Invertibility

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 13 August 2013

Invertibility

GLn(R) is a group with identity In.

Bn,Un,Tn,Sn are all subgroups of GLn(𝔽).

are called elementary matrices.

Any matrix A=GLn(R) is a product of elementary matrices.

Any matrix AMm×n(R) can be written in the form A=P(Ir000) Q where PGLm(R), QGLn(R) and r is the row rank of A.

Determinants

Laplace expansion det ( a11a12a1n a21a22a2n an1an2ann ) =ipidet ( ai1j1 ai1jp aipj1 aipjp ) ( aip+1jp+1 aip+1jn ainjp+1 ainjn ) where j1,,jn is a fixed permutation of 1,2,,n and the sum is over all possible divisions of 1,2,,n into two sets i1<<ipand ip+1<<in, and pi=+1 or -1 according as i1,..,in and j1,,jn are like or unlike derangements of 1,2,,m.

If A and B are n×n matrices then det(AB)= det(A)det(B).

a) Let B be the matrix obtained by switching two rows of A. Then det(B)=-det(A).
b) Let B be the matrix obtained by adding a multiple of a row of A to another row of A. Then det(B)=det(A).
c) Let B be the matrix obtained by multiplying a row of A by a constant cR. Then det(B)=cdet(A).

If two rows of A are the same then det(A)=0.

i=1n aikAhi =ShkdetA.

If det(A) is a unit in R then A-1=det (a)-1 (Aij).

Cramer’s rule for AX=B.

Put Thms VII-X of Hodge and Padoe as exercises.

λ-matrices

If A is a p×qλ-matrix of rank r and E1(λ),,Er(λ) are its invariant factors then there exist MGLp(K[λ]) and NGLq(K[λ]) such that MAN= ( E1(λ) E2(λ)0 Er(λ) 00 0 ) .

Two p×qλ-matrices A and B are equivalent if and only if they have the same invariant factors and if and only if they have the same elementary divisors.

Note that these proofs work for any Euclidean domain.

Cayley-Hamilton Theorem.

Note that the proof of Theorem II §10 Hodge and Padoe!
Theorem III (Hodge and Padoe) gives minimal polynomial.

When is the adjoint well defined?

a) If V is a vector space over and V has a symmetric positive definite bilinear form, then T is symmetric if and only if v,Tw=Tv,w.
b) If V is a vector space over and V has a Hermitian form, then T is Hermitian if and only if v,Tw=Tv,w.
c) If V is a vector space over F and V has a skew-symmetric form then T is skew-symmetric if and only if v,Tw=Tv,w.

a) If V is a vector space over and V has a symmetric positive definite bilinear form, then T is orthogonal if and only if v,w=Tv,Tw.
b) If V is a vector space over and V has a Hermitian form, then T is Unitary if and only if v,w=Tv,Tw.
c) If V is a vector space over F and and V has a skewed-symmetric form, then T is symplectic if and only if v,w=Tv,Tw.

a) If AMn() is symmetric then there exists POn() such that P-1AP is diagonal.
b) If AMn() is Hermitian then there exists PUn() such that P-1AP is diagonal and P-1APMn().
c) AMn() is normal if and only if there exists PUn() such that P-1AP is diagonal.
d) If AMn(𝔽) is skew-symmetric then there exists PSp2n(F) such that P-1AP is diagonal.

a) V is a vector space over with symmetric positive definite bilinear form and T:VV linear transformation. T is orthogonal if and only if T is a change of basis matrix between orthonormal bases.
b) V is a vector space over with Hermitian form and T:VV linear transformation. T is Hermitian if and only if T is a change of basis matrix between orthonormal bases.
c) V is a vector space over F with skew-symmetric form and T:VV linear transformation. T is symplectic if and only if T is a change of basis between symplectic bases.

