anonymous
  • anonymous
any one familiar with linear transformations? would be appreciated! attached file
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
anonymous
  • anonymous
@oldrin.bataku
anonymous
  • anonymous
the wording of the questions hurts my brain

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anonymous
  • anonymous
since \((0,1)\) is a basis for \(V\) it follows that \(F((0,1))\) should be a basis for \(F(V)\) I imagine...
anonymous
  • anonymous
would we relate this question we the rotation matrix |dw:1377910829146:dw|? not really sure
anonymous
  • anonymous
I suppose you could:$$\mathbf{F}=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}\\\mathbf{F}\left(\lambda\begin{bmatrix}0\\1\end{bmatrix}\right)=\lambda\mathbf{F}\begin{bmatrix}0\\1\end{bmatrix}=\lambda\begin{bmatrix}-\sin\theta\\\cos\theta\end{bmatrix}$$so \((-\sin\theta,\cos\theta)\) should be a basis I think
anonymous
  • anonymous
do you think thats what they are saying?
anonymous
  • anonymous
well that is just the matrix corresponding to the linear transformation so sure
anonymous
  • anonymous
alrighty, i'll go with that then
anonymous
  • anonymous
i had some idea of a) but b) is not my type of proofs
anonymous
  • anonymous
for the second it seems almost obvious from intuition but I'm not sure how they'd like us to prove it... I suppose you can show \(\mathcal{R}(T)\) is just a subspace spanned by the column vectors
anonymous
  • anonymous
how would you do that?
anonymous
  • anonymous
but aren't they asking two direction here? i'm confused
anonymous
  • anonymous
nup i've gotten no where
blockcolder
  • blockcolder
The range of T is \(R^m\); thus, for arbitrary \(v \in R^m\), there exists a \(u\in R^m\) such that T(u)=v. Write \(u\) as \(u=c_1e_1+c_2e_2+\cdots+c_ne_n\) where \(e_i\) are the standard basis in \(R^n\). Since T is a linear transformation, we have \(T(u)=c_1T(e_1)+c_2T(e_2)+\cdots+c_nT(e_n)=v\). We have written an arbitrary \(v\in R^m\) as a linear combination of the columns of the standard matrix of T. Thus, they span R^m.
anonymous
  • anonymous
is v an arbitrary vector?
blockcolder
  • blockcolder
Yep
anonymous
  • anonymous
don't u have to prove that the range of T is Rm...not state it?
blockcolder
  • blockcolder
I only proved the first direction of the if-and-only-if statement. The reverse will be my next post.
blockcolder
  • blockcolder
Now, let us assume that the columns of the standard matrix of T span R^m. Thus, for an arbitrary \(v_2 \in R^m\), we have \(v_2=d_1T(e_1)+d_2T(e_2)+\cdots+d_nT(e_n)\). Since T is a linear transformation, we can rewrite \(v_2\) as \(v_2=T(d_1e_1+d_2e_2+\cdots+d_ne_n)\). Call the vector in the parenthesis \(u_2\). Note that \(u_2 \in R^n\) since it is a linear combination of the standard basis in \(R^n\). Since we have shown that for an arbitrary \(v_2 \in R^m\), there exists a \(u_2 \in R^n\) such that \(T(u_2)=v_2\), the range of T is the entirety of \(R^m\).
anonymous
  • anonymous
ohh, right. niceeee
anonymous
  • anonymous
your a lifesaver! thanks for your help, that clears everything!
blockcolder
  • blockcolder
No problem. :)
anonymous
  • anonymous
with the first question, were we on the right track?
anonymous
  • anonymous
@blockcolder
blockcolder
  • blockcolder
yep

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