tukeywilliam Posted March 23, 2007 Share Posted March 23, 2007 How would you find the dimension of the subspace of [math]\mathbb{P}_{3}[/math] spanned by the subset [math]\{t,t-1,t^{2}+1\}[/math]? How would you find a basis and dimension of the given subspace [math]\mathbb{R}^{3}\}[/math] [math]\{[a,a-b,2a+3b]|a,b, \in \mathbb{R}\}[/math] as a subspace of [math]\mathbb{R}^{3}[/math]? Link to comment Share on other sites More sharing options...
timo Posted March 23, 2007 Share Posted March 23, 2007 How would you find the dimension of the subspace of [math]\mathbb{P}_{3}[/math] spanned by the subset [math]\{t,t-1,t^{2}+1\}[/math]? Find the maximum number of linear independent vectors (the vectors being t, t-1 and t²+1). How would you find a basis and dimension of the given subspace [math]\mathbb{R}^{3}\}[/math] [math]\{[a,a-b,2a+3b]|a,b, \in \mathbb{R}\}[/math] as a subspace of [math]\mathbb{R}^{3}[/math]? For the dimension I´d simply say it´s two since there´s two independent parameters describing the subspace - there´s possibly some more formal way of doing it. For constructing a basis, just pick two vectors from the subspace which are linearly independent. Link to comment Share on other sites More sharing options...
tukeywilliam Posted March 23, 2007 Author Share Posted March 23, 2007 For the first one I got the dimension to be 3. Is this correct? Link to comment Share on other sites More sharing options...
timo Posted March 23, 2007 Share Posted March 23, 2007 Yes, but it would be helpful (also for others who might be interested in reading it) to state how you came to that result. An example: From the three spanning vectors define new vectors a, b and c: a = t - (t-1) = 1 b = t c = (t²+1) - (t+1) + t = t² From the three spanning vectors you have, by linear combination, constructed three vectors 1, t and t² which are obviously linearly independent. Therefore the dimension is three. Link to comment Share on other sites More sharing options...
tukeywilliam Posted March 23, 2007 Author Share Posted March 23, 2007 Thanks. Also for the second one I found one basis to be: [math] a, 2a+3b [/math] because they are linearly independent? is this right? And we can probably say that if the determinant is 0, then the vectors are linearly dependent because inverse is 0. Link to comment Share on other sites More sharing options...
timo Posted March 23, 2007 Share Posted March 23, 2007 - I think you´re not understanding the 2nd question correctly. Find out what the base vectors should look like, first (we´re talking about the normal R³ here). - Using a determinant should work for deciding whether vectors are linearly independent. Link to comment Share on other sites More sharing options...
tukeywilliam Posted March 23, 2007 Author Share Posted March 23, 2007 do you mean: [math]\vec{e_1} = \begin{bmatrix}1\\0\\0\end{bmatrix}, \vec{e_2} = \begin{bmatrix}0\\1\\0\end{bmatrix}, \vec{e_3} = \begin{bmatrix}0\\0\\1\end{bmatrix} [/math]? Link to comment Share on other sites More sharing options...
akmather Posted April 25, 2007 Share Posted April 25, 2007 For problem 2: If every vector is of the form [math] \vec{x} = \begin{bmatrix}a\\a-b\\2a+3b\end{bmatrix}[/math], then [math] \vec{x} = \begin{bmatrix}a\\a-b\\2a+3b\end{bmatrix}=a\begin{bmatrix}1\\1\\2\end{bmatrix}+b\begin{bmatrix}0\\-1\\3\end{bmatrix}[/math] Hence a basis for your subspace is: [math] \left\{ \begin{bmatrix}1\\1\\2\end{bmatrix},\begin{bmatrix}0\\-1\\3\end{bmatrix} \right\}[/math] As the two vectors are certainly independent, the dimension of the space would be two. I know this is an old problem, but I didn't see any harm in putting down a solution at this point and I needed the latex refresher... Link to comment Share on other sites More sharing options...
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