My Math Forum Transformation is not invertible

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 April 27th, 2013, 11:41 AM #1 Newbie   Joined: Apr 2013 Posts: 11 Thanks: 0 Transformation is not invertible T is a linear transformation in ($\mathbb{F}$^3,$\mathbb{F}$^2) S is a linear transformation in ($\mathbb{F}$^2,$\mathbb{F}$^3) Prove that the transformation ST in ($\mathbb{F}$^3) is not invertible. Prove (or give example) that TS in ($\mathbb{F}$^2) can be invertible Thanks a lot in advance
 April 27th, 2013, 02:47 PM #2 Newbie   Joined: Apr 2013 Posts: 10 Thanks: 0 Re: Transformation is not invertible Note that the image of ST is contained in the image of S, and the image of S can be at most 2-dimensional, i.e. $rank(ST) \leq rank(S) \leq \dim(\mathbb{F}^2)=2$. Hence, ST has no chance of mapping onto $\mathbb{F}^3$. On the other hand, if we define T and S by T(x,y,z) = (x,y) and S(x,y) = (x,y,0) (note these are both linear), then TS(x,y)=(x,y) is the identity map on $\mathbb{F}^2$, which is clearly invertible.
 April 27th, 2013, 03:42 PM #3 Newbie   Joined: Apr 2013 Posts: 11 Thanks: 0 Re: Transformation is not invertible Thanks a lot greentunic. I am just wondering about one property. It says if T is a linear transformation in (F^3,F^2) S is a linear transformation in (F^2,F^3) Then ST is a linear transformation in (F^3) Isn't it? Then ST should map onto F^3 Am I wrong? Thanks again.
 April 28th, 2013, 04:49 PM #4 Newbie   Joined: Apr 2013 Posts: 10 Thanks: 0 Re: Transformation is not invertible ST is a linear map from $\mathbb{F}^3$ into $\mathbb{F}^3$, but it does not map onto $\mathbb{F}^3$. In other words, ST is not surjective.

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