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Consider the set of (column) vectors defined by$$X = \left \{x \in R^3 \mid x_1 + x_2 + x_3 = 0, \text{ where } x^T = \left[x_1,x_2,x_3\right]^T\right \}$$.Which of the following is TRUE?

  1. $\left\{\left[1,-1,0\right]^T,\left[1,0,-1\right]^T\right\}$ is a basis for the subspace $X$.
  2. $\left\{\left[1,-1,0\right]^T,\left[1,0,-1\right]^T\right\}$ is a linearly independent set, but it does not span $X$ and therefore is not a basis of $X$.
  3. $X$ is not a subspace of $R^3$.
  4. None of the above
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4 Answers

Best answer
40 votes
40 votes

$\begin{vmatrix} 1 & -1 & 0\\ 1 & 0 & -1 \end{vmatrix} \ne 0$

Rank = 2 because size of largest square submatrix whose determinant value not equal to $0$ is $2.$ i.e. Both are linearly independent vectors

Given, $x_1 + x_2 + x_3 = 0 \implies x_3 = - (x_1 + x_2)$

$A= \begin{vmatrix} 1 & -1 & 0\\ 1 & 0 & -1\\ x_1 & x_2 & -(x_1+x_2) \end{vmatrix} = x_2 + 1(-x_1 -x_2) + x_1 =0$

which means Rank$(A) \ne 3 \implies$ there exist at least one vector which is linearly dependent. We already now that $\begin{bmatrix}1 & -1& 0\end{bmatrix}$ and $\begin{bmatrix}1 & 0& -1\end{bmatrix}$  are linearly independent.

So, vector $\begin{bmatrix}x_1 & x_2& x_3\end{bmatrix}$  where $x_1 + x_2 + x_3 = 0$ is linearly dependent on $\begin{bmatrix}1 & -1& 0\end{bmatrix}$ and $\begin{bmatrix}1 & 0& -1\end{bmatrix}$

Since all vectors of form $\begin{bmatrix}x_1 & x_2& x_3\end{bmatrix}$ where $x_1 + x_2 + x_3 = 0$ is a linear combination of $\begin{bmatrix}1 & -1& 0\end{bmatrix}$ and $\begin{bmatrix}1 & 0& -1\end{bmatrix}$, I can say that $\begin{bmatrix}1 & -1& 0\end{bmatrix}$ and $\begin{bmatrix}1 & 0& -1\end{bmatrix}$ is the basis of vector space, $X.$

So, (A) is TRUE.  

 

Subspace of a vector space

Any subset of a vector space(collection of vectors) is called as subspace iff

  1.  that subset contains zero vector
  2. that subset is closed under scalar multiplication 
  3. that subset is closed under addition of any $2$ vectors in that subset.

$X = \begin{bmatrix}x_1 & x_2& x_3 \end{bmatrix}$ where$ x_1 + x_2 x_3 = 0$

1. $X$ contains zero vector $\begin{bmatrix}0 & 0& 0 \end{bmatrix}$ as $0 + 0 + 0 = 0$

2. Let $C$ be a scalar (any real number)

$C\begin{bmatrix}x_1 & x_2& x_3 \end{bmatrix} = \begin{bmatrix}Cx_1 & Cx_2& Cx_3 \end{bmatrix} \in X$ because $Cx_1 + Cx_2 + Cx_3 = C (x_1 + x_2 + x_3) = C.0 = 0$.

3. Let $Y=\begin{bmatrix}y_1 \\ y_2\\  y_3\end{bmatrix} \in X$ where $y_1 + y_2 + y_3 = 0$ and $z=\begin{bmatrix}z_1 \\ z_2\\  z_3\end{bmatrix} \in X$ where$ z_1 + z_2 + z_3 = 0.$

$Y + Z  = \begin{bmatrix}y_1+ z_1 \\ y_2+ z_2\\  y_3 + z_3\end{bmatrix} \in X$ as

$y_1 + z_1+ y_2 + z_2 + y_3  + z_3$

$= (y_1 + y_2 + y_3) + (z_1 + z_2 + z_3) =0.$

So, clearly $X$ is subspace of $R^3$

So (A) is the answer.

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9 votes
9 votes

Here, subspace $X \subseteq \mathbb{R}^{3}$ is defined as :-

$$X = \left \{x \in \mathbb{R}^3 \mid x_1 + x_2 + x_3 = 0, \text{ where } x^T = \left[x_1,x_2,x_3\right]^T\right \}$$.

which means sum of all the components of the vectors are zero which are in subspace $X$.

We can write  $ x_1 + x_2 + x_3 = 0$ as

                                                 $\begin{bmatrix} 1 &1 &1 \end{bmatrix}$$\begin{bmatrix} x_{1}\\x_2 \\ x_3 \end{bmatrix} = 0$

Now, assuming vector $x$ is in the nullspace of matrix $[1,1,1]$ and we have to find the basis for it.

