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Lecture

# Lecture 35.pdf

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University of Waterloo

Mathematics

MATH 136

Robert Sproule

Winter

Description

Wednesday. April 2 − Lecture 35 : Diagonalizable matrices.
Concepts:
1. Define diagonalizable n ×n matrix A.
2. Recognize that a matrixA is diagonalizable if the matrix A has n distinct
eigenvalues.
3. Diagonalize simple 2 by 2 or 3 by 3 matrices.
35.1 Theorem − Suppose A is an n × n matrix with n distinct eigenvalues. Let X be the
n × n matrix whose columns are the n eigenvectors which are respectively associated to
the n distinct eigenvalues of a matrix A. Then X is invertible.
Proof:
Given : Suppose {v ,1v ,2…, v }nare n distinct eigenvectors respectively associated to n
distinct eigenvalues {λ 1 λ 2 …, λ n.
Required to show : That A = [v v 1 v2] is invertible. Suffices to show : That the set
{v 1 v2, …, v n is linearly independent.
Suppose not. Then one vector, say v , is a linear combination of the others {v , …, v }.
1 2 n
− If this set {v 2 …, v n is not linearly independent we can reduce it to a smaller linearly
independent set, say {v 2 …, v }m
− Then v = 1 v +2 2v + 3 3 … + a m mhere a to 2 aremunique scalars associated to v . 1
Then we can write
… …
v1= a 2 2 a v3 3 + a m m ⇒ Av = 1 Av 2 a 2v + 3 3 + a mv m
…
⇒ λ v 1 1 λ v2 2 2λ v 3 3 3 + a m m m (*)
Case 1: λ is not zero. Then,
1
…
v1= a 2λ 2λ 1v 2 a (3 /3 )v1+ 3 + a mλ mλ )1 m
− Since the scalars a2to a mre unique for v we1have a(λ/λ i =ia1 for ii= 2 to m.
So λ = λ , for i = 2 to m. This contradicts λ ≠ λ if i ≠ j. So if λ is not zero then v
i 1 i j 1 1
cannot be a linear combination of the others.
Case 2: λ i1 zero. Then, since λ ’i are distinct, none of {λ ,2λ 3 …, λ }mcan be zero
− Since λ is zero then we have λ v = 0 = a λ v + a λ v + … + a λ v .
1 1 1 2 2 2 3 3 3 m m m
…
− Linear independence of {v , 2, v } mmplies 0 = a λ =2 2λ = 3 3 = am m. − So 0 = a = 2 = …3= a . m
− Since v = 1 v +2 2v + …3 3a v then vm m0. Contr1diction! The vector v cannot 1
be the 0-vector since v is an eigenvector.
1
− So if λ i1 zero then v ca1not be a linear combination of the others.
We conclude that {v , v , …, v } must be linearly independent, as required.
1 2 n
Recall the definition of similar matrices: Two matrices A and C are said to be similar if
−1
they satisfy the property C = B AB for some invertible matrix B.
35.2 Definition − A matrix A is said to be diagonalizable if and only if A is similar to a
diagonal matrix D.
35.2.1 Observations
Equivalently, A is diagonalizable if and only if there exists an invertible matrix P such
that the matrix P AP = D is a diagonal matrix.
− We showed earlier that if C= B AB then C and A have the same eigenvalues.
− We also showed (in lecture 32) that the eigenvalues of a triangular matrix are those
numbers which are on the diagonal.
− So if D is a diagonal matrix which is similar to the matrix A then the eigenvalues
of A are those numbers “sitting” the diagonal of D.
35.3 Definition − Finding the matrix P so that P AP is a diagonal matrix is a process
called diagonalizing a matrix A. In such cases, we say that “Pdiagonalizes A”. Of
course, we can only diagonalize a matrix A if the matrix A is diagonalizable.
Our objective is twofold:
1) Determine methods which allow to recognize those matrices Awhich are
diagonalizable.
2) If A is diagonalizable, find an efficient way of diagonalizing A.
35.3.1 Example − In the following example we study a matrix A which has n distinct
eigenvalues. We diagonalize the matrix A by explicitly constructing the matrix P.
Consider the matrix A
Verifying that A has as eigenvalues the numbers λ = 1, λ = 2 and λ = 3 is left as an
1 2 3
exercise.
Solving for the 3 systems of linear equations (A − 1I)x = 0, (A − 2I)x = 0, and
(A − 3I)x = 0, we get the 3 associated eigenspaces E = α(−1, −1, 1) = Null(A – 1I),
E2= α(0, −1, 0) = Null(A – 2I), E = α31, −1, 1) = Null(A – 3I). To each of these eigenvalues we respectively find an associated eigenvector from each of E, E and 1 2
E 3 say x =1(−1, −1, 1),
x2= (0, −1, 0) and x =3(1, −1, 1).
Using these as columns and respecting the order, we form the matrix
This matrix is invertible since the eigenvectors are associated to distinct eigenvalues.
By a method of your choice you will find that the inverse P of P, is
−1
Multiplying P AP we obtain the diagonal matrix D
Observe that the entries on the diagonal are precisely the eigenvalues of A.
This illustrates that the matrix A with 3 distinct eigenvalues is similar to a diagonal
matrix which contains it eigenvalues on the diagonal.
We formalize what was said in the beginning of the previous example in the
statement of the following theorem. It shows precisely how to diagonalize a
diagonalizable matrix.
35.3.2 Notation − The expression
diag{λ , λ , λ , …, λ }
1 2 3 n
is a convenient way of representinga diagonal matrix who

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