Linear algebra, prove that this operator is linear So Im stuck with this one, i know there are condition to be satisfied but i just don't get it,
so if you're willing, please do
$A:\Bbb{R}^3 \rightarrow \Bbb{R}^3$, 
$Ax=x_2e_1+(x_1-3x_2)e_2+5x_3e_3$, where $x,e_1,e_2,e_3$ are vectors
 A: What you need to check is that $A(x+y)=A(x)+A(y)$, $A(\lambda x) =\lambda A(x)$. Since $A$ is acting on the components of a vector only by linear operations, which are multiplication by a constant and addition, it is indeed linear. For example \begin{equation}\begin{split}A(x+y)&= (x_2+y_2) e_1 +(x_1+y_1-3(x_2+y_2))e_2+5(x_3+y_3)e_3\\&= x_2 e_1+(x_1-3x_2)e_2+5x_3 e_3+y_2 e_1+(y_1-3y_2)e_2+5y_3 e_3 =A(x)+A(y)\end{split}\end{equation}
A: There are two things that you must show:


*

*Given $x,y\in\Bbb R^3,$ we have $A(x+y)=A(x)+A(y).$

*Given $x\in\Bbb R^3$ and $c\in\Bbb R,$ we have $A(cx)=cA(x).$


The big thing to keep in mind is that for any $v\in\Bbb R^3,$ we can write $v$ uniquely in the form $v_1e_1+v_2e_2+v_3e_3$ for some $v_1,v_2,v_3\in\Bbb R.$
A: If $A$ is linear, then it coincides with the multiplication by a matrix, precisely the matrix
$$
\begin{bmatrix}Ae_1 & Ae_2 & Ae_3\end{bmatrix}
$$
(where I specify the matrix by its columns). Since
$$
Ae_1=e_2,\quad
Ae_2=e_1-3e_2,\quad
Ae_3=5e_3
$$
the matrix should be
$$
\begin{bmatrix}
0 & 1 & 0 \\
1 & -3 & 0 \\
0 & 0 & 5
\end{bmatrix}
$$
Now
$$
\begin{bmatrix}
0 & 1 & 0 \\
1 & -3 & 0 \\
0 & 0 & 5
\end{bmatrix}\,
\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}
=
\begin{bmatrix}
x_2 \\
x_1-3x_2 \\
5x_3
\end{bmatrix}
=
x_2e_1+(x_1-3x_2)e_2+5e_3=Ax.
$$
Therefore $A$ is the multiplication by a matrix, so it is linear.
A: If you would like, you can use the language of matrices, to
map ${\Bbb{R}}^3\to {\Bbb{R}}^3$ linearly, you can use matrix's multiplicaction:
$$
x=
\left(\!\begin{array}{c}
x_1\\
x_2\\
x_3
\end{array}\!\right)
\longrightarrow
Ax=\left(\!\begin{array}{ccc}
0&1&0\\
1&-3&0\\
0&0&-5
\end{array}\!\right)
\left(\!\begin{array}{c}
x_1\\
x_2\\
x_3
\end{array}\!\right).
$$
With this device is a routine to check $A(x+y)=Ax+Ay $ and $A(cx)=cAx$ that you need to grasp.
Check that  $Ax+Ay=A(x+y)$ all vectorized amounts to this:
$$
Ax+Ay=
\left(\!\begin{array}{ccc}
0&1&0\\
1&-3&0\\
0&0&-5
\end{array}\!\right)
\left(\!\begin{array}{c}
x_1\\
x_2\\
x_3
\end{array}\!\right)
+
\left(\!\begin{array}{ccc}
0&1&0\\
1&-3&0\\
0&0&-5
\end{array}\!\right)
\left(\!\begin{array}{c}
y_1\\
y_2\\
y_3
\end{array}\!\right)
$$
$$
=
\left(\!\begin{array}{ccc}
0&1&0\\
1&-3&0\\
0&0&-5
\end{array}\!\right)
\left(
\left(\!\begin{array}{c}
x_1\\
x_2\\
x_3
\end{array}\!\right)
+
\left(\!\begin{array}{c}
y_1\\
y_2\\
y_3
\end{array}\!\right)
\right)
$$
$$
=
\left(\!\begin{array}{ccc}
0&1&0\\
1&-3&0\\
0&0&-5
\end{array}\!\right)
\left(\!\begin{array}{c}
x_1+y_1\\
x_2+y_2\\
x_3+y_3
\end{array}\!\right)
$$
$$
=A(x+y)
$$
