Since you asked for an intuitive way to understand covariance and contravariance, I think this will do. First of all, remember that the reason of having covariant or contravariant tensors is because ...

Assume that L is regular and take w = 0^n1^n \in L. By the Pumping Lemma, we can write w = xyz where y \not = \epsilon, |xy| \leq n, and xy^iz \in L for all i \geq 0. Now, because |xy| \leq n ...

We are asked to solve this using Variation of Parameters (VoP), given: \tag 1 y''-3y'+2y=4x+4 Step 1 Find the homogenous solution to (1), so we have: \tag 2 y''-3 y'+ 2 y = 0 This yields: y_h = c_1e^x + c_2 e^{2x} ...

Find A by taking the inverse of A^{-1}. A=(A^{-1})^{-1}=\left[ \begin{array}{cc} 3&-2\\ -1&1 \end{array} \right]

The beginning looks fine. However, note that you need one Lagrange multiplier per constraint. Thus you need a vector \lambda. The function to minimize is then \operatorname{tr}(A^T-B^T)(A-B) - \lambda^T (B x -v). ...

d) In one step with x_0 = 0 you have x_1 = B u_0, so you can only reach the image of B i.e. \operatorname{im} B = \operatorname{span} \left(\begin{bmatrix} 1 & 0 & 0 \end{bmatrix}^\top,\begin{bmatrix} 0 & 1 & 2 \end{bmatrix}^\top \right) \\ = \{\begin{bmatrix} 1 & 0 & 0 \end{bmatrix}^\top a + \begin{bmatrix} 0 & 1 & 2 \end{bmatrix}^\top b\ |\ a,b \in \mathbb{R} \} ...