Linear Transformation defined by a Matrix and Invariant Subspaces I got stuck solving this problem:
Let $T:\mathbb{R}^3\to \mathbb{R}^3$ be the linear transformation defined by the matrix A in the standard basis of $\mathbb{R}^3$, $E=\{e_1,e_2,e_3\}$
$$A=\begin{bmatrix} 3 & 1 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$
and let $W=\ker(T-3I)$ be a subspace of $\mathbb{R}^3$, Show that there is no $T$ invariant subspace $U$ of $\mathbb{R}^3$ that satisfies $\mathbb{R}^3=W \oplus U$
Note: when solving this problem I encountered this problem:
Direct sum of subspaces of the three dimensional space
 A: kernel of $ (T-3*I)$  is any vector of the form $(0,x,0)$ which is spanned by $A=(0,1,0)$ is not it?
Edited:
we we have like this
A=[3 1 0;0 3 0;0 0 2]
A =
 3     1     0
 0     3     0
 0     0     2

and we have done
A-3*eye(3)
ans =
 0     1     0
 0     0     0
 0     0    -1

where
eye(3)
ans =
 1     0     0
 0     1     0
 0     0     1

clearly this matrix
A-3*eye(3)
ans =
 0     1     0
 0     0     0
 0     0    -1

has solution
$0*x_1+1*x_2+0*x_3=0$
$0*x_1+0*x_2+0*x_3=0$
$0*x_1+0*x_2+(-1)*x_3=0$
we  get
$x_3=0$
$x_2=0$
$x_1=k$
yes exactly it is spanned by $e_1=(1,0,0)$
we have
Let v be an eigenvector of T, i.e. T v = λv. Then W = span {v} is T invariant. As a consequence of the fundamental theorem of algebra, every linear operator on a complex finite-dimensional vector space with dimension at least 2 has an eigenvector. Therefore every such linear operator has a non-trivial invariant subspace. The fact that the complex numbers are algebraically closed is required here. Comparing with the previous example, one can see that the invariant subspaces of a linear transformation are dependent upon the underlying scalar field of V.
[V D]=eig(A)
V =
1.0000   -1.0000         0
     0    0.0000         0
     0         0    1.0000

D =
 3     0     0
 0     3     0
 0     0     2

see please what is output of eigenvalue decomposition
