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84 CHAPTER 5. SOME IMPORTANT LINEAR ALGEBRA

where P is n−1×n−1. Multiply it by

B =

(p qr Q

)where B is also an n×n matrix and Q is n−1×n−1.

You use block multiplication(a bc P

)(p qr Q

)=

(ap+br aq+bQpc+Pr cq+PQ

)Note that this all makes sense. For example, b = 1× n− 1 and r = n− 1× 1 so br is a1×1. Similar considerations apply to the other blocks.

Here is an interesting and significant application of block multiplication. In this the-orem, pM (t) denotes the characteristic polynomial, det(tI−M) . Thus the zeros of thispolynomial are the eigenvalues of the matrix, M.

Theorem 5.6.4 Let A be an m×n matrix and let B be an n×m matrix for m≤ n. Then

pBA (t) = tn−m pAB (t) ,

so the eigenvalues of BA and AB are the same including multiplicities except that BA hasn−m extra zero eigenvalues.

Proof: Use block multiplication to write(AB 0B 0

)(I A0 I

)=

(AB ABAB BA

)(

I A0 I

)(0 0B BA

)=

(AB ABAB BA

).

Therefore, (I A0 I

)−1( AB 0B 0

)(I A0 I

)=

(0 0B BA

)

Since the two matrices above are similar it follows that(

0 0B BA

)and

(AB 0B 0

)have

the same characteristic polynomials. Therefore, noting that BA is an n× n matrix and ABis an m×m matrix,

tm det(tI−BA) = tn det(tI−AB)

and so det(tI−BA) = pBA (t) = tn−m det(tI−AB) = tn−m pAB (t) . This proves the theorem.

84 CHAPTER 5. SOME IMPORTANT LINEAR ALGEBRAwhere Pisn—1xn—1. Multiply it by_{ P 4o-(2 8)where B is also ann Xn matrix and Q isn—-1xn—1.You use block multiplicationa b Pp q \_f apt+br aq+bQ(< P(e 4 )=( eter eo)Note that this all makes sense. For example, b = 1 xn—1 andr=n—1x1sobrisa1 x 1. Similar considerations apply to the other blocks.Here is an interesting and significant application of block multiplication. In this the-orem, py (t) denotes the characteristic polynomial, det (tl —_M). Thus the zeros of thispolynomial are the eigenvalues of the matrix, M.Theorem 5.6.4 Let A be an m xn matrix and let B be ann x m matrix for m <n. Then__ 4n-—mPpa(t)=t" ""pap(t),so the eigenvalues of BA and AB are the same including multiplicities except that BA hasn—m extra zero eigenvalues.Proof: Use block multiplication to write(FS) FB)(42)(8 )-(2 B)Therefore,I A\ '( AB 0 1 A\ (0 00 1 B O 0 I) \ B BA. . Loa, 0 O AB OSince the two matrices above are similar it follows that B BA and BO havethe same characteristic polynomials. Therefore, noting that BA is an n x n matrix and ABis an m X m matrix,t” det (t1 — BA) = t” det (t1 — AB)and so det (t! — BA) = pga (t) =" det (t! — AB) = t" pap (t). This proves the theorem.