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Specify a base for each of the following vector spaces and prove your claims:

  • a) $\{(x_1,x_2,x_3)\in \mathbb{R}^3:x_1=x_3\}$ over $\mathbb{R}^3$
  • b) $\{f=\sum^n_{i=0}a_it^i\in \mathbb{Z}_7[t]: a_0=0,\; \deg(f)\leq5\}$ over $\mathbb{Z}_7$

a) Let $u=(1,0,1),\; v=(2,1,2),\; w=(0,1,0)\quad u,v,w\in\mathbb{R}^3$ $$ \begin{pmatrix} 1 & 2 & 0\\ 0 & 1 & 1\\ 1 & 2 & 0 \end{pmatrix} \iff \begin{pmatrix} 1 & 2 & 0\\ 0 & 1 & 1\\ 0 & 0 & 0 \end{pmatrix} $$ Therefore, one possible Basis would be $$B_1=\left\{(1,0,1),(2,1,2)\right\}$$ and the dimension is $dim()=2$

b) In other words, $f=a_1t^1+\dots+a_5t^5$
Therefore, one Basis (the standardbasis) would be $$B_2=\{(1,0,0,0,0,0),(0,1,0,\dots,0),\dots,(0,\dots,0,1)\}$$ Notation: $(t^5,t^4,\dots,t^0)$


Is this correct? If yes, how do I prove that this is indeed a basis? (I guess I need to show that we can display the whole vector space with the help of linear combinations of the vectors of the basis, right?)

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  • $\begingroup$ a) Why don't you take the somewhat obvious basis $(1,0,1)$, $(0,1,0)$? Btw, you do not show that your answer is a base -- b) $(1,0,0,0,0)$ is a tuple and the given space does not contain tuples, it contains polynomials. $\endgroup$ – Hagen von Eitzen May 14 at 19:49
  • $\begingroup$ My notation is $(t^5,t^4,\dots,t^1,t^0)$ $\endgroup$ – Doesbaddel May 14 at 20:20
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To prove you have a basis, you need to show that any vector is your subspace is a linear combination of the set of vectors you’ve provided. For example, we can write

$$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \stackrel{x_1 = x_3}{=} \begin{bmatrix} x_1 \\ x_2 \\ x_1 \end{bmatrix} = \begin{bmatrix} x_1 \\ 0 \\ x_1 \end{bmatrix} + \begin{bmatrix} 0 \\ x_2 \\ 0 \end{bmatrix} = x_1\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + x_2\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}.$$

This means that $\{(1,0,1)^T, (0,1,0)^T\}$ spans your subspace. Now you just need to show that this set of vectors is linearly independent.


Similarly, if we represent $f = a_1t^1 + \dots + a_5t^5$ as $(a_5, \dots, a_1, 0)^T$, we can write $f$ as

$$f = \begin{bmatrix} a_5 \\ a_4 \\ a_3 \\ a_2 \\ a_1 \\ 0 \end{bmatrix} = a_5\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} + a_4 \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} + a_3 \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + a_2\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} + a_1 \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}.$$

That means these vectors span your subspace — the last step, just as above, to show that they are linearly independent. Note that these vectors are just a way of representing the polynomials $\{ t^5, t^4, t^3, t^2, t^1\}$.

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