Google Answers Logo
View Question
 
Q: Spectral Theorem Proof ( Answered 5 out of 5 stars,   0 Comments )
Question  
Subject: Spectral Theorem Proof
Category: Science > Math
Asked by: oaaltone-ga
List Price: $50.00
Posted: 12 May 2003 16:39 PDT
Expires: 11 Jun 2003 16:39 PDT
Question ID: 202898
Was given this extra-credit project. The professor said ANY outside
source could be used, including other professors, as long as the
source was cited when the answers are passed in. A higher quality, pdf
version of the questions is available at
http://www.aaltonen.us/download/math235extracredit.pdf

Here it is in plain text:

We start with a n x n mx A. A subspace E of Rn is an A-subspace if AX
is in E whenever X is in E.

1. Show that if E is an A-subspace of Rn , then Eperp is an
AT-subspace.

2. Show that if B1 , ... , Bn is a basis of the A-subspace E and Bn,
... , Bn is a basis of Eperp, and B is the mx ( B1 , ... , Bn ) then
B-1AB has the shape

C 0
0 D

where C is r x r and D is (n-r) x (n-r).


3. Show that in 2., if B is orthogonal and A is symmetric, then C and
D are symmetric.


We assume that if A is symmetric n x n mx and E != (0) is an
A-subspace. then A has a (unit) eigenvector in E.


4. Show , using Gram-Schmidt, that N( A) has an orthonormal basis, B1
, ... , Bs
From now on. we assume that A is symmetric., 

5. Suppose that A has rank r and X is an eigenvector of A with
eigenvalue a != 0. Show that A - aXXT is symmetric, and that it has
rank <= r - 1. (Use a previous exercise.)

6. Suppose that we have found orthonormal eigenvectors Bs+1, ... ,
Bs+k, of A with non-zero elgenvalues bs+i  … bs+k that the mx
A(k) = A - bs+1 B s+1 B s+1 T  … - bs+k B s+k B s+k T
has rank <= r - k. Show that A(k) is symmetric, that N( A(k)) is an
A(r)-subspace and that if A( k) != 0. then A(k) has a unit eigenvector
Bs+k+1 in N (A(k) )perp. Show that B1  … Bs+k+1 are orthonormal, and
that rank A( k+ 1) <= r - k - 1,
and that each mx BjBjT is a projection of rank 1 Conclude that A( n) =
0, and that
A = b1 B1 B1 T +. .. + bn Bn Bn T
where b1 = .... = bs = 0, bs+1 	bn * 0 and that if B is the mx (B1 ,
... , Bn ),
then B-1 A B is a diagonal mx. with diagonal entries b1 , ... , bn.

This result is called the "spectral theorem"

PART 2. Let A be the mx

1  0 -1
0  2  0
-1 0  1

1. Find the distinct eigenvalues X of A and their multiplicities 2.
Determine the dimensions of the eigenspaces N( A - &#955;I)

3. Find orthonormal bases of the eigenspaces in 2.. Combine these
bases into one
orthonormal basis B of R3 and verify that the matrix of A relative to
B is a diagonal matrix with entries the eigenvalues of A, each
repeated as many times as its multiplicity.

Clarification of Question by oaaltone-ga on 12 May 2003 16:40 PDT
I need the answer to this by Thursday 5/15 at the very latest. Thanks!

Request for Question Clarification by mathtalk-ga on 12 May 2003 17:03 PDT
Hi, oaaltone-ga:

I believe that part 2. of these questions requires that A be
symmetric.

Consider for example the standard basis e_1 = (1,0)' and e_2 = (0,1)'
where ' denotes for convenience the transposition of row to column. 
Then if A =

 0  1 
 0  0

we see that A e_1 = 0 and that the subspace spanned by e_1 is
therefore an A-space, to use the problems terminology.  Of course the
orthogonal complement of that space is the space spanned by e_2.  But
while the latter space is an A'-space, it is not an A-space.  Indeed A
e_2 = e_1, so that (as we already knew), the representation B^-1 A B
of A with respect to the standard basis is not block diagonal (with
respect to the sub-bases {e_1} and {e_2}) as the problems conclusion
asserts.

