# Define the linear transformation TA(v) = A(v) where (v) is the co-ordinate vector

I'm having some problems with this question. Could someone point me in the right direction? Thanks

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The questions seems clear and well-hinted. What happens when you do what it tells you to do? – Chris Eagle Feb 15 '13 at 20:13
For part b, i chose a vector (where v= (a,b) and Ta(v) = lambda(v) = -2(v)) where a= 1 and b=0) which was (-2,0) this doesn't make much sense though as A*(1,0) doesn't give (-2,0) which means that this vector is not the image of (v) under transformation Ta. – John Smith Feb 16 '13 at 4:14
You should probably start with part (a) – Chris Eagle Feb 16 '13 at 7:07

Hints:

1) $\,\det(\lambda I-A)=(\lambda +2)^2=0\Longleftrightarrow \lambda =-2\,$

2) Solve the homogeneous system

$$(-2I-A)\binom{x}{y}=\binom{0}{0}$$

Its solution gives you the eigenvectors of $\,A\,$

3) Solve the rest of the exercise...

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I understand that the eigenvectors give me my one dimensional invariant subspace. It only gives me one eigenvector which means only one invariant subspace. With this, I don't know how I'm supposed to solve the equation without having v1=v2. This would then lead to the two vectors being dependent which wouldn't work out for the basis. – John Smith Feb 17 '13 at 0:21