Two subspaces S and T of $R^4$ are spanned by {(15,1,0,1), (-1,15,0,1) } and { (2,0,1,0), (12, 46, 0, 4), (14, 46, 1, 4)} respectively. Find a basis {X} for $S\cap T$.

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I am suspicious that the question isn't correct, because I found a solution set for $S\cap T$ that is spanned by 2 vectors, not one. Can the dimension of $S\cap T$ really be 1 as the question implies?

BTW, my sol'n set is: (if that's correct)

$S=\lbrace \left( \begin{array}{c}a_1\\a_2\\a_3\\a_4\\a_5 \end{array} \right) =
\left( \begin{array}{c}1\\3\\0\\1\\0 \end{array} \right) a_4 +
\left( \begin{array}{c}1\\3\\-1\\0\\1 \end{array} \right) a_5 \mid a_4,a_5 \in R \rbrace$

(if someone can suggest a better way to typeset column vectors I'd like to know)

2. Can someone help with this? It is somewhat urgent. If the question is incorrect, I'd like to notify the instructor and let him know.

Otherwise, could you show where I went wrong?

I looked at it again, and can see no way for S intersect T to be spanned by 1 vector.

3. Can someone check if my answer is correct?

The only vector in $S \cap T$ is the zero vector (0,0,0,0), and hence the basis for $S \cap T$ is {(0,0,0,0)}.

My approach:

Let v = (a,b,c,d) be a vector in $S \cap T$.

To be in S, v must equal $(15x - y, x + 15y, 0, x + y)$for arbitrary x, y in R.
To be in T, v must equal $(2x + 12y + 14z, 46y+46z, x+z,4y+4z)$.

From this we can see that:

$15x-y=2x+12y+14z$
$x+15y=46y+46z$
$0=x+z$
$x+y=4y+4z$

Solving this homogeneous linear system, it seems like it only has the trivial solution, (x, y, z) = (0, 0, 0).

Therefore the v must be the vector (0, 0, 0, 0).