Dimension and Basis

[tukeywilliam]

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]?

[timo]

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.

[tukeywilliam]

For the first one I got the dimension to be 3. Is this correct?

[timo]

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.

[tukeywilliam]

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.

[timo]

- 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.

[tukeywilliam]

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]?

[akmather]

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...