You should also "uncorrect" the business (hand-written in the pdf
version) of changing the first subscript range on the basis elements
from B_1,...,B_r to B_1,...,B_n.  Note that r + (n-r) = n, and the
union of the basis for E and E^perp is always a basis for R^n.

regards, mathtalk-ga

Clarification of Question by oaaltone-ga on 12 May 2003 17:12 PDT
Hi mathtalk-ga,

I spoke with my professor RE #2, and he stated that A is indeed
symmetric. Thanks also for the clarification as to the
hand-corrections, I realize my mistake now.

Clarification of Question by oaaltone-ga on 12 May 2003 18:34 PDT
If it's not too much to ask, can anyone please post if they're going
to work on  the questions? I'd like to get an idea that someone is in
fact working on it.

Thanks in advance!

oaaltone-ga

Request for Question Clarification by mathtalk-ga on 13 May 2003 15:23 PDT
Hi, oaaltone-ga:

In order to be efficient in assisting you (esp. since your timeframe
is short), I'd like to have a better idea of your math background in
relation to these two multiple part problems.  If I were to spend a
lot of time explaining basics of linear algebra that are well-known to
you, it might be useful to another reader, but probably not to you. 
So please take a few minutes to explain what about this assignment you
do know how to tackle, and what part needs a bit of clarification.

Thanks in advance, mathtalk-ga

Clarification of Question by oaaltone-ga on 13 May 2003 17:32 PDT
Hi mathtalk-ga,

I'm finishing up an intro course to Linear Algebra, so I understand
most of the basic concepts. The assignment was an old worksheet from
the graduate level Linear Algebra class that the professor decided to
hand out as extra credit. To tell the truth, if the professor had
handed this out a couple weeks earlier and I had sufficient time to go
over a couple textbooks, I'd probably be able to figure the problems
out, but in this timeframe I do not believe it is possible. The reason
I went to Google Answers was because:

1) The professor encouraged outside research and collaboration.
2) The extra credit is worth enough to spend money of Google Answers
for quality work.

If you were to post your answers without going through too much
trouble to explain your strategies, I would most likely be able to
look over them and understand where the answers came from. If I had
any further questions, I could always post them later.

Thanks in advance,

oaaltone-ga

Request for Question Clarification by mathtalk-ga on 13 May 2003 18:09 PDT
Hi, oaaltone-ga:

Ah, that does put a lot of clarity on the situation.

Let me drill down on Problem 2 just a bit.  This is pretty standard
calculation stuff (find the three eigenvalues and eigenvectors, up to
multiplicity), and it looks like you've handwritten "Done" next to
part 2 of Problem 2 (in the margin).  Eigenvectors that correspond to
distinct eigenvalues of a symmetric matrix are automatically
orthogonal to one another, so one pretty quickly gets an orthonormal
basis here.  If you know how to find the eigenvalues and eigenvectors,
I don't want to beat that subject to death; but if it's something you
need help with, I won't be shy about it.

There are some other fine researchers here with strong backgrounds in
linear algebra.  I will make sure one of us gets you a satisfactory
answer within your timeframe.

Thanks, mathtalk-ga

Clarification of Question by oaaltone-ga on 13 May 2003 19:21 PDT
mathtalk-ga,

I've looked over Part 2 and I think I understand how to do it. It is,
as you say, a straight-forward calculation problem. I should be able
to do that part on my own, but then again, I wouldn't mind having some
answers from an expert to compare if it's not too much additional
trouble... My main concern is with Part 1 -- the meat of the
worksheet.

Thanks again for the help! If I could get some answers in the
timeframe specified, I would be extremely greatful!

oaaltone-ga

Clarification of Question by oaaltone-ga on 15 May 2003 08:14 PDT
Any updates? I need this answer by this answer by tonight! Thanks in advance!

oaaltone-ga

Request for Question Clarification by mathtalk-ga on 15 May 2003 09:24 PDT
Hi, oaaltone-ga:

I've finished writing up steps 1-5 of Part 1.  In step 6 there is a
reference to A(r), in the context of showing something is an
A(r)-space.  While r must surely be taken to mean the rank of A, as it
was used this way in previous problems, it's a bit hard to come up
with an interpretation of A(r) except as the foreseeable extension of
the construction A(k).

In confirming this notation against your PDF version, I noticed that
the last step of Part 1 is separately labelled as 7, so I will follow
that approach in my write-up.  However in step 7 the notation A(n)
rather than A(r) is used.  In short there's a bit of confusion in the
notation, which I will try to sort out for you in my answer.

