Linear transformation and find basis

fk378
Messages
366
Reaction score
0

Homework Statement


Define T: R2-->R2 by T(x)=Ax
Find a basis B for R2 with the property that [T]_B is diagonal.

A=
0 1
-3 4


The Attempt at a Solution


The eigenvalues of a diagonal matrix are its diagonal entries, so here the eigenvalues are 1, and -3. For eigenvalue=1 I get the basis [1,1] and for eigenvalue -3 I do not get a linearly dependent system. However, accidentally solved the system for eigenvalue=3 and got the basis [1,3] which is supposed to be correct. Can anyone see what I'm doing wrong?
 
Physics news on Phys.org
The eigenvalues are 3 and 1. Not -3 and 1. If for -3 you don't get a linearly dependent system doesn't that make you wonder what's going wrong? Start by fixing that. I really don't think you needed to post this question.
 
Last edited:
Oh so I would need to solve for the eigenvalues here, not just look at the diagonals? I think I confused this matrix with a triangular matrix.
 
Of course you need to solve for the eigenvalues. I don't know what sort of rule you made up to read them off of the opposite diagonal, but it's clearly not right. Like I said, the fact you didn't get a linearly dependent system should immediately say, that's not an eigenvalue!
 
fk378 said:

Homework Statement


Define T: R2-->R2 by T(x)=Ax
Find a basis B for R2 with the property that [T]_B is diagonal.

A=
0 1
-3 4


The Attempt at a Solution


The eigenvalues of a diagonal matrix are its diagonal entries, so here the eigenvalues are 1, and -3.
Yes, the eigenvalues of a diagonal matrix the entries on the main diagonal. But this is NOT a diagonal matrix- you are asked to change it into one. Furthermore, -3 and 1 are not on the main diagonal!

For eigenvalue=1 I get the basis [1,1] and for eigenvalue -3 I do not get a linearly dependent system. However, accidentally solved the system for eigenvalue=3 and got the basis [1,3] which is supposed to be correct. Can anyone see what I'm doing wrong?
What do MEAN by "the basis [1,3]"? That's a single vector. A basis for a two dimensional vector space consists of two vectors.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top