Let $T$ be the linear operator on $\mathbb{R}^3$ defined by
$$T(x_1, x_2, x_3)= (3x_1, x_1-x_2, 2x_1+x_2+x_3)$$
Is $T$ invertible? If so, find a rule for $T^{-1}$ like the one which defines $T$.
Prove that $(T^2-I)(T-3I) = 0.$
I've found that $T$ invertible, and that the rule is $$T^{-1}(x_1, x_2, x_3) = (\frac{1}{3}x_1, -x_2, -\frac{2}{3}x_1+x_2).$$
But to show that $(T^2-I)(T-3I) = 0$, am I supposed to use linearity/earlier part, or just the matrix?
The set $\mathcal{P}$of all real polynomials $f(x)$ is a linear subspace of the vector space $\mathcal{C}^{\infty}$ of real differentiable functions on the real line. Find the null space of the following mappings defined on $\mathcal{P}$.
$F_2(f(x)) = xf(x)$
$F_3(f(x)) = x^2f''(x)-2xf'(x)$
Wh...
The given vector $x$
$$
x = 4 e_1+ 4 e_3
$$
can be represented according the first or the second basis too.
\begin{align}
x
&= x_1^{(0)} b_1^{(0)} + x_2^{(0)} b_2^{(0)} + x_3^{(0)} b_3^{(0)}
= (x_1^{(0)},x_2^{(0)},x_3^{(0)})^T = [x]_0 \\
&= x_1^{(1)} b_1^{(1)} + x_2^{(1)} b_2^{(1)} + x_3^{(1)} ...