You may want to mention to your instructor that step 5 requires the
additional assumption that eigenvector "X" is of unit length.

regards, mathtalk-ga

Clarification of Question by oaaltone-ga on 15 May 2003 10:14 PDT
Hi, mathtalk-ga,

Thanks for the update. I just looked over my notes from class, and the
professor had confirmed that 5 does indeed require the assumption that
X is a unit vector. As far as notation discrepancies, your guess is as
good as mine! I'll try to get in touch with my professor and ask, but
classes are over already, and I don't know where he'll be today.

oaaltone-ga
Answer  
Subject: Re: Spectral Theorem Proof
Answered By: mathtalk-ga on 16 May 2003 01:50 PDT
Rated:5 out of 5 stars
 
Hi, oaaltone-ga:

Sorry for the delay in posting this tonight, but we had an extended
power outtage due to severe weather.

Here are my notes on Part 1, intermingled with the problem statement. 
I have adapted the notation for clarity in this text based medium,
using capital letters only for matrices, lowercase letters in a "upper
range" like u,v,w for vectors, and lowercase letters toward the
beginning of the alphabet like a,b,c for scalars.  Subscripts are
shown by an underline character "_" and superscripts, esp. "perp" for
the orthogonal complement, in combination with "^" hat characters.

I will immediately post answers to Part 2 as a follow-up.

regards, mathtalk-ga

Part 1.

Let A be an n by n matrix.  We say that subspace E of R^n is an
A-subspace iff:

u in E implies Au in E

For convenience we shall refer to the orthogonal complement of E,
E^perp, by F, so that the direct sum E + F = R^n.  Note that there are
always at least two trivial A-subspaces, {0} and R^n, assuming n > 0.


We will denote the Euclidean inner product on R^n between vectors u,v
by u*v, where vectors are identified with single column matrices. 
Another notational convenience will be A' for the transpose of A. Note
that by the properties of transpose and matrix multiplication:

u*v = u'v

Au * v = u'A'v = u * A'v

1.  Show that if E is an A-subspace of R^n, then F, the orthogonal
complement of E in R^n, is an A'-subspace.

Proof:  Let u belong to F.  We are asked to show this implies A'u
belongs to F.  For this it suffices to show that A'u is orthogonal to
each vector in E.

Let v belong to E. Now:

A'u * v = u * A''u = u * Av = 0

since Av is in E and u belongs to the orthogonal complement F.  Thus
A'u is again in F, which proves that F is an A'-subspace.

2.  Let {b_1,...,b_r} be a basis for A-subspace E.

Let {b_{r+1},...,b_n} be a basis for F, the orthogonal complement of
E.  By a standard construction the union of these is a basis for R^n.

Then the matrix B = [b_1 | b_2 | ... | b_n] is a "change of basis"
matrix which translates from "B" coordinates to standard coordinates.

Assuming that A is symmetric, A = A', show that the similarity
transformation B^-1 A B produces a block-diagonal result:

C  0

0  D

where C is r by r and D is (n-r) by (n-r).

Proof: Let {e_1,e_2,...,e_n} be the standard basis of unit vectors in
R^n.  One way to elucidate the structure of M = B^-1 A B is to "probe"
the columns of M by forming matrix-vector products M e_k, which yields
the k'th column of M.

In particular we will show that:

 (i) If 1 <= k <= r, then M e_k = (c_{1,k},...,c_{r,k},0,...0)' for
some real constants c_{1,k},...,c_{r,k}.

(ii) If r+1 <= k <= n, then M e_k = (0,...,0,d{1,k},...,d{n-r,k})' for
some real constants d_{1,k},...,d_{n-r,k}.

From these two demonstrations it follows that M = B^-1 A B has the
block-diagonal structure claimed for it, with C = [c_{i,j}], D =
[d_{i,j}].

Now B e_k = b_k by virtue of our construction of B above, and
similarly:

B^-1 b_k = e_k

Let's tackle case (i), in which k <= r.  Thus b_k is in A-subspace E,
so that A b_k is again in E.  Therefore it can be uniquely expressed
aa a linear combination of the first r basis elements b_1,...,b_r:

A b_k = c_{1,k} b_1 + ... + c_{r,k} b_r

Thus M e_k = B^-1 A B e_k = B^-1 A b_k, and:

M e_k = B^-1 (c_{1,k} b_1 + ... + c_{r,k} b_r)

      = c_{1,k} B^-1 b_1 + ... + c_{r,k} B^-1 b_r
      
      = c_{1,k} e_1 + ... + c_{r,k} e_r
      
      = (c_{1,k},...,c_{r,k},0,...,0)'

just as we desired to show.

