Provide a counterexample to show that the following statement is false: "If U₁, U₂, W are subspaces of a vector space V such that U₁ W = U₂ ☺ W, then U₁ = U₂."

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Compute the curl of each of the following vectors: V = (yz, xz, xy) F = (x, y, z) = = ( Note: Y 2 (x² + y² +2²)³/²¹ (x² + y² + x2)³/²¹ (x² + y² + x2)³/2 • If the answer is a scalar, you can just type it in the box (using the Calcpad if you like, or using / for fractions, ^ for exponents, etc.) For multiplication, you can either leave a space, or use *. So x*x=xx = x². Note that this is not the same as xx without a space; that will get read as an entirely different variable! • If you need to enter a vector, enter an ordered list of components, so for A you can enter either A = (A₂, Ay, A₂) = {Ax, Ay, Az} . Note that the system isn't great with multiplying through by overall factors, so it's better not to write e.g. • If you need to enter O, enter it in three components, as (0, 0, 0) or {0,0,0}. V x V = ▼ xr= vx= (2A, 2A, 2A₂).

Suppose that two vectors z and y in R" are orthogonal, meaning their inner product satisfies z ·g = 0. Show that if both vectors are assumed to be non-zero, then {T, g} are linearly independent. Show that this linear independence fails if at least one of the vectors is the zero vector.