[james303]

I've got the the following basis

<1,-4,0,-1,-2,0> ; <0,-1,1,0,-1,0> ; <0,-2,0,-1,-1,1>


and I've got to show that they are similar to:

<1,-2,-2,-1,0,0> ; <0,1,-1,0,1,0> ; <0,1,1,1,0,-1>




I can see that on the first row the 2nd basis can be multiplied by (-1) and that would produce the one on the second row, but I'm not how to work out the other 2

I would really appreciate help with this

[little_wizard123]

Tbh I can't remember much about linear algebra, so I'm assuming it just uses row operations:

1,-4,0,-1,-2,0
0,-1,1,0,-1,0
0,-2,0,-1,-1,1

Take the third row from the second row to give you the final third row. The second row is multiplied by -1. The first row comes from the original first row minus 2x the original second row.

[WarriorInAWig]

You have 6 vectors. Call the first three A, B, C and the last three D, E, F. Then you want to show that A=TD, B=TE, C=TF for T which is some invertible matrix. Off you go!

[nuodai]

(Original post by james303)
I've got the the following basis

<1,-4,0,-1,-2,0> ; <0,-1,1,0,-1,0> ; <0,-2,0,-1,-1,1>


and I've got to show that they are similar to:

<1,-2,-2,-1,0,0> ; <0,1,-1,0,1,0> ; <0,1,1,1,0,-1>




I can see that on the first row the 2nd basis can be multiplied by (-1) and that would produce the one on the second row, but I'm not how to work out the other 2

I would really appreciate help with this

Call your old basis and your new basis . You can more or less express the new basis in terms of the old one by inspection. As you observed, . It's also fairly easy to see that . Since these both only involve , you know that your expression for has to involve , and you can see that the last component of is 0 so for it to match up you must have . Then and so on. It comes out pretty quickly.

The coefficients of the new basis in terms of the old then gives you the columns of a 3x3 matrix; but you need a 6x6 matrix. Since these bases span a 3-dimensional subspace of a 6-dimensional space, you can add three vectors (e.g. standard basis vectors, to make your life easier) to extend to a basis for . These can be the same for both the old and new bases (since the two bases you have span the same subspace), and so this will only affect your change-of-basis matrix by adding some 1s on the diagonal to extend the 3x3 matrix to a 6x6 matrix.