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Lecture

MATB24H3 Lecture Notes - Orthogonal Matrix, Symmetric Matrix, Row And Column Vectors

Department
Mathematics
Course Code
MATB24H3
Professor
Sophie Chrysostomou

This preview shows pages 1-3. to view the full 12 pages of the document. University of Toronto at Scarborough
Department of Computer & Mathematical Sciences
MAT B24S Fall 2011
MAT B24
Lectures 17-18.
THEOREM: Properties of Axfor an Orthogonal Matrix ALet Abe an
orthogonal n×nmatrix and xand yare column vectors in Rn.Then:
1. (Ax)·(Ay) = x·yPreservation of dot product.
2. ||Ax|| =||x|| Preservation of length.
3. the angle between nonzero vectors xand yequals the angle between
Axand Ay
proof.
1. (Ax)·(Ay) = (Ax)TAy= (xTAT)Ay=xT(ATA)y=xTIy=xTy=x·y
2. Since (Ax)·(Ax) = x·xthen ||Ax||2=||x||2or ||Ax|| =||x||.
3. If the angle between the nonzero vectors Axand Ayequals θthen
cos θ=(Ax)·(Ax)
||Ax|| · ||Ay|| =x·x
||x||||y||
Showing that θis also the angle between the nonzero vectors xand y.
1

Only pages 1-3 are available for preview. Some parts have been intentionally blurred. THEOREM: Orthogonality of Eigenspaces of a Real Symmetric Matrix:
Let Abe a real symmetric matrix and λ1and λ2be distinct eigenvalues of
A. Then the eigenspaces Eλ1and Eλ2are orthogonal.
proof. : To show that Eλ1and Eλ2are orthogonal we must prove v1and
v2are orthogonal for all v1Eλ1and v2Eλ2.
Let v1Eλ1and v2Eλ2. Then
Av1=λ1v1and Av2=λ2v2.
Also,
λ1(v1·v2) = (λ1v1)·v2= (Av1)·v2= (Av1)Tv2= (vT
1AT)v2= (vT
1A)v2=vT
1(Av2)
=vT
1(λ2v2) = λ2(vT
1v2) = λ2(v1·v2)
Thus
λ1(v1·v2) = λ2(v1·v2)or (λ1λ2)(v1·v2) = 0
Since λ16=λ2then λ1λ26= 0 thus v1·v2= 0
This brings us to another theorem:
THEOREM: Fundamental Theorem of Real Symmetric Matrices: Every
real symmetric matrix Ais diagonalizable. The diagonalization C1AC =D
can be achieved by using a real orthogonal matrix C.
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Only pages 1-3 are available for preview. Some parts have been intentionally blurred. EXAMPLE: Let A=
4 2 2
2 4 2
2 2 4
. Find an orthogonal matrix Cthat diag-
onalizes A.
solution. det(AλI) = (λ2)2(λ8). Therefore Ahas only two eigen-
values: λ1= 2 and λ2= 8.
You may verify that Eλ1=sp
1
1
0
,
1
0
1
and Eλ1=sp
1
1
1
We ﬁnd an orthonormal basis for each of these eigenspaces using the
Gram-Schmidt process.
So
1/2
1/2
0
,
1/6
1/6
2/6
is an orthonormal basis for Eλ1
and
1/3
1/3
1/3
is an orthonormal basis for Eλ2
Thus: C=
1/21/6 1/3
1/21/6 1/3
0 2/6 1/3
and C1AC =
200
020
008
3