third dashed green line rises vertically out of the plane. A thick amber curve lies on top of the gray circle at first; as a slider is dragged the amber curve stretches into a long ellipse whose long axis lies exactly along the red dashed line and whose short axis stays exactly the width of the original circle along the blue dashed line. The stretching happens only in the red direction, never turning the ellipse off those two axes. — Looking down at the plane $z equals 0$ . The dashed red and blue lines are the eigenlines for $λ equals 3$ and $λ equals 1$ ; the dashed green line rising out of the plane is the third, for $λ = - 1$ . The faint gray circle is the unit circle, and the thick amber curve is what becomes of it when each eigenline is stretched by its own eigenvalue raised to the power $k$ . At whole-number $k$ that is exactly $A$ applied that many times; in between, the slider interpolates — which means something here only because the two eigenvalues acting in this plane, $3$ and $1$ , are both positive. The $λ = - 1$ eigenline is drawn for completeness but nothing moves along it, which is just as well: no real matrix is a half-power of a flip.

Explore the figure

At $k equals 0$ the amber curve sits exactly on the gray circle — no transformation has happened yet.
Drag $k$ up slowly. The circle stretches into an ellipse, and the stretching happens along the red dashed line only. That is the $λ equals 3$ direction, and $3 to the power k$ is how far it has been pulled.
Look at the blue dashed direction as you drag. The ellipse's width across it never changes. Its eigenvalue is $1$ , and $1^{k} = 1$ for every $k$ : that direction is not merely un-turned, it is untouched.
Notice what the ellipse never does. Its long axis stays welded to the red line and never swings around, at any $k$ . In the eigenbasis, $A$ has no ability to rotate anything — it can only stretch along the axes it is already given.

Writing the frame down

Put the eigenvectors in the columns of a matrix:

$P = (\begin{matrix} 1 & 1 & 0 \\ 1 & - 1 & 0 \\ 0 & 0 & 1 \end{matrix}), D = (\begin{matrix} 3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{matrix})$

$P$ is exactly the coordinate matrix of section 2.7: hand it a list of eigenbasis coordinates and it returns the point in standard coordinates. Then

$A = P D P^{- 1}$

and each factor has a job you can name. Read right to left, following what happens to a vector: $P^{- 1}$ converts it into eigenbasis coordinates, $D$ scales each coordinate by its eigenvalue, and $P$ converts back to the standard grid. Doing all three is doing $A$ — because in the middle, where the coordinates are the right ones, $A$ is the diagonal matrix.

[\begin{matrix} \frac{1}{2} & \frac{1}{2} & 0 \\ \frac{1}{2} & - \frac{1}{2} & 0 \\ 0 & 0 & 1 \end{matrix}]

Translating into the eigenbasis ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The inverse of the eigenvector matrix, which is the translator from standard coordinates into eigenbasis coordinates. Its first row reads off the amount of $open parenthesis 1, 1, 0 close parenthesis$ in a vector, its second the amount of $open parenthesis 1, minus 1, 0 close parenthesis$ , and its third the amount of $open parenthesis 0, 0, 1 close parenthesis$ .

minus 2

The eigenvectors really are a basis ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The determinant of $P$ is $minus 2$ , and the only thing that matters is that it is not zero. By the Invertible Matrix Theorem the columns are independent, so $P^{- 1}$ exists and the factorization is legitimate. A matrix whose eigenvectors failed to be independent would have a singular $P$ and could not be diagonalized at all — which is exactly what goes wrong for the matrices this book has been careful to avoid.

Two matrices $A$ and $B$ are called similar when $B = P^{- 1} A P$ for some invertible $P$ . The definition looks like a technicality and is not: similar matrices are the same transformation written in two different bases. Everything basis-independent survives the translation — determinant, rank, eigenvalues, the characteristic polynomial — while the entries themselves, which are only ever coordinates, do not.

Powers, for free

Here is the payoff that makes diagonalization more than a change of viewpoint. Compute $A^{2}$ from the factorization:

$A^{2} = P D P^{- 1} \cdot P D P^{- 1} = P D (P^{- 1} P) D P^{- 1} = P D^{2} P^{- 1}$

The inner $P^{- 1} P$ collapses to the identity, and the same cancellation happens however many factors you stack:

$A^{k} = P D^{k} P^{- 1}$

Raising a diagonal matrix to a power is just raising each diagonal entry to that power, so $D^{k} = diag (3^{k}, 1^{k}, (- 1)^{k})$ and no matrix multiplication is needed at all. That formula is what the figure is drawing: a point at eigen-coordinates $open parenthesis c sub 1, c sub 2, c sub 3 close parenthesis$ goes to $(3^{k} c_{1}, c_{2}, (- 1)^{k} c_{3})$ , so the circle in the $z equals 0$ plane becomes an ellipse with semi-axes $3 to the power k$ and $1$ along the two eigenlines.

It also tells you the long run without any computation. As $k$ grows, $3 to the power k$ swamps everything else, so almost every vector's image swings toward the $λ equals 3$ eigenline and grows by roughly a factor of $3$ per step. What licenses that reading is the strict gap $| 3 | > | 1 | = | - 1 |$ : when one eigenvalue is strictly larger in absolute value than every other, it decides the long-term behavior — a fact the next section turns into a working prediction. The strictness is not decoration. Drop the $3$ and the remaining eigenvalues $1$ and $minus 1$ tie in absolute value, and the long run never settles down to one direction at all.

Example 1 A power, without multiplying matrices

Use the factorization to compute $A to the power 4 open parenthesis 3, 1, 2 close parenthesis$ .

First convert to eigenbasis coordinates by writing $open parenthesis 3, 1, 2 close parenthesis$ as a combination of the eigenvectors. The third coordinate must come entirely from $open parenthesis 0, 0, 1 close parenthesis$ , so its coefficient is $2$ . For the first two, solve $c_{1} (1, 1, 0) + c_{2} (1, - 1, 0) = (3, 1, 0)$ : adding the two equations $c_{1} + c_{2} = 3$ and $c_{1} - c_{2} = 1$ gives $2 c sub 1 equals 4$ , so $c sub 1 equals 2$ and $c sub 2 equals 1$ . Hence

$(3, 1, 2) = 2 (1, 1, 0) + 1 (1, - 1, 0) + 2 (0, 0, 1)$

Now apply $A to the power 4$ , which scales each piece by its eigenvalue to the fourth power: $3^{4} = 81$ , $1^{4} = 1$ , and $(- 1)^{4} = 1$ . So

$A^{4} (3, 1, 2) = 162 (1, 1, 0) + 1 (1, - 1, 0) + 2 (0, 0, 1) = (163, 161, 2)$

Four matrix multiplications avoided, and the answer's shape is informative in a way the raw multiplication would have hidden: the first two coordinates are nearly equal, because the $λ equals 3$ component has grown by $81$ while everything else has stood still. The vector has been dragged almost onto the dominant eigenline, exactly as step 2 of the exploration suggested.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Its chapter sequence is aligned to Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL); this work is original, copies nothing from it, and is not affiliated with or endorsed by its authors. License: CC-BY-NC-SA-4.0.