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Determinant of linear transformations

  • Thread starter succubus
  • Start date
  • #1
33
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I thought this problem was pretty straightforward, but I can't seem to match the answers in the back of the book.


The problem is: Find the determinant of the following linear transformation.

T(v) = <1, 2, 3> x v (where the x means cross product)

from the plane V given by x + 2y + 3z = 0

So I find the basis vectors

<-2, 1, 0> and <-3, 0, 1>


And I perform the transformation by

T( <-2, 1, 0>) = <-3, 6, 5>
T(<-3, 0, 1>) = < 2, 8, 6>


And so I get the 2 column vectors to be

| -3 2 |
| 6 8 |
| 5 6 |

I know this is so off, but what am I doing wrong exactly?

I can get the others but this one is giving me fits and I know it has to be so simple......
 

Answers and Replies

  • #2
726
1
Let u = <-2, 1, 0> and w = <-3, 0, 1>

I haven't checked your calculations, so I'm assuming they're correct.

The matrix representation you have for T is incorrect. First, you need to order your basis. Your basis will either be {u,w} or {w,u}. Let's say you choose the first ordered basis.

What you did was say that T(u) = < -3, 6, 5> and so <-3, 6, 5> will be a column vector in the matrix. This is incorrect. You have to express <-3, 6, 5> as a linear combination of the basis vectors. So, T(u) = <-3, 6, 5> = a*u + b*w, where a and b are some scalars. Then <a,b> will be the first column vector for your matrix representation. Do the same thing with T(w) for some scalars c and d and so <c,d> will be your second column vector for the matrix, then just calculate the determinant.
 
  • #3
33
0
I see what you're saying. I had a brain fart. Thanks again.
 

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