If you have two hyperplanes in R^5 given as...
H1 = (1, -2, -5, 0, 1)^T.x = 0
H2 = (0, 1, 0, 1, -1)^T.x = 0
(where ^T is transpose)
Then S, which is a subspace of the 5th dimension, is simply the intersection of H1 and H2, ie. H1 n H2. I got this intuitively since the planes pass through the origin.

1) How would you prove this?

2) For any x the system holds as a solution to Ax = 0. But is it possible to row-reduce the system Ax = 0 to row-echelon form and use this to find a basis for S?

Thankyou for any assistance or guidance on this matter.

Putting the two hyperplanes into a matrix we have:

(1 -2 -5 0 1)
(0 1 0 1 -1)

The solution space for this matrix is found by putting:

x₅ = s
x₄ = t
x₃ = u

Then

x₂ = s - t
x₁ = -s +5u +2x₂ = -s + 5u +2(s-t) = s - t + 5u

So, we can write:

<b>x^T</b> = s(-1,1,0,0,1) + t(-1,-1,0,1,0) + u(0,5,1,0,0)

Thus, the intersection of the two hyperplanes is a three dimensional space spanned by the vectors (-1,1,0,0,1)^T, (-1,-1,0,1,0)^T, (0,5,1,0,0)^T, which together form a basis for the solution space.

Thankyou James for your efforts. But don't go away just yet!

Another thing, could you please explain how you came to this conclusion

xT = s(-1,1,0,0,1) + t(-1,-1,0,1,0) + u(0,5,1,0,0)

Remarks:
You row-reduced the 2x5 matrix and obtained three vectors in R^5 namely (-1,1,0,0,1)^T, (-1,-1,0,1,0)^T and (0,5,1,0,0)^T. Since these three vectors are linearly independant (and are solutions to Ax = 0) you have proved that all you need to form a basis for the intersection of the given hyperplanes is these three vectors. Hence they form a basis for the intersection (ie. solution space). -->please tell me if this is a correct statement<--

Also, does this tell me that the solution space has dimension 3?

Further, how can you conclude that the solution space is a subspace of R^5. Is it that the bases are closed under vector addition and scalar multiplication, or is it more detailed than this?

Finally, say you had the vector (5, 3, -2, 0, 1)^T how would you show that it is a solution?

Wait, I see.

x1 = s - t + 5u
x2 = s - t
x3 = u
x4 = t
x5 = s

= (x1, x2, x3, x4, x5)^T = (s-t+5u, s-t, u, t, s)^T
= (s, s, 0, 0, s)^T + (-t, -t, 0, t, 0)^T + (5u, 0, u, 0, 0)^T

Since x4 = t and x5 = s we can discard them since they can be constructed from x1 and x2.

= s(1, 1, 0, 0, 1)^T + t(-1, -1, 0, 1, 0) + u(5, 0, 1, 0, 0)^T

However slightly different to your answer.
hmmm?

Oops. I made a couple of mistakes.

The solution is, as you say:

x₁ = s - t + 5u
x₂ = s - t
x₃ = u
x₄ = t
x₅ = s

In vector form that's::

<b>x^T</b> = s(1,1,0,0,1) + t(-1,-1,0,1,0) + u(5,0,1,0,0)

which agrees with your answer.

<i>You row-reduced the 2x5 matrix and obtained three vectors in R^5 ...Since these three vectors are linearly independant (and are solutions to Ax = 0) you have proved that all you need to form a basis for the intersection of the given hyperplanes is these three vectors. Hence they form a basis for the intersection (ie. solution space). -->please tell me if this is a correct statement<--</i>

Yes, that's right.

<i>Also, does this tell me that the solution space has dimension 3?</i>

Yes. The number of basis vectors for a space is always equal to its dimension.

<i>Further, how can you conclude that the solution space is a subspace of R^5. Is it that the bases are closed under vector addition and scalar multiplication, or is it more detailed than this?</i>

It's no more complicated than that.

<i>Finally, say you had the vector (5, 3, -2, 0, 1)^T how would you show that it is a solution?</i>

Simple. Plug it into the original equations.

(1 -2 -5 0 1).(5 3 -2 0 1) = 0 ????

Doesn't work, so (5 3 -2 0 1) isn't a solution.