The case (ii) is precisely the same sort of calculation involving the
last (n-r) elements of the basis formed by the columns of B, once it
is acknowledged that F is an A-subspace too.

The only point that needs to be carefully examined is the fact that F
is also an A-subspace.  But this follows from the symmetry of A = A'
and our work in the previous step where we showed that F is an
A'-subspace.  [Recall from the example considered in the request for
clarification that if A were not symmetric, F could be an A'-subspace
but _not_ an A-subspace.]

3.  Show that in 2., if B is orthogonal and A is symmetric, then C and
D are symmetric also.

Proof:  The symmetry of block submatrices C,D is equivalent to the
symmetry of M overall, since:

     --     --
M' = | C'  0 |
     | 0   D'|
     --     --

so that M = M' if and only if both C = C' and D = D'.

Now the definition of an orthogonal matrix is that B^-1 = B', the
inverse is the transpose.  So a simple computation, using the
orthogonality of B and the symmetry of A, gives that:

M = B^-1 A B = B' A B

M' = (B' A B)' = B' A' B = B' A B = M

Thus M is symmetric under the given hypotheses.

Underlined quotation:

  We assume that if A is symmetric n by n matrix and
  E not {0} is an A-subspace, then A has a (unit) 
  eigenvector in E.

[Note:  The insertion of this assumption at this particular point
seems a little strange.  As we have seen, the symmetry of A was
required for all steps in this problem after the first.  On the other
hand no symmetry assumption or assumption about existence of a unit
eigenvector is needed in step 4 following, if the notation N(A) is
properly understand as "nullspace of A".]

4.  Show, using Gram-Schmidt, that N(A) has an orthonormal basis,
{u_1,...,u_s}.

Proof:  Any (sub)space has a basis, so suppose that the nullspace of
A:

N(A) = { v | Av = 0 }

has basis {v_1,...,v_s}.  

The Gram-Schmidt construction applies to any inner-product space.  To
kick it off one simply normalizes the first vector of the (ordered)
basis:

u_1 = (1/|v_1|) v_1

where we denote the norm (length) of a vector v by |v|.

We process subsequent basis vectors v_k by projecting them onto to the
subbasis formed by previous output vectors u_1,...,u_{k-1}, obtaining
a "residual" r_k:

r_k = v_k - (v_k*u_1)u_1 - ... - (v_k*u_{k-1})u_{k-1}

Because of the linear independence of the basis vectors v_k, noting
that u_1,...,u_{k-1} are linear combinations of v_1,...,v_{k-1}, we
know r_k is not zero.  Therefore we can normalize r_k to get u_k:

u_k = (1/|r_k|) r_k

Keep this up.  The resulting sequence of vectors:

u_1, u_2, ... , u_s

are normalized (unit length) and mutually orthogonal by their
construction.  For example, r_k above is perpendicular (orthogonal) to
each u_i, i < k, because of the projection used to define r_k, with
u_i being already orthogonal to each u_j with j not equal to i:

r_k*u_i = v_k*u_i - 0 ... - (v_k*u_i)(u_i*u_i) ... - 0

        = v_k*u_i - (v_k*u_i)

because after normalization u_i*u_i = 1.

Therefore the Gram-Schmidt process converts the ordered basis {v_k}
into a linearly independent orthonormal sequence {u_k} of equal size,
which must then also be a spanning set and basis by the invariance of
the dimension of a vector space.

[Alternatively one could explicitly show that each v_k can be
expressed as a linear combination of the vectors u_1,...,u_k, so that
the spanning property of the vectors u_1,...,u_s is inherited from
that of the vectors v_1,...,v_s.]

Underlined quotation:

  From now on, we assume that A is symmetric.

[Note:  We must add to the statement of the next step the condition
that |u| = 1.]

5.  Suppose that A has rank r and u is a _unit_ eigenvector of A with
nonzero eigenvalue a.  Show that A - auu' is symmetric, and that it
has rank at most r-1.  (Use a previous exercise.)

Proof:  If A has rank r, then the nullity of A, the dimension of its
nullspace, must be (n-r).  This is often proven by elementary row
operations and counting the number of leading ones in a reduced row
echelon form of A.  We return to this observation at the end, possibly
what was meant by "a previous exercise".

