For what value of $\lambda, (0,\lambda,\lambda^2)^t$ belongs to column space of $A$ $$A=\begin{pmatrix}1+\lambda&1&1\\1&1+\lambda&1\\1&1&1+\lambda\end{pmatrix}$$
I need to know for what value of $\lambda, (0,\lambda,\lambda^2)^t$ belongs to column space of $A$, could anyone help me to solve this one?
 A: Solving the linear system corresponding to $\left( 0,\lambda,\lambda^2 \right)^t$ being in the column space, i.e. this vector can be written as a linear combination of the columns, will naturally lead to a few cases. You can distinguish these cases beforehand if you start from the determinant.
Since $\det A = \lambda^2(\lambda+3)$, you have:


*

*for $\lambda \in \mathbb{R} \setminus \left\{ -3,0 \right\}$, $\det A \ne 0$ so the column space is $\mathbb{R^3}$ and thus...

*for $\lambda = 0$, $\left( 0,\lambda,\lambda^2 \right)^t = \left( 0,0,0 \right)^t$, so...

*for $\lambda = -3$, the column space is ... and $\left( 0,\lambda,\lambda^2 \right)^t = \left( 0,-3,9 \right)^t$, so...


Can you fill in the gaps / dots?
A: You have to compute the values of $\lambda$ for which the following linear system $$\begin{pmatrix}1+\lambda&1&1\\1&1+\lambda&1\\1&1&1+\lambda\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\\lambda\\\lambda^2\end{pmatrix}$$ is consistent. You should take it from here now.
A: A slightly different (but same in principles) approach: note that if $\text{rank}(A) = 3 $, i.e $\text{det}(A) \neq 0$ then every vector of $\mathbb{R}^{3}$ would be a linear combination of column vectors of $A$. Finally, discuss one by one the values of $\lambda$ for which $\text{det}(A) = 0$. 
A: The "column space" of a matrix is the space spanned by the columns of the matrix so any vector in the column space is of the form $\alpha(1+ \lambda, 1, 1)+ \beta(1, 1+ \lambda, 1)+ \gamma(1, 1, 1+ \lambda)= (\alpha+ \beta+ \gamma+ \alpha\lambda, \alpha+ \beta+ \gamma+ \beta\lambda, \alpha+ \beta+ \gamma+ \gamma\lambda)$ so we must have $\alpha+ \beta+ \gamma+ \alpha\lambda= 0$, $\alpha+ \beta+ \gamma+ \beta\lambda= \lambda$, and $\alpha+ \beta+ \gamma+ \gamma\lambda= \lambda^2$.
Subtracting the first equation from the second, $(\beta- \alpha)\lambda= \lambda$ or $\beta- \alpha= 1$.  Subtracting the first equation from the third, $(\gamma- \alpha)\lambda= \lambda^2$ or $\gamma- \alpha= \lambda$.  
There exist solutions to those equations for all $\lambda$.
