Thread: Find subspace

Find the basis for the solution space in R^5 of this linear system. Find this subspace dimension:

[1 2 3 4 5
2 -1 2 -1 2
0 5 4 9 8
1 7 7 13 13]

;; 1x+ 2y + 3z + 4s +5d = 0
;; 2x + -1y + 2z + -1s + 2d = 0
;; 0x + 5y + 4z + 9s + 8d = 0
;; 1x + 7y + 7z + 13s + 13d = 0

row reduction =

1 0 7/5 2/5 9/5

0 1 4/5 9/5 8/5

0 0 0 0 0

0 0 0 0 0



so x3 x4 x5 are free

x1 = -7/5 x3 - 2/5 x4 - 9/5 x5
x2 = -4/5 x3 - 9/5 x4 - 8/5 x5

therefore linearly independent and form a basis for their span of 3 dimensions
is this correct ?

Elimination gives me:



could you check your work (and perhaps mine) ?

You also write :
;; 1x+ 2y + 3z + 4s +5d = 0

while it is 1 2 3 4 6 in your "matrix"..

Originally Posted by Mollier

Elimination gives me:



could you check your work (and perhaps mine) ?

You also write :
;; 1x+ 2y + 3z + 4s +5d = 0

while it is 1 2 3 4 6 in your "matrix"..

sorry it should be a 5

Originally Posted by Mollier

Elimination gives me:



could you check your work (and perhaps mine) ?

You also write :
;; 1x+ 2y + 3z + 4s +5d = 0

while it is 1 2 3 4 6 in your "matrix"..

When the 5 - 6 confussion is corrected and the arithmetic fixed accordingly this is essentially correct. There are two free variables and so the solution space is two dimensional.

CB

harveyo: Perhaps you could re-do the elimination now?
You could also try looking at how I posted my matrix, and copy-paste it into your own post. It makes it easier to read.

i feel my elimination is correct! i do not know how to post the array properly

You can click on the matrix I wrote and the code will pop up in a new window.

It looks like this:

[--math]
\begin{array}[pos]{ccccc}
1 & 2 & 3 & 4 & 6\\
0 & -5&-4 &-9 &-10\\
0 & 0 & 0 & 0 &-2\\
0 & 0 & 0 & 0 & 0
\end{array}
[/math--]

Without the "--" in .

Now you just copy-paste that into your own reply and it should work. Give it a try.