Since A and auu' = (auu')' are both symmetric, it is clear that A -
auu' is symmetric.

We want to show that the nullspace of A - auu' is strictly larger than
the nullspace of A.  To begin with let's show that the nullspace of A
is contained in the nullspace of A - auu'.  For this we need a
"lemma":

If Av = 0, and Au = au with a nonzero, then u*v = 0.

Pf. of "lemma":

au*v = Au*v = u*Av = u*0 = 0 (using symmetry of A)

Then since a is nonzero, u*v = 0.

Therefore with this lemma, any vector v belonging to the nullspace of
A will again belong to the nullspace of A - auu':

Av = 0 implies

  (A - auu')v = Av - auu'v = Av - 0u = 0

But also u, which is not in the nullspace of A (why?), belongs to the
nullspace of A - auu'.  Here the assumption that u has unit norm comes
into play, because u'u = 1:

(A - auu')u = Au - au = au - au = 0

Since the dimension of the nullspace of A - auu' is therefore strictly
larger than that of A, we deduce from rank + nullity = n that the rank
of A - auu' is less than rank of A, which is r.  Thus rank of A - auu'
is at most r-1. QED

[Note: The general thrust of steps 6 and 7 is to extend the result of
step 5 by induction to obtain A - SUM a_i u_i (u_i)' with rank 0, i.e.
to represent A as the sum of such symmetric rank one updates as
indicated.  My presentation tries to emphasize this continuity in the
thread of ideas.]

6.  Suppose that we have found (unit) orthonormal eigenvectors 
u_{s+1},...,u_{s+k} with non-zero eigenvalues a_{s+1},...,a_{s+k}
respectively, so that the matrix:

A_k = A - SUM a_i u_i (u_i)' FOR i = s+1 to s+k

has rank at most r-k.  Show that A_k is symmetric, that N(A_k) is an
"A_r" -subspace [sic], and that if A_k is not zero then A_k has a unit
eigenvector u_{s+k+1} orthogonal to N(A_k).

Proof:  The symmetry of A_k follows easily from the same argument
offered in step 5, that A and each individual subtrahend a_i u_i
(u_i)' are symmetric.

The claim that N(A_k) is an "A_r"-subspace is a bit hard to
understand.  It is of course true that N(A_k) is an "A_k"-subspace;
that merely reflects the trivial observation that every vector in
N(A_k) is mapped to zero by A_k and so is trivially back in N(A_k). 
Furthermore, since the construction we have already explored in step 5
provides for the strict inclusion of N(A_k) in N(A_m) for any m > k
later in the induction proof, it follows also that N(A_k) is an
"A_r"-space for the same trivial reason.

Of greater difficulty would be showing that N(A_k) is an A-space.  I
will not pursue this tangent at present, but if it turns out to be of
interest, you are welcome to request a clarification on that issue.

Finally let's suppose A_k is nonzero.  Then from the symmetry of A_k,
as shown in the proof of step 2, it follows that not only is N(A_k) an
A_k-space, its orthogonal complement N(A_k)^perp is also an A_k-space.
 From the fact that A_k is nonzero it follows that N(A_k) is not all
of R^n, and thus that N(A_k)^perp is more than just {0}.  We are
therefore justified in applying the underline assumption given between
steps 3 and 4 to the effect that A_k has a unit eigenvector in
N(A_k)^perp.

That we label such a unit eigenvector of A_k as u_{s+k+1} is perhaps
more significant than may first appear, for as we are engaged in an
induction proof, there is an implied obligation to show that u_{s+k+1}
is an eigenvector of the original matrix A and corresponds there to a
nonzero eigenvalue a_{s+k+1}.

But u_{s+k+1} is orthogonal to each of u_1,...,u_{s+k} by virtue of
these latter vectors belonging to N(A_k).  Hence in particular:

a_i u_i (u_i)' u_{s+k+1} = 0  for each i < s+k+1

and therefore:

A u_{s+k+1} = A_k u_{s+k+1} = a_{s+k+1} u_{s+k+1}

QED

7.  Show that u_1,...,u_{s+k+1} are orthonormal, and that
rank(A_{k+1}) is at most r-k-1, and that each matrix u_i (u_i)' is a
projection of rank 1.  Conclude that A_n = 0 [sic] and that:

A = SUM a_i u_i (u_i)' FOR i = 1 to n

where a_1 = ... = a_s = 0 and a_{s+1},...,a_n nonzero, and that if:

B = (u_1 | ... | u_n)

then B^-1 A B is a diagonal matrix with diagonal entries a_1,..., a_n.

