Q. Let A be a real n×n matrix. (a) Show that A =0 if AA^T =0....

Q. Let A be a real n×n matrix.

(a) Show that A =0 if AA^T =0.

(b) Show that A is symmetric if and only if A^2= AA^T

Solutions

Expert Solution

(a) Let A be an n×n matrix with real entries, where n≥2. Let AAT=[bij], where AT is the transpose of A. If b11+b22+⋯+bnn=0, show that A=0

Let

A=(aij)⟹AT=(aji)⟹AAT=(bij)=

so that

0==

Now by the given condition b11+b22+...+bnn=0

i.e. trace of ATA=0

i.e. ||A||=0 if and only if A=0

(b) let matrix An×n is real matrix,such AAT=A2, The transpose of matrix A is written AT

the method for solving An equivalent condition for a real matrix to be skew-symmetric applies. Briefly speaking, for real square matrices, 〈X,Y〉=trace(YTX) defines an inner product and symmetric matrices are orthogonal to skew-symmetric matrices. Now, write A=H+K, where H=12(A+AT) is the symmetric part and K=12(A−AT) is the skew-symmetric part. Then AAT=A2 implies that (H+K)K=0 and in turn HK=−K2=KTK. Taking trace on both sides, we get 〈K,H〉=〈K,K〉. Since 〈K,H〉=0, it follows that 〈K,K〉=0 and K=0, i.e. A is symmetric.

Colby Messinger answered 1 year ago

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