Eigenvectors and Linear Transformations

The factorization $A = P D P^{- 1}$ is a statement about coordinates: the transformation $x \mapsto A x$ is the same map as the scaling $u \mapsto D u$ , written in a different basis. Making this precise needs the matrix of a linear transformation between abstract vector spaces.

The matrix of a transformation relative to bases

Let $V$ be $n$ -dimensional with basis $B = {b_{1}, \dots, b_{n}}$ , let $W$ be $m$ -dimensional with basis $C$ , and let $T : V \to W$ be linear. Every $x \in V$ has a coordinate vector $[x]_{B} \in R^{n}$ , and its image $T (x)$ has a coordinate vector $[T (x)]_{C} \in R^{m}$ . These two are linked by a matrix.

If $x = r_{1} b_{1} + \dots + r_{n} b_{n}$ , then linearity gives $T (x) = r_{1} T (b_{1}) + \dots + r_{n} T (b_{n})$ , and applying the coordinate map (itself linear) yields

[T (x)]_{C} = r_{1} [T (b_{1})]_{C} + \dots + r_{n} [T (b_{n})]_{C} = M [x]_{B},

where the columns of $M$ are the coordinate vectors of the images of the basis vectors.

The matrix

M

makes the square commute: coordinatize by

B

, multiply by

M

, and the result is the

C

-coordinates of

T (x)

A transformation from a space to itself

When $W = V$ and $C = B$ , the matrix is written $[T]_{B}$ and called the matrix for $T$ relative to $B$ , or the $B$ -matrix for $T$ . It satisfies

[T (x)]_{B} = [T]_{B} [x]_{B}, for all x \in V .

Differentiation sends each basis monomial of P2 to its derivative; the images 0, 1, 2t supply the columns of the B-matrix.

Diagonal matrix representations

On $R^{n}$ a linear transformation usually appears first as a matrix map $x \mapsto A x$ in the standard basis. If $A$ is diagonalizable, there is a basis of eigenvectors, and in that basis the matrix of the transformation is diagonal.

The two maps $x \mapsto A x$ and $u \mapsto D u$ are the same transformation described in two coordinate systems. Diagonalizing $A$ is nothing more than finding a basis in which the transformation acts as independent scalings along the axes.

Example. For $A = [7 - 4 21] = P D P^{- 1}$ with $P = [1 - 1 1 - 2]$ and $D = [5003]$ , the basis $B$ of the columns of $P$ makes the $B$ -matrix of $x \mapsto A x$ equal to $D$ : along $b_{1}$ the map scales by $5$ , along $b_{2}$ by $3$ .

The same transformation in two frames: it shears the standard grid (left) but scales the eigenvector grid along its own axes (right).

Similarity as change of basis for maps

The proof of the diagonal-matrix-representation theorem never used that $D$ was diagonal. The same computation shows that if $A = P C P^{- 1}$ for any $C$ , then $C$ is the $B$ -matrix of $x \mapsto A x$ when $B$ is the basis of columns of $P$ . Conversely, for any basis $B$ with column matrix $P$ , the $B$ -matrix is $[T]_{B} = P^{- 1} A P$ . So the matrices similar to $A$ are precisely the matrix representations of the transformation $x \mapsto A x$ .

Similarity

A = P C P^{- 1}

reads as: change into

B

-coordinates, apply

C

, change back. The two routes around the square agree.

The Jordan form

If $A$ is defective, no basis makes its matrix diagonal, but a nearly diagonal representation still exists.

Situation	Best $B$ -matrix	Basis $B$
$A$ diagonalizable	diagonal $D$	$n$ eigenvectors
$A$ defective	Jordan form (block-triangular)	eigenvectors + generalized eigenvectors
General $C$ similar to $A$	$C = P^{- 1} A P$	columns of $P$

The same change of basis applies to real matrices whose simplest representation is a rotation rather than a scaling, found by way of complex eigenvalues.¹

Lay, Linear Algebra and Its Applications, 5th ed., §5.4 — Eigenvectors and Linear Transformations: the matrix of $T$ relative to bases, the $B$ -matrix for $T : V \to V$ , Theorem 8 (diagonal matrix representation), and similarity as the set of all matrix representations of $x \mapsto A x$ , including the Jordan-form example. ↩