Math 311 Lecture 35
RECALL. D A is symmetric iff AT = A.
E For the standard column vector inner product,
[(u, v)] = uT[v.
F L:VéV is diagonalizable iff V has a basis of
eigenvectors for L.
LEMMA. (Av, u) = (v, ATu).
PROOF. (Av, u) = (Av)T[u = (vT[AT)[u = vT[(ATu) = (v, ATu).
DEFINITION. A matrix is orthogonal iff A-1 = AT. A matrix
A or a linear transformation L is an isometry iff it
preserves inner products, i.e., for all u, vLV,
(Au, Av) = (u,v) or (Lu, Lv) = (u,v) respectively.
THEOREM. A is orthogonal iff its columns are orthonormal
(orthogonal unit vectors) iff its rows are orthonormal iff
A is an isometry.
PARTIAL PROOF. Assume A is orthogonal. W A-1 = AT and so
ATA = A-1A = I. We want orthonormal columns.
If vi, vj are columns of A, then viT is the ith row of AT.
W(vi, vj) = viT[vj = the (i,j)th entry of ATA = the (i,j)th
entry of I (since ATA = I) = [0 if i j and 1 if i = j].
If i j, then (vi, vj) = 0 and vi and vj are orthogonal.
If i = j, then (vi, vi) = 1 and so vi has length 1.
Finally we show that A is an isometry. For any u, vLV,
(Au, Av) = (u, ATAv) = (u, A-1Av) = (u, Iv) = (u, v). E
THEOREM. A symmetric matrix has a basis of eigenvectors.
PROOF. Omitted.
THEOREM. Eigenvectors for distinct eigenvalues of a
symmetric matrix are orthogonal.
PROOF. Suppose A is symmetric and O and P are distinct
eigenvalues with eigenvectors v and u.
WA=AT, Av = Ov, Au = Pu and, since O P, (OP) 0 .W
O(v,u)=(Ov,u)=(Av,u)=(v, ATu)=(v, Au)=(v, Pu)=P(v,u).
WO(v, u) P(v, u) = 0. W(OP)(v, u) = 0. W(v, u) = 0.
Hence v and u are orthogonal. E
THEOREM. Let A be a symmetric n[n matrix; let D be the
diagonal matrix of eigenvalues for A. Then A = PDPT
where P is an orthogonal matrix whose ith column is an
eigenvector for the ith eigenvalue on D’s diagonal.
PROOF. Let A be symmetric. By the second theorem, there
is a basis S = {v1, v2, ..., vn} of eigenvectors for A. Let D
be the diagonal matrix with diagonal entries O1, O2, ..., On
where Oi = the eigenvalue of vi. The eigenvectors of
different eigenvalues are already orthogonal. Apply
Gramm-Schmidt to orthonormalize eigenvectors which
have the same eigenvalue. As before, apply the
“change-of-basis” theorem to get A = PDP-1 where P =
PU
á
S = [v1| v2| v3| ... | vn] = the orthogonal matrix with
columns vi. P orthogonal î P -1 = PT î A = PDPT. E
When P is orthogonal, the inverse P -1 = PT is easy to
compute.
RECALL. For diagonalizable matrices, an eigenvalue of
degree k has k independent eigenvectors.
CFind a diagonal matrix D and an orthogonal matrix P
such that A = PDP-1 where A = .
022
202
220
Characteristic polynomial = |OIA| = (O+2)2(O4).
Eigenvalues: O = 4 with degree 1, O = -2 with degree 2.
List the positives first, then the negatives, then the 0’s.
Eigenvectors for O = 4: [1,1,1]T.
Eigenvectors for O = -2: [-1,1,0]T, [-1,0,1]T.
For O = 4, the normal vector is (1/O3¯)[1,1,1]T.
For O = -2, apply Gramm-Schmidt. We’ll omit T until the
end.
{w1, w2} = {[-1,1,0], [-1,0,1]}.
v1 = (1/O2¯ )[-1,1,0]
u2 = w2 (w2, v1)v1 = [-1,0,1]
½([-1,0,1], [-1,1,0])[-1,1,0]
= [-1,0,1]
½
1
[-1,1,0]= [-½,-½,1] ~ [-1,-1,2].
v2 = (1/O6¯ )[-1,-1,2]T.
Wfor O = -2, we have (1/O2¯)[-1,1,0]T, (1/O6¯ )[-1,-1,2]T.
Answer: D = , P =
40 0
0−20
00−2
1
3
−1
2
−1
6
1
3
1
2
−1
6
1
302
6
Hw 33 Answers
3DJH.
. [8, 2, 1]T
. D No E Yes
. Consider a transition matrix A =
0.20
0.3.3
1.5.7
D Suppose x = [0, 1, 0]T. Find
x(1) = [.2, .3, .5]T
x(2) = [.06, .24, .70]T
x(3) = [.048, .282, .67]T
F [3/53, 15/53, 35/53]T = [.0566, .2830, .6604]T
. For each matrix, find the steady-state probability
vector, if any. Write “none” if there are none.
D F
2
3
1
3
9
17
4
17
4
17