aa) Let V be a vector space over with positive definite symmetric bilinear form. Then there exists an orthonormal basis.
ab) Let AMn() be a symmetric positive definite matrix. Then there exists QGLn() such that QAQt=In.
ba) Let V be a vector space over with a Hermitian form. Then there exists an orthonormal basis.
bb) Let AMn() be a Hermitian matrix. Then there exists QGLn() such that QAQ*=In.
ca) Let V be a vector space over with a symmetric bilinear form. Then there exists an orthogonal basis such that vi,vi=±1 or 0.
cb) Let AMn() be symmetric. Then there exists QGLn() such that QAQt= ( Ip0 -Iq 00 )
da) Let V be a vector space over F with a skew-symmetric form. Then there exists a symplectic basis.
db) Let A be a skew-symmetric matrix AM2n(𝔽). Then there exists QGL2n(𝔽) such that QAQt= (0In-In0).

Example.

1) Schwartz Inequality.
2) Triangle Inequality.
3) Conics and Quadrics.
4) Positive definite Exs.

Let T:VV be a linear transformation.

Let AMm×n(F). Then there exists QGLm(F) and PGLn(F) such that QAP=(Ir000).

Let B and B be bases and let T:VV be a linear transformation. Then let A and A be the matrices of T with respect to B and B respectively. Let P be the change of basis matrix Pb=b. Then A=PAP-1.

Ex. Matrix in Jordan form, with chTmT.

mT(t) exists and is unique.

Let mT(t)= p1(t)e1 p2(t)e2 pk(t)ek be the prime factorization of the minimal polynomial of T and let fi(t)=mT(t)pi(t)e1. Let Ei=fi(T), then Ei2 = Ei. EiEj = 0ifij. 1 = E1+E2++En.

λ is an eigenvalue of T if and only if λ is a root of the characteristic polynomial of T, chT(t).

a) Let AMn(F) such that chA(t) factors into linear factors in F. Then there exists PGLn(F) such that PAP-1 is triangular.
b) Let AMn(F) and suppose chF(t) has n distinct roots in F. Then there exists PGLn(F) such that PAP-1 is diagonal.

Ex. Does the converse hold?

Proofs for §X.

Let mT(t)= p1(t)e1 p2(t)e2 pk(t)ek be the prime factorization of the minimal polynomial of T and let fi(t)=mT(t)pi(t)e1. Let Ei=fi(T), then Ei2 = Ei. EiEj = 0ifij. 1 = E1+E2++En.

Proof.

Let qi(x)m(x)pi(x)ei.
Then the qi(x) have gcd=1.
Since F[x] is a Euclidean domain there exists ai(x) such that

1=q1(x)a1(x) ++qs(x)as(x).

Let fi(x)=qi(x)ai(x).
Then 1=f1(x)++fn(T).
So a) 1=f1(x)++fn(T).

To show:

b) fi(T)fj(T) =0 if ij.

Proof.
fi(x)fj(x) = qi(x) ai(x) qj(x) aj(x) = m(x)pi(x)ei m(x)pj(x)ej ai(x)aj(x).

So m(x)|fi(x)fj(x).
So fi(T)fj(T)=0.

To show:

c) fi(T)2=fi(T).

Proof.
fi(T)·1 = fi(T) (fj(T)) = fi(T)2.

To show:

d) fi(T)V0.

Proof.
V=jfi (T)V.

If fi(T)V=0 then

V=ji fj(T)V.

So

qi(T)V = jiqi (T)qj(T) aj(T)V = 0

since m(X)|qi(x)qj(x).
But m(x) is the minimal polynomial.
So qi(T)V=0.
So fi(T)0.

To show:

e) fi(T)V=nullsp(pi(T)ei).

Proof.

ea) fi(T)Vnullsp(pi(T)ei).
Let vfi(T)V.
Then pi(T)eiV = pi(T)ei fi(T)v = m(T) ai(T)v = 0. So vnullsp(pi(T)ei).
eb) Let vnullsp(pi(T)ei).
Then v=f1(T)v1++fs(T)vs.
So pi(T)eiv =0= pi(T)ei f1(T)r1 ++ pi(T)ei fs(T)vs.

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