Diagonalization

References are to Anton–Rorres, 7th Edition

Let T: R^{n} → R^{n} be a linear transformation, where we treat the vectors in R^{n} as column vectors. By Theorem 4.3.3, there is a unique n×n matrix A such that T is expressed in terms of A by

y = T (x) = Ax  (1)

In the reverse direction, we write A = [T]_{E}, the matrix of T with respect to the standard basis E = {e_{1}, e_{2}, …, e_{n}}. Explicitly, A is the matrix

A = [T (e_{1})|T (e_{2})| … |T (e_{n})]  (2)

whose j-th column is the image T (e_{j}) of e_{j} under T. Expanding this column, we have

T (e_{j}) = a_{1j}e_{1} + a_{2j}e_{2} + … + a_{nj}e_{n}  (3)

Change of basis We wish to simplify the matrix by changing to a different basis B = {v_{1}, v_{2}, …, v_{n}} of R^{n}. Any vector x ∈ R^{n} can be expressed uniquely as a linear combination

x = x′_{1}v_{1} + x′_{2}v_{2} + … + x′_{n}v_{n}  (4)

for suitable coefficients x′_{i} ∈ R.

Definition 5 The coordinate vector [x]_{B} of x with respect to the new basis B is the vector

x′ = [x]_{B} = [ x′_{1} ; x′_{2} ; ⋮ ; x′_{n} ]

Definition 6 The matrix [T]_{B} of T with respect to the new basis B is the n×n matrix D whose entries d_{ij} satisfy the analogue

T (v_{j}) = d_{1j}v_{1} + d_{2j}v_{2} + … + d_{nj}v_{n}  (7)

of equation (3) for each j.

Let y′ = [y]_{B} be the new coordinate vector of y = T (x). Then the analogue of equation (1) is

y′ = Dx′  (8)

The transition matrix We need to know how to compute the new matrix D from A, and the new coordinate vector x′ from x; also vice versa. Obviously, we have to use the v’s in some way, so we encode them in a matrix.

Definition 9 The transition matrix from the basis B to the basis E is the n×n matrix

P = [v_{1}|v_{2}| … |v_{n}]

whose j-th column is the vector v_{j}.

Note the direction; P expresses the new basis B in terms of the old standard basis E, not the other way round. To relate x and x′, we take the i-th coordinate in equation (4), noting that we have encoded the i-th coordinate of the vector v_{j} as the entry p_{ij} in P,

\begin{aligned}x_{i} & = x′_{1}p_{i1} + x′_{2}p_{i2} + … + x′_{n}p_{in} \\ & = p_{i1}x′_{1} + p_{i2}x′_{2} + … + p_{in}x′_{n} \end{aligned}

which we recognize as the entry in row i of the matrix product P x′. Thus

x = P x′  (10)

We also need to express x′ in terms of x. This is now easy to do. We note that because B is a basis, the columns of P are linearly independent, the column rank of P is n and P is therefore invertible. We multiply (10) on the left by P^{−1} to get

x′ = P^{−1}x  (11)

We note that P^{−1} is the transition matrix from E to B.

Then equation (11) (for y) and equations (1) and (10) give

y′ = P^{−1}y = P^{−1}Ax = P^{−1}AP x′

Comparison with equation (8) shows that the new matrix D is given by

D = [T]_{B} = P^{−1}AP  (12)

Similarity The two matrices A and D both describe the same linear transformation T, using different coordinates.

Definition 13 Two n×n matrices A and D are similar if there exists an invertible matrix P such that equation (12) holds. The matrix A is diagonalizable if it is similar to a diagonal matrix.

Eigenvectors We should choose a basis B of R^{n} that is related in some way to T. In the most favorable situation (which is not that rare), we can find a basis B of R^{n} that consists entirely of eigenvectors of T (or A). Specifically, we have

T (v_{i}) = Av_{i} = λ_{i}v_{i}  (14)

for each i, where λ_{i} denotes the associated eigenvalue. On comparing equations (14) and (7), we deduce that the new matrix D is a diagonal matrix, with diagonal entries d_{ii} = λ_{i} and d_{ij} = 0 for i ≠ j.

Theorem 15 (=Thm. 7.2.1) The following conditions on the matrix A are equivalent:

(a) A is diagonalizable;

(b) R^{n} has a basis of eigenvectors of A;

(c) There is a basis B of R^{n} such that D = [T]_{B} is a diagonal matrix.

Proof We have just shown that (b) implies (a). The argument works just as well in reverse to show that (a) implies (b): if a matrix P exists as in equation (12) with D diagonal, we take v_{j} as the j-th column of P and then equation (7) shows that v_{j} is an eigenvector. Finally, equation (7) shows that (b) and (c) are equivalent.