Are these subspaces of R^n?

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November 12th 2011, 09:50 AM
gotmejerry

Are these subspaces of R^n?

Which ones of the following subsets of $\mathbb{R}^n$ are subspaces.

a) $L=\{x= \[\alpha_1,\ldots,\alpha_n\]^\top\ | \sum_{i=1}^n (-1)^n\alpha_i =0\}$

b) $M=\{x= \[\alpha_1,\ldots,\alpha_n\]^\top\ | \alpha_i - \alpha_{i-1} = constant\\ i= 2,3,\dots,n\}$

c) $N=\{x= \[\alpha_1,\ldots,\alpha_n\]^\top\ | \left\frac{\alpha_i}{\alpha_{i-1}} \right = constant\\ i= 2,3,\dots,n\}$

d) $P=\{x= \[\alpha_1,\ldots,\alpha_n\]^\top\ | \sum_{i=1}^n 2^{n-1}\alpha_i =0\}$

I should decide their dimensions and give a basis for each. I think a) and d) are subspaces, how can I find the dimension of these spaces and how can I give a basis?

Thank you!
November 12th 2011, 10:39 AM
FernandoRevilla

Re: Are these subspaces of R^n?

Quote:

Originally Posted by gotmejerry

Which ones of the following subsets of $\mathbb{R}^n$ are subspaces. a) $L=\{x= \[\alpha_1,\ldots,\alpha_n\]^\top\ | \sum_{i=1}^n (-1)^n\alpha_i =0\}$

(i) Obviously $(0,\ldots,0)^T\in L$ (ii) Consider $\alpha=(\alpha_1,\ldots,\alpha_n)^T\in L$ and $\beta=(\beta_1,\ldots,\beta_n)^T\in L$ Then, $\sum_{i=1}^n(-1)^n(\alpha_i+\beta_i)=\sum_{i=1}^n(-1)^n\alpha_i+\sum_{i=1}^n(-1)^n\beta_i=0+0=0$ , so $\alpha+\beta\in L$ (iii) ....

What have you tried for the rest?.
November 12th 2011, 11:00 AM
gotmejerry

Re: Are these subspaces of R^n?

I did it this way. So when the linear combination of 2 random elements of the set gives an element which in the set too, it is a subspace.

For c) the 0-vector isn't in the set so it cannot be a subspace.

For d) I did the same what you did with a). It is a subspace.

For b) I got from the lin.comb of 2 elements (a+b)*c when c is the constant and a,b are scalars, so I think I should have got c, so it is not a subspace.

But my main problem is I cannot decide their dimensions and cannot give a basis(Speechless)
November 12th 2011, 11:37 AM
FernandoRevilla

Re: Are these subspaces of R^n?

Quote:

Originally Posted by gotmejerry

But my main problem is I cannot decide their dimensions and cannot give a basis(Speechless)

For example $L\equiv \alpha_1+\ldots +\alpha_n=0$ or equivalently:

$\begin{bmatrix} \alpha_1\\{\alpha_2}\\ \vdots\\{\alpha_n}\end{bmatrix}=\begin{bmatrix} \alpha_1\\{\alpha_2}\\ \vdots\\{-\alpha_1-\ldots -\alpha_{n-1}}\end{bmatrix}=\alpha_1\begin{bmatrix} 1\\{0}\\ \vdots\\{-1}\end{bmatrix}+\alpha_2\begin{bmatrix} 0\\{1}\\ \vdots\\{-1}\end{bmatrix}+\ldots$

Conclude.
November 12th 2011, 12:09 PM
gotmejerry

Re: Are these subspaces of R^n?

My conclusion, its dimension is n-1. What about the d)?
November 12th 2011, 01:48 PM
FernandoRevilla

Re: Are these subspaces of R^n?

Quote:

Originally Posted by gotmejerry

My conclusion, its dimension is n-1.

Right.

Quote:

What about the d)?

You wrote:

Quote:

Originally Posted by gotmejerry

d) $P=\{x= \[\alpha_1,\ldots,\alpha_n\]^\top\ | \sum_{i=1}^n 2^{n-1}\alpha_i =0\}$

In such case , $P=L$ unless you meant $\sum_{i=1}^n 2^{i-1}\alpha_i =0$
November 12th 2011, 01:54 PM
gotmejerry

Re: Are these subspaces of R^n?

And are my assumptions correct that b) and c) are not subspaces?
November 12th 2011, 03:03 PM
FernandoRevilla

Re: Are these subspaces of R^n?

Quote:

Originally Posted by gotmejerry

And are my assumptions correct that b) and c) are not subspaces?

$M$ is subspace of $\mathbb{R}^n$ and:

$\begin{bmatrix}\alpha_1\\{\alpha_2}\\ \vdots\\{\alpha_n}\end{bmatrix}=\begin{bmatrix} \alpha_1\\{\alpha_1+k}\\ \vdots\\{\alpha_1+(n-1)k}\end{bmatrix}=\alpha_1\begin{bmatrix}1\\{1}\\ \vdots\\{1}\end{bmatrix}+k\begin{bmatrix}0\\{1}\\ \vdots\\{n-1}\end{bmatrix}$

$N$ is not a subspace of $\mathbb{R}^n$ .