ordered bases

**Jskid** · March 15th 2011, 07:10 PM

Let $S={\vec v_1, \vec v_2}$ and $T={\vec w_1, \vec w_2}$ be bases for $P_1$ where $\vec w_1 = t-1$ and $\vec w_2 = t+1$ . If the transition matrix from T to S is $\left[ {\begin{array}{cc}<br /> 1 & 2 \\<br /> 2 & 3 \\<br /> \end{array} } \right]<br />$ determine S.

So the columns of the matrix are the coordinates fo the T-basis vectors with respect to the S-basis, but that's not good because S is what we're trying to find

**FernandoRevilla** · March 16th 2011, 02:18 AM

There is a little problem about terminology. I don't know if your teaher means:

$(i)\;[p(t)]_S=\begin{bmatrix}{1}&{2}\\{2}&{3}\end{bmatrix}[p(t)]_T\textrm{\;\;or\;\;}(ii)\;[p(t)]_T=\begin{bmatrix}{1}&{2}\\{2}&{3}\end{bmatrix}[p(t)]_S$

Supposing $(i)$ we have:

$[\vec{w_1}]_T=\begin{bmatrix}{1}\\{0}\end{bmatrix},\; [\vec{w_2}]_T=\begin{bmatrix}{0}\\{1}\end{bmatrix}$

So,

$[\vec{w_1}]_S=\begin{bmatrix}{1}&{2}\\{2}&{3}\end{bmatrix}\be gin{bmatrix}{1}\\{0}\end{bmatrix}=\begin{bmatrix}{ 1}\\{2}\end{bmatrix}$

Then, $\vec{w_1}=t-1=\vec{v_1}+2\vec{v_2}$ etc .

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