Math 206 HWK #23 Due Thursday May 2

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Suggestions: After you read 8.5, try the first set of problems from 8.5. Then read 7.1 (and the tail of 8.5) and do the problems from 7.1. Lastly, read the rest of 8.5, read 7.2 and do the problems for 7.2 and the last couple problems from 8.5. Print this page and keep it handy, for the notes given below, as you do your reading and work on the problems. (Scroll down to see.)

Reading

• Read 8.5 (Similarity) up to the discussion of eigenvalues that starts at the bottom of p408.
• Then read 7.1. (Eigenvalues, Eigenvectors - for an n x n matrix A). Look back to Exple 6 p100 at the point where it's suggested to do so.
• Then, at p408 in 8.5, read Exple 4 and the paragraph or so that precede(s) it. This example has to do with eigenvalues and eigenvectors for a linear transformation T (which match up with eigenvalues and eigenvectors for a representing matrix for T).
• Read 7.2 (Diagonalization).
  

Problems

• 8.5 p411: 2, 5, 6, 7, 9c
  Click to view solutions
  
  
• 7.1 p344: 1a-2a-3a, 1e-2e-3e, 4a-5a-6a, 10ac
  Click to view solutions
  
  
• 7.2 p354: 3, 5, 13, 15
  Click to view solutions
  
  
• 8.5 p411: 13
  Click to view solutions
  
  

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Notes to summarize the reading/classnotes for 8.5.

• The transition matrix for changing from one basis to another is the same as the matrix that would be used to carry out the identity transformation using the old basis for inputs and the new basis for outputs. In other words, if B = {u₁, u₂, ..., u_n} is one basis for some n-dimensional vector space V, and B' is another basis for V, then the transition matrix used to change from B-coordinates to B' coordinates is the n x n matrix whose jth column contains the B' coordinates for the original jth basis vector u_j. In other words, this matrix, denoted [I]_B,B' in the text, is
  [I]_B,B' = [ [u₁]_B' | [u₂]_B' | ...| [u_n]_B' ].
  
  
• Question: Given two different bases for the same n-dimensional vector space V, and given some linear operator T:V-->V, how are the matrix that represents T relative to one basis and the matrix that represents T relative to the other basis related?
  • Why do we care? If we can understand the answer, then we may be able to choose a good basis. Good basis in this case would mean one that makes the representing matrix as simple as possible, for instance a diagonal matrix.
  • Answer: Say that the two bases are B and B'. Denote the two representing matrices by [T]_B and [T]_B', respectively. Let P be the transition matrix from B' to B. Let Q = P^-1 = the transition matrix from B to B'.Then
    [T]_B' = P^-1 [T]_B P which says the same as [T]_B' = Q [T]_B P
    [T]_B = Q^-1 [T]_B' Q which says the same as [T]_B = P [T]_B' Q
    
  • Why this answer? Think of B as a language, B' as another language. Given the B'-language version of a vector v, one way to find the B'-language version of T(v) is simply to multiply by the matrix [T]_B'. Another way to do the same thing would be to (1) change languages (2) carry out the transformation T in the new language (3) change languages back. In matrix form this would mean: given the B'-language version of v, we would (1) first multiply by P; (2) then multiply by [T]_B; (3) then multiply by Q. Thus multiplication by [T]_B' and multiplication by Q [T]_B P both accomplish exactly the same thing. In other words both have exactly the same effect on all the x's in Rⁿ. This gives us that they must actually be the same matrix.
  • See Exple 1 p405
• Definition. If A and B are square matrices (of the same size), then we say that B is similar to A iff there is an invertible matrix P such that B = P^-1 A P.
  • It is pretty easy to show that B similar to A ==> A similar to B.
  • So usually we just say A and B are similar.
  • The question and answer just above could be restated as follows: all of the matrices that could be used to represent some linear operator T:V-->V (where V is some n-dimensional vector space) are similar to each other.
  • When two matrices are similar, they have a lot in common! See p 406.
    

Notes to guide the reading for 7.1.

• Why do eigenvectors matter? One important reason is the following: when we are trying to use a matrix machine to carry out a linear operator T:V --> V, we would like to have the matrix be a nice one, for instance a diagonal matrix. Finding eigenvectors lets us decide whether this is possible. When it is possible, using eigenvectors lets us choose a basis for V that will make the matrix for T a diagonal matrix. So eigenvectors help us get good/nice/manageable matrices for linear transformations.
• What is an eigenvector for an n x n matrix A? What is an eigenvalue for A? What is an eigenspace?
  • It is a vector x in Rⁿ that (a) is not the zero vector and (b) has the property that Ax is a scalar multiple of x.
  • If x is an eigenvector for which Ax = lx, then l is called an eigenvalue for A and we say that x is an eigenvector corresponding to the eigenvalue l. (l is pronounced "lambda", by the way) So eigenvectors come equipped with corresponding eigenvalues and vice versa.
  • If l is an eigenvalue for A, then the set of eigenvectors for A corresponding to l almost forms a subspace of Rⁿ, called the eigenspace corresponding to l. It's fails to be a subspace only because it lacks a zero vector. It is otherwise a subspace because it's essentially the solution space for a homogeneous system.
  • Does a matrix always have eigenvalues? Well, yes and no. Yes, if we allow complex scalars (which would make sense in the context of complex vector spaces rather than real vector spaces). No, if we only allow real scalars, as we have been doing. So if complex scalars are ruled out (ruling out complex eigenvalues) some matrices have (real) eigenvalues and some don't. If there are real eigenvalues there are at most n of them.
• Equivalent conditions for eigenvalues.
  The condition "l is an eigenvalue for A" has the following important equivalent conditions.
  1. The homogeneous system (A-lI)x=0 has nontrivial solutions.
  2. A-lI is not invertible
  3. det(A-lI) = 0
  4. det(lI-A) = 0

  
  So how do we find eigenvalues, eigenvectors, eigenspaces?
  
