Image of linear system I found the determinant and null space of the matrix in previous exercises, but I am having trouble understanding how to find the image of following matrix.
"Given the matrix
$$A_{a} = 
 \begin{pmatrix}
  a & 1 & 1 \\
  a & 0 & a \\
  0 & 1 & a
 \end{pmatrix}$$
for $a\in \mathbb{R}$, determine the image
$$B(A_{a})=\{y\in \mathbb{R}^3 \mid A_ax=y\;,\;x\in \mathbb{R}^3 \}"$$
To me, it seems the equation $A_ax=y$ have so many unknowns; a, x and y?
Solutions and hints are very welcome.
 A: The image is spanned by $A(e_1)=ae_1+ae_2$, $A(e_2)=e_1+e_3$ and $A(e_3)=e_1+ae_2+ae_3$ according to the matrix $A$. Here $e_1,e_2,e_3$ is a basis of $\mathbb{R}^3$, and $a\in \mathbb{R}$. For $a\neq 0, \frac{1}{2}$ the image is $\mathbb{R}^3$, because $\det(A)=a(1-2a)$, so that $A$ is a surjective linear map.
A: If the determinant is differen from zero, it means that the the matrix has three eigenvalues (different from zero) with independent eigenvectors. You have to concentrate then on the values $a$ that make the determinant equal to zero. In this case it is useful to determine the eigenvectors and to see if their eigenvalue is different from zero or not. If the eigenvector is asociated with an eigenvalue different from zero, it is included in the image of the matrix. In the oposite case, the eigenvector is not included in the image of the matrix. I hope this answer to be useful for you
A: A general method is with Gaussian elimination. We can do
\begin{align}
\begin{pmatrix}
  a & 1 & 1 \\
  a & 0 & a \\
  0 & 1 & a
\end{pmatrix}
&\to
\begin{pmatrix}
  a & 1 & 1 \\
  0 & -1 & a-1 \\
  0 & 1 & a
\end{pmatrix} && R_2\gets R_2-R_1
\\
&\to
\begin{pmatrix}
  a & 1 & 1 \\
  0 & -1 & a-1 \\
  0 & 0 & 2a-1
\end{pmatrix} && R_3\gets R_3+R_2
\\
\end{align}


*

*If $a\ne0$ and $a\ne 1/2$, the rank of the matrix is $3$, so the image is $\mathbb{R}^3$.

*If $a=1/2$, the rank is $2$ and the image is generated by the first two columns of the matrix, so a basis consists of
$$
\begin{pmatrix}1/2 \\ 1/2 \\ 0\end{pmatrix},
\qquad
\begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix}
$$

*If $a=0$, the rank is $2$ and the image is generated by the last two columns of the matrix, so a basis consist of
$$
\begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix},
\qquad
\begin{pmatrix}1 \\ 0 \\0\end{pmatrix}
$$

A: You want to find $\{Ax: x \in \mathbb{R}^3\}$.  First note that $Ax$ is a linear combination of the columns of $A$, where the coefficients of this combination are the components of $x$.  So the question is: what is the subspace obtained from all possible linear combinations of the columns of $A$?
Observe from the lower-left 2 by 2 submatrix that if $a$ is nonzero, then the last two rows are linearly independent.  Looking at the second column and first column, we see that the only way the first row is a linear combination of the second and third rows is if it is equal to their sum, in which case (by the entries in the third column) $2a=1$, or $a=1/2$.  
Thus, if $a \ne 0$ and $a \ne 1/2$, the matrix has rank 3. In this case, $A$ is invertible and so is a bijective map from $\mathbb{R}^3$ to itself.  The image $\{Ax: x \in \mathbb{R}^3\}$ is then all of $\mathbb{R}^3$.  
If $a=0$, then the matrix $A$ is equal to $\left(\begin{array}{ccc} 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 1 & 0\end{array}\right)$.  The column space of $A$ is then equal to the set of all linear combinations of the vectors $(0,0,1)^T$ and $(1,0,0)^T$. 
If $a=1/2$, the matrix $A$ is equal to $\left(\begin{array}{ccc} 1/2 & 1 & 1 \\ 1/2 & 0 & 1/2 \\ 0 & 1 & 1/2\end{array}\right)$.   The third column is equal to the sum of the first column and half the second column. So the third column can be ignored, since the column space is the span of the first two columns of $A$. Thus, the column space is the subspace spanned by $(1,1,0)^T$ and $(1,0,1)^T$. 