Since, all the components of vector $x$ are zero. So, we can write $x_3 = -x_2 - x_1$

So,

$\begin{bmatrix} x_{1}\\x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} x_{1}\\x_2 \\ -x_2-x_1 \end{bmatrix} =x_1\begin{bmatrix} 1\\0 \\ -1 \end{bmatrix} + x_2\begin{bmatrix} 0\\1 \\-1 \end{bmatrix}$ where $x_1,x_2 \in \mathbb{R}$

So, vectors $\left \{ \begin{bmatrix} 1\\0 \\-1 \end{bmatrix}, \begin{bmatrix} 0\\1 \\-1 \end{bmatrix} \right \}$ should be in the subspace $X.$

Now, we have to check whether these vectors are linearly independent or not.

So,

$k_1\begin{bmatrix} 1\\0 \\-1 \end{bmatrix} + k_2\begin{bmatrix} 0\\1 \\-1 \end{bmatrix} = \begin{bmatrix} 0\\0 \\0 \end{bmatrix}$

$\Rightarrow$ $k_1 = 0$ and $k_2 = 0$ which implies both vectors $\left \{ \begin{bmatrix} 1\\0 \\-1 \end{bmatrix}, \begin{bmatrix} 0\\1 \\-1 \end{bmatrix} \right \}$ are linearly independent.

Since, $x \in \mathbb{R}^{3}$ and it is in nullspace which implies no. of columns($n$) = $3$ (it can be seen from the matrix $[1,1,1])$. 

Now, say, matrix $[1,1,1]$ is $A$. So, $rank(A)= r = 1$.

So, $dimension(N(A)) = n-r = 3-1 =2 $. It means basis for nullspace should have $2$ vectors. Basis contains linearly independent vectors which must span the whole vector space.

Hence, we can say that here basis has $2$ linearly independent vectors which are  $\left \{ \begin{bmatrix} 1\\0 \\-1 \end{bmatrix}, \begin{bmatrix} 0\\1 \\-1 \end{bmatrix} \right \}$

So, Basis for $X$ is $\left \{ \begin{bmatrix} 1\\0 \\-1 \end{bmatrix}, \begin{bmatrix} 0\\1 \\-1 \end{bmatrix} \right \}.$

Basis is not unique. We can have many bases for a given vector space but no. of vectors in all the bases should be same. Since, answer contains different basis. So, we have to change the basis for subspace $X$.

Since, Basis for $X$ is $\left \{ \begin{bmatrix} 1\\0 \\-1 \end{bmatrix}, \begin{bmatrix} 0\\1 \\-1 \end{bmatrix} \right \}.$ It means vectors $\begin{bmatrix} 1\\0 \\-1 \end{bmatrix}\; and \; \begin{bmatrix} 0\\1 \\-1 \end{bmatrix}$ span the whole subspace $X$. So, on doing linear combination of both vectors, we will get another vector as :-

$1*\begin{bmatrix} 1\\0 \\-1 \end{bmatrix} + (-1)\begin{bmatrix} 0\\1 \\ -1 \end{bmatrix} = \begin{bmatrix} 1\\-1 \\0 \end{bmatrix}.$

So, vectors $\begin{bmatrix} 1\\-1 \\0 \end{bmatrix}\; and \; \begin{bmatrix} 1\\0 \\-1 \end{bmatrix}$ will be in subspace $X.$ Now, we have to check whether these are linearly independent or not.

So, $k_1\begin{bmatrix} 1\\-1 \\ 0 \end{bmatrix} +k_2\begin{bmatrix} 1\\0 \\ -1 \end{bmatrix} = \begin{bmatrix} 0\\0 \\ 0 \end{bmatrix}$ which implies $k_1=0$ and $k_2=0.$

It means vectors $\begin{bmatrix} 1\\-1 \\0 \end{bmatrix}\; and \; \begin{bmatrix} 1\\0 \\-1 \end{bmatrix}$ are linearly independent and form the basis for subspace $X.$

So, Another Basis for $X$ is $\left \{ \begin{bmatrix} 1\\-1 \\0 \end{bmatrix}, \begin{bmatrix} 1\\0 \\-1 \end{bmatrix} \right \}.$

Hence, Answer is $(A).$

edited by
7 votes
7 votes

Option $(a)$, here is the answer,

http://math.stackexchange.com/a/1843452/153195

edited by
4 votes
4 votes

B I think is the answer we require two constant c1 and c2 which satisfy the R3 which is 
 

let X=

1 0 0
0 1 0
0 0 1

A=

1
-1
0

B=

1
0
-1

write it as echo-lean  form ABX it comes to be inconsistent and hence it does not span.
 

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