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One powerful application of diagonalization is in computing powers of a matrix. We conclude in addition that the eigenvalues of Ak are the kth powers of the ...
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Chapter 5: Diagonalization and Eigenvalues and Eigenvectors Section 2: Diagonalization of a matrix Ivan Contreras, Sergey Dyachenko and Bob Muncaster University of Illinois at Urbana-Champaign April 9 2018 Ivan Contreras, Sergey Dyachenko and Bob MuncasterUniversity of Illinois at Urbana-ChampaignApplied Linear Algebra——– April 9 2018 1 / 8
One of the most important uses of eigenvalue analysis is in the process of diagonalizing a square matrix. For example, which of these two systems would you rather attempt to solve:
A
x 1 x 2 x 3
Λ
y 1 y 2 y 3
The second is diagonal and clearly the easier of the two. It turns out that the two are linked by
x 1 x 2 x 3
S
y 1 y 2 y 3
It is, in fact, the case that the diagonal entries in Λ are the eigenvalues of A and the columns of S are the corresponding eigenvectors of A.
The formulas in this theorem allow you to switch from a problem involving A and x to an equivalent problem involving Λ and y : Ax = b ⇐⇒ S ΛS−^1 x = b ⇐⇒ ΛS−^1 x = S−^1 b ⇐⇒ Λy = c
where y = S−^1 x, c = S−^1 b Note that x = Sy which gives a way to recover x once y is known Now that we see the value of diagonalizing a matrix, can we always do it?
Ex:
A =
⇒ |A − λ I (^) | =
∣∣ 4 −^ λ^^1 0 4 − λ
∣∣ = ( 4 − λ )^2 ⇒ λ = 4, 4
Now for the corresponding eigenvectors:
(A − λ I ) x =
a b
⇒ b = 0 ⇒ x = a
In this case we get only one linearly independent eigenvector, and so the theorem cannot be used.
Fact: n × n matrices that DO NOT have n linearly independent eigenvectors cannot be diagonalized (the next best specialized form of a matrix is its Jordan normal form and that can always be found - see the text). So we need n eigenvectors AND linear independence to be able to diagonalize. So here is a useful result:
Theorem: Let A have eigenvalues λ i and corresponding eigenvectors xi , i = 1, ..., n. If the λ ’s are all different, then the eigenvectors are linearly independent. Proof: Suppose that x 1 , ..., xp are linearly independent but x 1 , ..., xp , xp+ 1 are not. This means that there are coefficients ci , i = 1, ..., p + 1, not all zero, such that c 1 x 1 + · · · + cp xp + cp+ 1 xp+ 1 = 0, cp+ 1 6 = 0 Multiply by A and use Axi = λ i xi to get c 1 λ 1 x 1 + · · · + cp λ p xp + cp+ 1 λ p+ 1 xp+ 1 = 0
Theorem: Any matrix with distinct eigenvalues can be diagonalized.
Ex: P =
2
1 1 2 2
1 2
⇒ Px =
2 x^1 +^
1 1 2 x^2 2 x^1 +^
1 2 x^2
x 1 + x 2 2
and so P is an orthogonal projection onto the vector with 1’s as entries. This means that
P
From these we read off the eigenvalues 1 and 0 and their corresponding eigenvectors, and see that
S =
Ex: Since
we can see the eigenvalues and eigenvectors directly and conclude that
While the previous result seemed to require k to be a positive integer, it also works for negative integers provided we view A−k^ as the inverse of Ak^. This follows from the following observation.
Since det A = λ 1 λ 2 · · · λ n, we can see that an invertible matrix never has zero as an eigenvalue. Then
Ax = λ x ⇒ x = λ A−^1 x ⇒ A−^1 x =
λ
x
We conclude that the eigenvalues of A−^1 are the reciprocals of the eigenvalues of A, with the same eigenvectors. In terms of diagonalization we can see this equally well by
A−^1 = (S ΛS−^1 )−^1 = (S−^1 )−^1 Λ−^1 S−^1 = S diagonal(
λ 1
λ 2
λ n
The diagonalization factorization formula gives a simple way to relate the eigenvalues of A and AT^ : A = S ΛS−^1 ⇒ AT^ = (S−^1 )T^ ΛT^ ST Since Λ is diagonal, ΛT^ = Λ. Also we already know that the transpose of an inverse is the inverse of the transpose: (S−^1 )T^ = (ST^ )−^1. Thus AT^ = (ST^ )−^1 ΛST^ = RΛR−^1 where R = (ST^ )−^1 This gives us a diagonalization of AT^ with the same Λ as for A. This establishes:
Theorem: A and AT^ have the same eigenvalues. Thus det A = λ 1 · · · λ n = det AT^.
Another way to see this is to recall that a matrix and its transpose have the same determinant: 0 = det(A − λ I ) = det((A − λ I )T^ ) = det(AT^ − λ I T^ ) = det(AT^ − λ I )
