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 =

P−1AP 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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