This result is called the "spectral theorem".

Proof:  

Recall that s = nullity of A = n - rank of A = n-r.  We have
previously shown the construction of u_1,...,u_s as an orthonormal
basis for N(A) by the Gram-Schmidt process, and have shown the
extension of the orthogonality property by the remaining unit vectors
above in the proof of step 6 in connection with the status of
u_{s+k+1} as an eigenvector for A.

Therefore the sequence of vectors u_1,...,u_n forms an orthonormal
basis, and the matrix B obtained by concatenating these columns is an
orthogonal matrix.

Note that as u_i (u_i)' is clearly of rank 1 since u_i is nonzero, we
need only show that it is also a projection, i.e. squaring this matrix
gives itself:

u_i (u_i)' u_i (u_i)' = u_i (u_i)'

But this is a direct application of the associativity of matrix
multiplication and the unit length of u_i:

(u_i)' u_i = 1

We have also shown in step 6, citing the argument used in step 5, that
the sequence of nullspaces of A = A_0, A_1, ... , A_r is strictly
increasing.  Since the nullity of A is s, it follows that the nullity
of A_{k+1} is at least s + k + 1, so rank of A_{k+1} is at most n -
(s+k+1) = r - k - 1.

Indeed, since the rank of A_r is then at most 0 and also at least 0,
we have that A_r = 0 (and incidentally that all of the rank
inequalities must be tight, i.e. matched to equalities).

But if A_r = A - SUM a_i u_i (u_i)' = 0, then:

A = SUM a_i u_i (u_i)'

as desired.  

It remains only demonstrate the diagonality of B^-1 A B and the
correspondance of its diagonal elements with the specified eigenvalues
a_1,...,a_n.  But the structure of B^-1 A B is constrained by steps 2
and 3 and the induction process to have a block diagonal "shape" in
which the incremental blocks after the first s zeroes are 1 by 1
blocks with nonzero entries.  If you like we can recall the
computation of step 2, in which standard basis vectors e_i are used to
"test" the contents of the entries of B^-1 A B.  Here B e_i = u_i has
been shown to be an eigenvector of A with:

A u_i = a_i u_i

for i = 1 to n.

We must point out that the notation developed in step 6 would make it
proper here to speak of A_r = 0 rather than A_n = 0 as step 7 has
denoted it.  But there can be little doubt of the author's intention;
the notational "error" appears motivated by the related indexing of
the eigenvalues and eigenvectors of A.

QED

The name "spectral theorem" sounds a little fancy, perhaps.  The term
spectrum has come to be identified with the eigenvalues of a linear
operator, or more generally the scalars a for which a linear operator
(A - aI) fails to be invertible.  The result here for matrices is a
"dressed down" version of a more sophisticated theorem about "bounded
normal" linear operators on a (possibly infinite dimensional) Hilbert
space.  A common feature is the "representation" of the linear
operator as a sum (or integral) involving projection operators.


Search Strategy

personal knowledge of material

Clarification of Answer by mathtalk-ga on 16 May 2003 02:22 PDT
In Part 2 we are asked to "illustrate" the spectral theorem with
respect to a matrix:

      1  0 -1 
A =   0  2  0 
     -1  0  1 

The first step is to identify the eigenvalues of this matrix, which we
can do by computing and solving the characteristic polynomial:

det (A - xI) = (1-x)(2-x)(1-x) - (2-x)
             = x(x-1)(2-x)

x = 0,1,2

An eigenvector corresponding to x = 0 is (1,0,1)', although this is
not yet of unit length.

Likewise we also have two eigenvectors corresponding to x = 2 in
(1,0,-1)' and (0,1,0)' which happen to be orthogonal.

