answer:Yes, it is possible. By inserting the steps of the known proofs into the description, you can create a complete formal proof that includes all the necessary steps.

answer:The open mapping theorem guarantees the existence of a delta > 0 such that for any z in G+L satisfying |z| < delta, we can find x in G and y in L with |x| + |y| < 1 such that z = x + y. Now, consider an arbitrary z neq 0 in G+L. Define z' = frac{delta z}{2|z|}, which implies |z'| < delta. By the open mapping theorem, we have z' = x' + y' with x' in G, y' in L and |x'| + |y'| < 1. Scale x' and y' by frac{2|z|}{delta} to obtain x and y such that z = x + y, where x = frac{2|z|}{delta}x' and y = frac{2|z|}{delta}y'. This gives us |x| + |y| leq frac{2|z|}{delta}(|x'| + |y'|) leq frac{2|z|}{delta}. Since |z'| < delta, we can choose c = frac{delta}{2}, resulting in |x| + |y| leq frac{1}{c}|z|. This inequality holds regardless of the size of z.

answer:Given a function, {eq}displaystyle f(x) = x^3 - 3x^2 + 12 {/eq} We need to determine the absolute maximum value of the function on the closed interval {eq}displaystyle x in [-2,4] {/eq}. To find the absolute maximum, we need to check the boundary values as well as the critical points of the function. Let us first determine the critical points, which can be found by solving {eq}displaystyle f'(x) = 0 {/eq}. The derivative of the function is simply, {eq}displaystyle f'(x) = 3x^2 -6x = 3x(x-2) {/eq} Hence the critical points of the function are, {eq}displaystyle f'(x) = 3x(x-2) = 0 Rightarrow x = { 0, 2 } {/eq} Now let us check the value of the function at these points. (1) Boundary values (i) {eq}displaystyle x = -2 {/eq} The value of the function at this point is simply, {eq}displaystyle f(-2) = (-2)^3 - 3(-2)^2 + 12 = -8 {/eq} (ii) {eq}displaystyle x = 4 {/eq} The value of the function at this point is simply, {eq}displaystyle f(4) = (4)^3 - 3(4)^2 + 12 =28 {/eq} (2) Critical points (i) {eq}displaystyle x = 0 {/eq} The value of the function at this point is, {eq}displaystyle f(0) = (0)^3 - 3(0)^2 + 12 = 12 {/eq} (ii) {eq}displaystyle x = 2 {/eq} The value of the function at this point is, {eq}displaystyle f(2) = (2)^3 - 3(2)^2 + 12 = 8 {/eq} Hence the absolute maximum value of the function occurs at {eq}displaystyle x = 4 {/eq} and is equal to {eq}displaystyle f(4) = 28 {/eq}

answer:The null space of the matrix M is the set of all vectors mathbf{v} = (x_1, x_2, x_3, x_4)^T such that Mmathbf{v} = mathbf{0}. To find the basis, we reduce the matrix to row echelon form and then express the free variables in terms of the others. Matrix M: M = left( begin{array}{cccc} 9 & 5 & 1 & 2 0 & 9 & 0 & -10 end{array} right) Reducing M to row echelon form: begin{align*} M &rightarrow left( begin{array}{cccc} 1 & 0 & frac{1}{9} & frac{68}{81} 0 & 1 & 0 & -frac{10}{9} end{array} right) end{align*} The free variables are x_3 and x_4. We can express x_1 and x_2 in terms of x_3 and x_4: begin{align*} x_1 &= -frac{x_3}{9} - frac{68x_4}{81} x_2 &= frac{10x_4}{9} end{align*} A basis for the null space of M is given by: left{ left( begin{array}{c} -68 90 1 0 end{array} right), left( begin{array}{c} -1 0 0 1 end{array} right) right} These vectors span the null space and are linearly independent.