Linear Algebra Standard Basis of $\mathbb{R}^3$ I have no idea what to do in this question, other then the fact that the formula $T$ is suppose to have some kind of $x$, $y$, $z$.
The standard basis is $E_1 = (1,0,0)$, $E_2 = (0,1,0)$, and $E_3 = (0,0,1)$. So if $X = (x,y,z) \in \mathbb{R}^3$, it has the form $$X = (x,y,z) = x(1,0,0) + y(0,1,0) + z(0,0,1) =  x E_1 + y E_2 + z E_3.$$
Now use linearity of $T$ (which says $T(aX + bY) = aT(X) + bT(Y)$ for any numbers $a,b \in \mathbb{R}$ and vectors $X,Y \in \mathbb{R}^3$, and this extends to any finite sum of vectors) to get
$$
T(X) = T(xE_1 + y E_2 + z E_3) = xT(E_1) + yT(E_2) + zT(E_3).
$$
You are given $T(E_1)$, $T(E_2)$, and $T(E_3)$, so substitute in to get the answer.
Use matrix, this application would not be more than the multiplication of the transpose of the image, ie, the matrix of T in the canonical bases. Where E is the canonical base, TE = Im (T). Note that the transpose of the canonical is herself. It is relatively simple, just imagine what their eyes are two dimensions and the third touch, movement, ie move your body is a linear application from R3 to R3, if you cut the arm of R3 to R2. The first thing is to understand what is the linear algebra. They are rotations of a abelian body relative origin. Body parallel to that form the space, which is always a abelian because infinite body is always finite. Poincare you recommend reading her books disclosure abstracion to understand mathematics. We facilitate life. Translations of google