Let v∈V .
Suppose
v=a1w1+a2w2+...+akwk ----->(1)
v=b1w1+b2w2+...+bkwk ----->(2)
Subtracting (2) from (1), we have;
0=(a1−b1)w1+(a2−b2)w2+...+(ak−bk)wk
Since S is a basis for V, then we have that
a1−b1=a2−b2=...=ak−bk=0
Hence, a1=b1,a2=b2,...,ak=bk
Comments