If W1,w2,w3 are independent vectors, show that the sums V1=W2+W3, V2=W1+W3 and V3=W1+W2 are independent . (Write C1V1+C2V2+C3V3=0 in terms of the W¡¦s. Find and solve equation for the C¡¦s)
To show that the sums V1=W2+W3, V2=W1+W3, and V3=W1+W2 are independent vectors, we need to show that the only solution to the equation C1V1+C2V2+C3V3=0 is C1=C2=C3=0.
First, let's express V1, V2, and V3 in terms of the given vectors W1, W2, and W3:
V1 = W2 + W3
V2 = W1 + W3
V3 = W1 + W2
Now, substitute these expressions into the equation C1V1+C2V2+C3V3=0:
C1(W2 + W3) + C2(W1 + W3) + C3(W1 + W2) = 0
Expand and simplify:
(C1W2 + C2W1 + C3W1) + (C1W3 + C2W3 + C3W2) + (C3W1 + C3W2) = 0
Now, group the terms by their coefficients:
(C1W2 + C2W1 + C3W1) + (C1W3 + C3W2 + C2W3) + (C3W1 + C3W2) = 0
Rearrange the terms:
(C2W1 + (C1W2 + C3W1)) + (C3W2 + (C1W3 + C2W3)) + (C3W1 + C3W2) = 0
Now, we can see that the coefficients for each vector W1, W2, and W3 must all equal zero for the entire equation to be equal to zero. Therefore, the following equations must hold:
C2 + C1 + C3 = 0 (1)
C3 + C1 + C2 = 0 (2)
C3 + C3 = 0 (3)
From equation (3), we can conclude that C3 = 0.
Now, substitute C3 = 0 into equations (1) and (2):
C2 + C1 = 0 (4)
C1 + C2 = 0 (5)
From equations (4) and (5), we can see that C1 = -C2.
Therefore, the solutions to equations (4) and (5) are C1 = C2 = 0.
Substituting these values back into equations (1) and (2), we find C3 = 0 as well.
Thus, the only solution to C1V1+C2V2+C3V3=0 is C1=C2=C3=0.
Therefore, we have shown that the vectors V1=W2+W3, V2=W1+W3, and V3=W1+W2 are independent vectors.