HW 4: If {x, y} is linearly independent, and if z is in Span{x, y}, then {x, y, z} is linearly dependent.
TRUE. Since z is in Span{x, y}, then z is a linear combination of x, y, thus {x, y, z} is linearly dependent.
We can prove that {x, y, z} is linearly dependent if z is in Span{x, y} using the following steps:
1. Since z is in Span{x, y}, we can write z as a linear combination of x and y. Let’s say that z = a\*x + b\*y for some scalars a and b.
2. We want to show that there exist scalars c and d, not both zero, such that c\*x + d\*y + e\*z = 0 for some nonzero scalar e. This would show that {x, y, z} is linearly dependent.
3. We can substitute z = a\*x + b\*y into the equation from step 2 to get c\*x + d\*y + e\*(a\*x + b\*y) = 0.
4. We can simplify this equation to get (c + e\*a)\*x + (d + e\*b)\*y = 0.
5. Since {x, y} is linearly independent, we know that c + e\*a = 0 and d + e\*b = 0. This is because if c + e\*a were nonzero, then we could solve for e and get a linear combination of x and y that equals zero, contradicting the linear independence of {x, y}.
6. Solving for e in each equation gives us e = -c/a and e = -d/b.
7. Since e must be the same in both equations, we can set -c/a = -d/b and solve for d to get d = -(b/a)\*c.
8. Therefore, we have c\*x + d\*y + e\*z = c\*x – (b/a)\*c\*y + (-c/a)(a\*x + b\*y).
9. Simplifying this equation gives us (ac – bc + ac)/a\*x + (-bc – ab + bc)/b\*y = 0.
10. We can factor out c from this equation to get (2ac – ab – bc)/(ab)\*(c\*x + d\*y) = 0.
11. Since c and d are not both zero (if they were, then {x, y, z} would be trivially linearly dependent), we must have 2ac – ab – bc = 0.
12. We can simplify this equation to get a = (b/c) and substitute this into the equation for z from step 1 to get z = (b/c)\*x + b\*y. This shows that z is a linear combination of x and y, so {x, y, z} is linearly dependent.
Therefore, we have shown that if {x, y} is linearly independent and z is in Span{x, y}, then {x, y, z} is linearly dependent.
More Answers:
Matrix Multiplication: Debunking The False Claim That Ab = [A1B1 A2B2] For 2 X 2 Matrices A And B.How To Prove That If Ab = Ac And A Is An Invertible Square Matrix, B = C
Proving The Statement: If A System Has Infinitely Many Solutions, The Matrix Isn’T Invertible.