Determining if a vector is in the column space of a matrix Hi there I'm having some trouble with the following problem:
I have a $3\times3$ symmetric matrix
$$
A=\pmatrix{1+t&1&1\\ 1&1+t&1\\ 1&1&1+t}.
$$
I am trying to determine the values of $t$ for which the vector $b = (1,t,t^2)^\top$ (this is a column vector) is in the column space of $A$.
I think I'm fairly aware of how to go about it, forming the augmented matrix $[A|b]$ and basically using row ops to find a solution with which I could solve for the value(s) of $t$.  But I've been trying this and have no luck.  May I be missing something?
Thank you
 A: In case you are familiar with determinants, you can see that the matrix is invertible, unless $t \in \{0,-3 \}$. If $A$ is invertible its column space is all of $\mathbb R^3$, and the two remaining cases $t=0,-3$ are easy to check separately.
A: Gaussian elimination is not difficult in this case:
\begin{align}
\left[\begin{array}{ccc|c}
1+t & 1 & 1 & 1 \\
1 & 1+t & 1 & t \\
1 & 1 & 1+t & t^2
\end{array}\right]
&\to
\left[\begin{array}{ccc|c}
1 & 1 & 1+t & t^2 \\
1 & 1+t & 1 & t \\
1+t & 1 & 1 & 1 
\end{array}\right]
\\
&\to
\left[\begin{array}{ccc|c}
1 & 1 & 1+t & t^2 \\
0 & t & -t & t-t^2 \\
0 & -t & -2t-t^2 & 1-t^2(1+t) 
\end{array}\right]
\end{align}
If $t\ne0$ we can go on:
\begin{align}
\left[\begin{array}{ccc|c}
1 & 1 & 1+t & t^2 \\
0 & t & -t & t-t^2 \\
0 & -t & -2t-t^2 & 1-t^2-t^3 
\end{array}\right]
&\to
\left[\begin{array}{ccc|c}
1 & 1 & 1+t & t^2 \\
0 & 1 & -1 & 1-t \\
0 & 0 & -3t-t^2 & 1+t-2t^2-t^3
\end{array}\right]
\end{align}
If $t=-3$, the last row becomes
$$
\begin{array}{ccc|c}
0 & 0 & 0 & 7
\end{array}
$$
so the system has no solution.
If $t\ne-3$, the system has a solution.
If $t=0$, we get, from the place we stopped at,
$$
\left[\begin{array}{ccc|c}
1 & 1 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1
\end{array}\right]
$$
and the system has no solution.
A: Swapping two columns of A will not change its column space.  Consider the matrix
$$
A' = \left( \begin{array}{ccc}
1 & 1 & 1+t \\
1 & 1+t & 1 \\
1+t & 1 & 1 \end{array} \right),
$$
which is obtained by interchanging the first and last column of A.  Forming the augmented matrix $[A', b]$ and subtracting the first row from the second row and the first row times $1+t$ from the third row yields
$$
\left( \begin{array}{ccc|c}
1 & 1 & 1+t & 1 \\
0 & t & -t & t-1 \\
0 & -t & -t^2-2t & t^2-t-1
\end{array}\right).
$$
Notice that if $t=0$, the second row will make the system inconsistent, so $t \ne 0$.  Adding the second row to the third and then dividing the second row by $t$ gives 
$$
\left( \begin{array}{ccc|c}
1 & 1 & 1+t & 1 \\
0 & 1 & -1 & 1-\frac{1}{t} \\
0 & 0 & -t^2-3t & t^2-2
\end{array} \right).
$$
Now, if $t=-3,0$, the third row will make the system inconsistent, so $t \ne -3,0$.  Thus, we can divide the third row by $-t^2 -3t$ to obtain the echelon form matrix
$$
R = \left(\begin{array}{ccc|c}
1 & 1 & 1+t & 1 \\
0 & 1 & -1 & 1-\frac{1}{t} \\
0 & 0 & 1 & \frac{t^2-2}{-t^2-3t}
\end{array} \right).
$$
Thus, $[1, t, t^2]^T$ is in the column space of A whenever $t \ne 0$ and $t \ne -3$.
A: Here is the matrix
$$ A = \left[ \begin{matrix} 1+t & 1 & 1 \\ 1 & 1+t & 1 \\ 1 & 1 & 1+t \end{matrix} \right] $$
and denote 
$$ b = \left[ \begin{matrix} 1 \\ t \\ t^2 \end{matrix} \right] $$
Then $b$ is in the column space of $A$ if and only if there is 
$$ v = \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] $$
such that $Av = b$.
You can bring the matrix to row echelon form via Gauss-Jordan elimination. It is a tidious process in this case. Using Wolfram Alpha one obtains
$$ x = \frac{2t}{t(t+3)}, \quad y = \frac{2t-1}{t(t+3)}, \quad z = \frac{t^3 + 2 t^2 - t -1}{t(t+3)} \ . $$
So for any $t \ne 0$ and $t \ne -3$ you have a solution, that is: $(1,t,t^2)^T$ is in the column space of $A$. It remains to check the cases $t=0$ and $t=-3$ to see if you have either $$ 0 \cdot x + 0 \cdot y + 0 \cdot z = 0 $$ (in that case, they may be infinite number of solutions) or $$ 0 \cdot x + 0 \cdot y + 0 \cdot z = 1 $$  in which there is no solution.
