# Basis question

• Apr 23rd 2009, 03:27 AM
scorpion007
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)
• Apr 23rd 2009, 06:09 PM
scorpion007
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.
• Apr 23rd 2009, 11:23 PM
scorpion007
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).