  

• We find eigenvalues first. To find eigenvalues, we usually solve the equation in condition 3 (or 4). This equation is called the characteristic equation) for l. So we find eigenvalues by solving the characteristic equation. See Example 6 on p100; Examples 2, 3, 4 pp339-340.
• Then we find eigenvectors. How? Once we have found a particular eigenvalue l, we find a basis for the corresponding eigenspace by simply solving the corresponding system from condition 1. See Example 6 p100, Example 5 p341.
• Displaying a basis for an eigenspace describes the eigenspace.
• Note the new, longer Long Theorem on p344. Ignore (r),(s),(t) for now, though.

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Notes to guide the reading for 7.2

• This section begins with two related problems concerning a square (n x n, say) matrix A.
  • Eigenvector Problem for A: Does Rⁿ have a basis consisting entirely of eigenvectors for A?
  • Diagonalization Problem for A is Does there exist a diagonal matrix D that is similar to A? In other words, do there exist a diagonal matrix D and an invertible matrix P such that D = P^-1AP?
• Why do these problems matter? See "Why do eigenvectors matter?" above. They are driven by the problem of trying to find a basis that will let us represent a linear transformation by a diagonal matrix (a nice matrix!).
• Theorem connecting the two problems.
  • Theorem. For a given A, either the "Eigenvector Problem" and the "Diagonalization Problem" both have a solution or neither has a solution. (i.e. both answers are yes or both answers are no).
  • More precisely, for a given A, whether the Diagonalization Problem can be solved depends on whether one can find enough independent eigenvectors for A. How many is enough? Enough to make a basis for Rⁿ.
• So how do we find eigenvectors? That's what's covered in 7.1.
• If we do find enough independent eigenvectors how do we use them to diagonalize A? See the procedure outlined on p349. See Exple 1 p349.
• Shortcuts for knowing we have (enough) independent eigenvectors.
  • Thm 7.2.3 If A has n distinct real eigenvalues, then it will be possible to diagonalize A.
    • Why? See Thm 7.2.2 below.
    • What if not? When an an x n matrix A fails to have n distinct real eigenvalues then it might or might not be possible to diagonalize, with the deciding factor still being whether or not there are n independent eigenvectors. See Exple 1 p349 and Exple 2 p340.
  • Thm 7.2.2 If A has n distinct real eigenvalues and we choose one eigenvector for each eigenvalue, then the resulting set of eigenvectors will be independent (so there will be enough). That's why it will be possible to diagonalize A in this case. See Exple 3 p352, Exple 4 p352.
  • 7.2.2 is a little more general.
    • Whenever we have distinct eigenvalues, however many, if we take one eigenvector for each eigenvalue, then the corresponding set of eigenvectors will be independent.
    • What 7.2.2. does not say: if chosen eigenvectors do not correspond to distinct eigenvalues then the eigenvectors won't form an independent set. Sometimes they do. (Sometimes they don't.) See Exple 1 p349 and Exple 2 p350.
  • A more general theorem. Given several distinct eigenvalues, if we find a basis for each of the corresponding eigenspaces and then "merge" these bases into a single set of eigenvectors, then the resulting set of eigenvectors will always be independent. See the Remark on p352.
    • For instance, say that the (only) eigenvalues are 1 and 2, and that the eigenspace corresponding to the eigenvalue 1 has dimension 2 and the eigenspace corresponding to the eigenvalue 2 has dimension 3. Then if we find a basis for the first eigenspace (so 2 vectors) and find a basis for the second eigenspace (so 3 vectors) and merge these two basis (so 5 vectors all together) then the 5 chosen eigenvectors form an independent set. If the matrix A happened to be 5 x 5 then we can diagonalize. But if A is bigger than that we can't.
  • Thm 7.2.4. Yet another criterion for diagonalizability:
    • A is diagonalizable iff, for each eigenvalue l₀, the dimension of the corresponding eigenspace is the same as the number of times that (l-l₀) appears as a factor of the characteristic polynomial.
    • The former is always less than or equal the latter. The criterion requires it be equal.
    • For instance, if the characteristic polynomial is (l-1)²(l-2)³ then the eigenspace corresponding to the eigenvalue 1 should have dimension 2 (it might only have dimension 1) and the eigenspace corresponding to the eigenvalue 2 should have dimension 3 (it might only have dimension 1 or 2).
    • See the alternative solution for Exple 2 p350.
    • See p353 "Geometric and Algebraic Multiplicity", if you like, for some alternative language that is used in this connection.
• Iterative processes often involve using higher and higher powers of the same matrix. Computing these powers can be a pain. If the matrix can be diagonalized, computing these powers becomes much, much easier. See pp353-354, especially Exple 5 and the remark that follows it.
  
  

• Alexia Sontag, Mathematics
• Wellesley College
• Date Created: January 4, 2001
• Last Modified: May 3, 2002
• Expires: June 30, 2002