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Diagonalizable operatorsMath 130 Linear AlgebraD Joyce, Fall 2012

Some linear operators T : V V have the niceproperty that there is some basis for V so that thematrix representing T is a diagonal matrix. Wellcall those operators diagonalizable operators. Wellcall a square matrix A a diagonalizable matrixif it isconjugate to a diagonal matrix, that is, there existsan invertible matrix P so that P1AP is a diagonal

matrix. Thats the same as saying that under achange of basis, A becomes a diagonal matrix.

Reflections are examples of diagonalizable oper-ators as are rotations if C is your field of scalars.

Not all linear operators are diagonalizable. Thesimplest one is R2 R2, (x, y) (y, 0) whose ma-

trix is A =

0 10 0

. No conjugate of it is diagonal.

Its an example of a nilpotent matrix, since somepower of it, namely A2, is the 0-matrix. In general,nilpotent matrices arent diagonalizable. There aremany other matrices that arent diagonalizable aswell.

Theorem 1. A linear operator on an n-dimensional vector space is diagonalizable if andonly if it has a basis ofn eigenvectors, in which casethe diagonal entries are the eigenvalues for thoseeigenvectors.

Proof. If its diagonalizable, then theres a basis forwhich the matrix representing it is diagonal. Thetransformation therefore acts on the ith basis vectorby multiplying it by the ith diagonal entry, so its aneigenvector. Thus, all the vectors in that basis areeigenvectors for their associated diagonal entries.

Conversely, if you have a basis ofn eigenvectors,then the matrix representing the transformation isdiagonal since each eigenvector is multiplied by itsassociated eigenvalue. q.e.d.

Well see soon that if a linear operator on an n-dimensional space has n distinct eigenvalues, thenits diagonalizable. But first, a preliminary theo-rem.

Theorem 2. Eigenvectors that are associated todistinct eigenvalues are independent. That is, if1, 2, . . ., k are different eigenvalues of an oper-ator T, and an eigenvector vi is associated to eacheigenvalue i, then the set of vectors v1, v2, . . ., vkis linearly independent.

Proof. Assume by induction that the first k 1 ofthe eigenvectors are independent. Well show all kof them are. Suppose some linear combination of

all k of them equals 0:

c1v1 + c2v2 + + ckvk = 0.

Take T kI of both sides of that equation. Theleft side simplifies

(T kI)(c1v1 + + ckvk)

= c1T(v1) kc1v1 + + ckT(vk) kckvk

= c1(1 k)v1 + + ck(k k)vk

= c1(1 k)v1 + + ck1(k1 k)vk1

and, of course, the right side is 0. That gives us alinear combination of the first k 1 vectors whichequals 0, so all their coefficients are 0:

c1(1 k) = = ck1(k1 k) = 0

Since k does not equal any of the other is, there-fore all the cis are 0:

c1 = = ck1 = 0

The original equation now says ckvk = 0, and sincethe eigenvector vk is not 0, therefore ck = 0. Thusall k eigenvectors are linearly independent. q.e.d.

Corollary 3. If a linear operator on an n-dimensional vector space has n distinct eigenvalues,then its diagonalizable.

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Proof. Take an eigenvector for each eigenvalue. Bythe preceding theorem, theyre independent, andsince there are n of them, they form a basis of then-dimensional vector space. The matrix represent-ing the transformation with respect to this basisis diagonal and has the eigenvalues displayed downthe diagonal. q.e.d.

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