How to describe Image of matrix function f by eigenvectors given
$$
\textbf{A}=\begin{bmatrix}
4& 1& 2\\
 2 & 2 & 4\\
1&1&2
\end{bmatrix}
$$
$f({\bf x})=A{\bf x}, {\bf x}\in R^3$
I know eigenvectors : $v_1=[0,-2,1]^T$, $v_2=[-2,2,1]^T$, $v_3=[2,2,1]^T$
eigenvalues: 0,2,6
the kernel is solution of Ax=0, $x=[0, -2t,t]^T, t \in R$
the image is $[4x_1+x_2+2x_3, 2x_1+2x_2+4x_3, x_1+x_2+2x_3]^T,x_1,x_2,x_3\in R$
but I can't figure how to use $\{v_1,v_2,v_3\}$ to describe image. the kernel = $v_1*t$ or $(v_2+v_3)*t , t\in R$.
 A: Given that
$$
A=\left[ \begin{array}{ccc}
4 & 1 & 4 \\
2 & 2 & 4 \\
1 & 1 & 2 \\
\end{array} \right]
$$
Note that $A$ has rank two.
(Obviously, the second row is twice the third row. In other words,
$R_2 = 2 R_3$ or that $0 R1 + R_2 - 2 R_3 = 0$. Hence, the rows of $A$
are linearly dependent)
A simple calculation gives the eigenpairs of $A$ as:
$$
\lambda_1 = 0, \ \ \mathbf{v}_1 = \left[ \begin{array}{c}
0 \\
-2 \\
1 \\
\end{array} \right]
$$
$$
\lambda_2 = 2, \ \ \mathbf{v}_2 = \left[ \begin{array}{c}
-2 \\
 2 \\
1 \\
\end{array} \right]
$$
$$
\lambda_3 = 6, \ \ \mathbf{v}_1 = \left[ \begin{array}{c}
2 \\
 2 \\
1 \\
\end{array} \right]
$$
$A$ has three distinct eigenvalues $0, 2, 6$ with corresponding
eigenvectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$.
Obviously
$$
\mathcal{B} = \left\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \right \}
$$
forms a basis of $\mathbf{R}^3$.
If $f(\mathbf{x}) = A \mathbf{x}$, then we have
$$
f(\mathbf{x}) = \left[ \begin{array}{c}
 4 x_1 + x_2 + 2 x_3 \\
2 x_1 + 2 x_2 + 4 x_3 \\
 x_1 + x_2 + 2 x_3
\end{array} \right] = c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3
$$
since $\mathcal{B}$ forms a basis of $\mathbf{R}^3$.
A simple calculation gives
$$
c_1 = 0, \ c_2 = - {x_1 \over 2} + {x_2 \over 4} + {x_3 \over 2}, \ \
c_3 = {3 x_1 \over 2} + {3 x_2 \over 4} + {3 x_3 \over 2}
$$
