Find the dimension of the subspace of all vectors in R3

[warwick]

whose first and third entries are equal.

a a a
b c d
a a a

I can set a3 = 1a1 + 0a2 if b = d. Can I now delete a3 and find that the basis has only two vectors?

 

[daon]

Here's my 2-cents without using any linear algebra... I hope I interpreted the question correctly.

I'm assuming the first and third entries of EACH VECTOR are equal. Two different vectors may have different 1st coordinates. Therefore a, c and e being distinct real numbers, b, d, f any real numbers we can have 3 distinct vectors like:

<a,b,a>
<c,d,c>
<e,f,e>

So for x,y,z coordinates we have the following relationship:
x = x
y = y
z = x

The last relationship defines a line in the x-z plane. x and y are free to vary. We get a tilted plane... (a line) x (a line) = a plane. (2-dimensional)

I'm sure if you row-reduced an arbitrary 3x3 matrix under these conditions you'll get the same answer...

 

[pka]

You want the subspace spannecd by the set {<1,0,1>,<0,1,0>}