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Chapter 7: Q43E (page 375)

Consider two realn脳nmatrices A and B that are 鈥渟imilar over C鈥�: That is, there is a complex invertiblen脳nmatrix S such that.B=S-1ASShow that A and B are in fact 鈥渟imilar over R鈥�: That is, there is a real R such thatB=R-1AR. Hint: WriteS=S1+iS2, where S1 and S2 are real. Consider the function,f(z)=det(S1+zS2)where z is a complex variable. Showthatf(z)isa nonzero polynomial. Conclude that there is a real number x such that. f(x)鈮�0Show thatR=S1+xS2does the job.

Short Answer

Expert verified

f(z) is a nonzero polynomial concluding that real number x such thatf(x)鈮�0andR=S1+xS2

HenceB=(S1+xS2)鈭�1A(S1+xS2)

Step by step solution

01

Step 1:Finding S=S1+iS2

where S1 and S2 are real matrices. Let

f(z)=det(S1+zS2)鈭赌z鈭�鈩�

We know that f is a non-zero polynomial because f(i)=det.S鈮�0Clearly,

鈭�x鈭�鈩�,f(x)鈮�0,soS1+xS2is invertible.Now we have,B=S鈭�1AS鈬�SB=ASso

S1鈭�1B=AS1S2鈭�1B=AS2

So, therefore we haveB=(S1+xS2)鈭�1A(S1+xS2)

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