So we only need to arrange for these mutually orthogonal vectors to
have unit length, and by scaling:

u_1 = (1/sqrt(2)) (1,0,1)'

u_2 = (1/sqrt(2)) (1,0,-1)'

u_3 = (0,1,0)'

Therefore the orthogonal matrix B would be:

     1/sqrt(2)   1/sqrt(2)     0
B =     0           0          1
     1/sqrt(2)  -1/sqrt(2)     0

A certain amount of tedious computation is needed to verify:

        0   2/sqrt(2)  0
A B =   0      0       2
        0  -2/sqrt(2)  0


           0  0  0
B^-1 A B = 0  2  0
           0  0  2

an "orthogonal diagaonlization" of A (since B^-1 = B').

regards, mathtalk-ga

Clarification of Answer by mathtalk-ga on 16 May 2003 02:30 PDT
OOPS... I made a copying error somehow in the follow-up section.

The paragraph that discusses the characteristic polynomial in Part 2
should have read:

The first step is to identify the eigenvalues of this matrix, which we
can do by computing and solving the characteristic polynomial:
 
det (A - xI) = (1-x)(2-x)(1-x) - (2-x) 
             = x(x-2)(2-x) 
 
x = 0,2,2 

The rest of the discussion then is consistent.  Sorry for the
confusion.

regards, mt

Request for Answer Clarification by oaaltone-ga on 16 May 2003 08:37 PDT
mathtalk-ga,

Wow. Amazing. I've been looking over your answer for the last hour or
so and you've done a _great_ job of explaining this. I am only
confused with Part 2. Is your answer the answer to all 3 questions
lumped together into one answer or is it only the answer to question 1
of part 2?

oaaltone-ga

Clarification of Answer by mathtalk-ga on 16 May 2003 11:09 PDT
Hi, oaaltone-ga:

I thought I was answering all parts of the question in "lumped
together" fashion to a level of detail suitable for you to check your
own computational work on this problem.  The one "variable" in the
answer has to do with the eigenspace corresponding to eigenvalue 2. 
Since this is two-dimensional, there are many ways to pick the basis
for that portion of the problem.  You'd always have two eigenvectors,
but they can be chosen in infinitely many ways.  I just picked a
particular pair of simple form; your choice might differ.

Let's see if I overlooked something in Part 2:

PART 2. Let A be the mx 
 
1  0 -1 
0  2  0 
-1 0  1 
 
1. Find the distinct eigenvalues x of A and their multiplicities.

[Eigenvalues are 0 and 2, with eigenvalue 2 having multiplicity 2.]

2. Determine the dimensions of the eigenspaces N( A - xI)

[This is really the same as the multiplicity question, at least for
real symmetric matrices.  One calls the multiplicity of a root of the
characteristic polynomial its "algebraic multiplicity" and the
dimension of the corresponding eigenspace its "geometric
multiplicity".  The spectral theorem tells us that these
multiplicities are equal, ie. that we have a complete basis of
eigenvectors, so I didn't dwell on the distinction.  Nonetheless I
produced one eigenvector for x = 0 and two orthogonal (hence linearly
independent) eigenvectors for x = 2.  So this settles the question
about dimensions of eigenspaces.]
 
3. Find orthonormal bases of the eigenspaces in 2.. Combine these
bases into one orthonormal basis B of R3 and verify that the matrix of
A relative to B is a diagonal matrix with entries the eigenvalues of
A, each
repeated as many times as its multiplicity.

[I did provide the normalized orthogonal eigenvectors and assembled
them as columns into an orthogonal matrix B.  As discussed above, you
might have picked a different pair of unit eigenvectors for eigenvalue
2, which would lead to a different matrix B.  In particular we could
swap the order of the two vectors that I chose, which actually makes
the matrix B a bit more elegant looking.  I didn't write more than a
summary of the computation of B^-1 A B, so if some additional details
are needed, please let me know.  Observe that the diagonal entries of
B^-1 A B contain 0 once and 2 twice, as agrees with the multiplicities
of these eigenvalues.]
oaaltone-ga rated this answer:5 out of 5 stars
Thanks for the help, I'm going to take your answer and review it
thoroughly. It will be very helpful!

Comments  
There are no comments at this time.

Important Disclaimer: Answers and comments provided on Google Answers are general information, and are not intended to substitute for informed professional medical, psychiatric, psychological, tax, legal, investment, accounting, or other professional advice. Google does not endorse, and expressly disclaims liability for any product, manufacturer, distributor, service or service provider mentioned or any opinion expressed in answers or comments. Please read carefully the Google Answers Terms of Service.

If you feel that you have found inappropriate content, please let us know by emailing us at answers-support@google.com with the question ID listed above. Thank you.
Search Google Answers for
Google Answers  


Google Home - Answers FAQ - Terms of Service - Privacy Policy