线性相关证明题If η is a solution of AX=b ,which and R(A)=r.Letζ1ζ2ζ
线性相关证明题
If η is a solution of AX=b ,which and R(A)=r.
Letζ1ζ2ζ3ζ4ζ5 be a basis of the solution space of AX=0.
Show η ζ1ζ2ζ3ζ4ζ5 that are linearly independent.
If η is a solution of AX=b ,which and R(A)=r.
Letζ1ζ2ζ3ζ4ζ5 be a basis of the solution space of AX=0.
Show η ζ1ζ2ζ3ζ4ζ5 that are linearly independent.
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Suppose η,ζ1,ζ2,ζ3,ζ4,ζ5 are linearly dependent.
Then there are k1,k2,...,k6 of which not all are 0 such that k1η+k2ζ1+k3ζ2+k4ζ3+k5ζ4+k6ζ5=0.
So A(k1η+k2ζ1+k3ζ2+k4ζ3+k5ζ4+k6ζ5)=0.
Because Aη=b,Aζi=0.
So k1b=0,namely k1=0.
So k2ζ1+k3ζ2+k4ζ3+k5ζ4+k6ζ5=0.
Because ζ1,ζ2,ζ3,ζ4,ζ5 are a basis of the solution space of AX=0,
so they are linearly independent.
So k2=k3=...=k6=0
That is impossible.
So η,ζ1,ζ2,ζ3,ζ4,ζ5 are linearly independent
Then there are k1,k2,...,k6 of which not all are 0 such that k1η+k2ζ1+k3ζ2+k4ζ3+k5ζ4+k6ζ5=0.
So A(k1η+k2ζ1+k3ζ2+k4ζ3+k5ζ4+k6ζ5)=0.
Because Aη=b,Aζi=0.
So k1b=0,namely k1=0.
So k2ζ1+k3ζ2+k4ζ3+k5ζ4+k6ζ5=0.
Because ζ1,ζ2,ζ3,ζ4,ζ5 are a basis of the solution space of AX=0,
so they are linearly independent.
So k2=k3=...=k6=0
That is impossible.
So η,ζ1,ζ2,ζ3,ζ4,ζ5 are linearly independent
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