Linear Algebra pre-requisites
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We say that b1,…,bn are linearly independent if c1b1+c2b2+…+cnbn=d1b1+d2b2+…+dnbn implies c1=d1,c2=d2…,cn=dn or, equivalently,b1,…,bn are linearly independent if c1b1+c2b2+…+cnbn=0 implies c1=c2=…=cn=0.
Example 1: Lets look at the vectors b1=(1,0,0),b2=(0,1,0),b3=(0,0,1) . Observe that if ab1+bb2+cb3=(1,2,3), then a=1,b=2, and c=3 and there is no other values for a,b, and c to give the vector (1,2,3). Next let's for the same values of b1,b2, and b3, the only values of a,b, and c such that ab1+bb2+cb3=(0,0,0) is a=0,b=0, and c=0. Therefore vectors b1=(1,0,0),b2=(0,1,0),b3=(0,0,1) are linearly independent.
Example 2: b1=1,b2=t,b3=t2.
1+2t+3t2≠1+2t+4t2.
If a+bt+ct2=1+2t+3t2 then a=1,b=2,c=3.
Application: Partial Fractions
4(x2+1)(x−3)=Ax+Bx2+1+Cx−3
If you don't like denominators, get rid of them:
4=(Ax+B)(x−3)+C(x2+1)4=Ax2+Bx−3Ax−3B+Cx2+C4=(A+C)x2+(B−3A)x−3B+C
I.e., 0x2+0x+4=(A+C)x2+(B−3A)x−3B+C
Thus 0=A+C,0=B−3A,4=−3B+C.
C=−A,B=3A,4=−3(3A)+−A⇒4=−10A.
Hence A=−25,B=3(−25)=−65,C=25.
Thus,
4(x2+1)(x−3)=−25x−65x2+1+25x−3=−2x−65(x2+1)+25(x−3)
Alternatively, can plug in x=3 to quickly find C and then solve for A and B. We can also use matrices to solve linear equations.