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Lecture

# Orthogonal Diagonalization of symetric matrix

This preview shows page 1. to view the full 4 pages of the document. Lecture 10 notes byY. Burda
1Towards orthogonal diagonalization of asym-
metric matrix
By now weknowthat if aquadratic form qhas coeﬃcientmatrix Arelative
to the standard basis, then relativeto adiﬀerentbasis Bit has coeﬃcient
matrix B=PTAP ,where P=[I]E,B.
If the new basis is orthonormal, then Pis orthogonal and B=PTAP =
P1AP is similar to A.
Our hopeis that if for some reason wecould ﬁndan orthonormal basis
of eigenvectors of A,then relativeto this basis the coeﬃcientmatrix of q
will bediagonal, i.e. theform itself would beequal to asum of squares with
some coeﬃcients.
Whyshould webeable to ﬁnd an orthonormal basis of eigenvectors for
A?It turns out that the fact that Ais symmetric is just enough for that.
The next theoremis the ﬁrst indication that it mightbepossible:
Theorem. Suppose Ais areal symmetric matrix. Then eigenvectors with
diﬀerent eigenvalues areorthogonal to each other.
Proof. Suppose Av1=λ1v1and Av2=λ2v2,where λ16=λ2and v1,v2are
non-zero. Then v1·Av2=v1·(λ2v2)=λ2v1·v2.On the other hand
v1·Av2=vT
1Av2=vT
1ATv2=(Av1)Tv2=(Av1)·v2=λ1v1·v2.Thus
λ1v1·v2=λ2v1·v2.Since λ1and λ2are distinct, v1·v2=0.
Now wehavethe following algorithm that works if Ais diagonalizable:
Choose bases for eigenspaces:
v1
1,...,v1
n1for Eλ1,
v2
1,...,v2
n2for Eλ2,
...,
vk
1,...,vk
nkfor Eλk
By applying Gram-Schmidtprocess to eachof thesebases, wecan ﬁnd
orthonormal bases foreachof these spaces:
w1
1,...,w1
n1for Eλ1,
w2
1,...,w2
n2for Eλ2,
...,
wk
1,...,wk
nkfor Eλk
1
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