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 ...

Usually, the notation e_j, or sometimes \hat{e}_j, is used to denote the standard basis vectors of \mathbb{R}^n; e_j is a vector that has a 1 in the j-th component and 0 in the remaining ...

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} ...

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} \} ...

Bad notation is bad.... Show that x^{(1)}(t) and x^{(2)}(t) are linearly dependent at each point in the interval 0 ≤ t ≤ 1. What you're being asked to prove here is that given t\in [0,1], the ...

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