Question

Read the following statements and find out which of these statements is/are correct:

I. If \(\alpha, \beta, \gamma\) are three linearly independent vectors in \(V(F)\) then \(\alpha+\beta\), \(\beta+\gamma\) and \(\gamma+\alpha\) are also linearly independent.

II. The vectors \(2x^3+x^2+x+1,\ x^3+3x^2+x-2,\ x^3+2x^2-x+3\) of \(V_4(R)\) are linearly independent.

• A. Only I is correct
• B. Only II is correct
• C. Both I and II are incorrect
• D. Both I and II are correct

Hint:

For statement I, check if the matrix relating the new combinations to the old vectors is invertible. For statement II, test the rank of the coefficient vectors using any nonzero 3x3 minor.

Answer 1

The Correct Option is D

Step 1: Check statement I. Suppose \(a(\alpha+\beta) + b(\beta+\gamma) + c(\gamma+\alpha) = 0\) for scalars \(a,b,c\). Collecting terms: \[(a+c)\alpha + (a+b)\beta + (b+c)\gamma = 0\]

Step 2: Since \(\alpha,\beta,\gamma\) are linearly independent, each coefficient must vanish: \(a+c=0\), \(a+b=0\), \(b+c=0\).

Step 3: Adding all three equations gives \(2(a+b+c)=0\), so \(a+b+c=0\). From \(a+c=0\) we get \(a=-c\). Substituting in \(a+b=0\) gives \(b=c\). Then \(b+c=0\) becomes \(2c=0\), so \(c=0\), and hence \(a=0,\ b=0\). Only the trivial solution exists, so \(\alpha+\beta,\beta+\gamma,\gamma+\alpha\) are linearly independent. Statement I is correct.

Step 4: Check statement II. Write the three polynomials as coefficient vectors (coefficients of \(x^3,x^2,x,1\)): \(v_1=(2,1,1,1)\), \(v_2=(1,3,1,-2)\), \(v_3=(1,2,-1,3)\).

Step 5: Form the \(3\times 4\) matrix with these rows and reduce it. Using \(v_2\) as pivot row: \(v_1 - 2v_2 = (0,-5,-1,5)\) and \(v_3-v_2=(0,-1,-2,5)\). Eliminating further, \((0,-5,-1,5) - 5(0,-1,-2,5) = (0,0,9,-20)\), a nonzero row. So the reduced matrix has 3 nonzero pivot rows, giving rank 3, which equals the number of vectors. Hence \(v_1, v_2, v_3\) are linearly independent. Statement II is correct.

Since both statements are true, the answer is \(\boxed{\text{Both I and II are correct}}\).