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: x<sub>5</sub> = s x<sub>4</sub> = t x<sub>3</sub> = u Then x<sub>2</sub> = s - t x<sub>1</sub> = -s +5u +2x<sub>2</sub> = -s + 5u +2(s-t) = s - t + 5u So, we can write: <b>x<sup>T</sup></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)<sup>T</sup>, (-1,-1,0,1,0)<sup>T</sup>, (0,5,1,0,0)<sup>T</sup>, 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 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: x<sub>1</sub> = s - t + 5u x<sub>2</sub> = s - t x<sub>3</sub> = u x<sub>4</sub> = t x<sub>5</sub> = s In vector form that's:: <b>x<sup>T</sup></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.