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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
asked in Linear Algebra by Veteran (52k points)
edited by | 2.9k views
0
@arjun sir
What is a subspace ? Is this in present syllabus ?
+1

Quoting from wikipedia:

If W is a vector space itself (which means that it is closed under operations of addition and scalar multiplication), with the same vector space operations as V has, then W is a subspace of V.

0
Can some one please explain question....?
What is  x={x Belongs R3} Here R3 means
+3
$R^3$ refers to the set of all possible 3-tuples $(x, y, z)$, where $x, y, z \in R$, $R$ denotes the set of real numbers.
0
why are they giving questions on vector space and all when it is not in syllabus???

4 Answers

+16 votes

Small correction : in 1 st line rank = 2 because size of largest square submatrix whose determinant value not equal to 0 is 2 ... 

So A) is the answer ... 

answered by Loyal (7.3k points)
edited by
+1
Thank you very much for the detailed solution.
0

@Vicky rix  

condition for a Basis: i.t should span and not contain any redundant vector

                                  -vectors should be linearly independent.

we can prove 1st option as it contain linear independent vectors along with that  it spans X(sub space of X), directly with this we can eliminate option2(which is contradict of 1st option) and option3(we are successful in finding validity of 1,means statement 3 is trivial subspace of X)

can we do with this approach?

0
i am not getting how it will not span x  is answer a or b
0

@Kaluti

both vectors are linearly independent and span the whole subspace $X$. It means take the linear combination of both vectors and resultant vector will also be in the subspace $X$ which means it will satisfy the given condition.

0
It means when we take determinant of those two vectors with the subspace it will always come to be zero it means linear combination of those two vectors lies in that subspace am i correct
0
determinant is defined for square matrices. Here we have 2 vectors which have $3$ components.So, you can't say anything through determinant.

basis for a vector space means it contains right no.(neither less nor more) of linearly independent  vectors which span the whole vector space. It means you can take any linear combination of these linearly independent vector, resultant vector will also be in that space.
0
But including those two vectors as well as the third vector which satisfy the subspace if i include that then determinant will be zero right
0
0
so answer will be A right
0
yes
0
thanks
+6 votes

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

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

answered by Active (1.8k points)
edited by
0
Here from option we can conclude that the two given column vectors are linearly independent.

Let A be the first vector and B be second , k1,k2 are constant...
From k1A+k2B=0
we are getting k1=k2=0.

But how to analyse that these vectors span X or not?
0
What is a subspace ?
0
+4 votes

B I think is the answer we requreire 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 incostitant and hence it does not span.
 

answered by Active (3.7k points)
edited by
0
what is subspace ??(given in option)
0
$X = \{x_1, x_2, x_3: x_1 + x_2 + x_3 = 0\}$

It implies,

$X = \{-x_2 - x_3, x_2, x_3: x_2, x_3 \in R^3\}$

which can also be expressed as,

$\{x_2(-1, 1, 0) + x_3(-1, 0, 1): x_2, x_3 \in R^3\}$.

Equivalently, it also holds as,

$\{x_2(1, -1, 0) + x_3(1, 0, -1): x_2, x_3 \in R^3\}$.
0
i am not getting how it will not span x
0 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 $\left \{ \begin{bmatrix} 1\\0 \\-1 \end{bmatrix}, \begin{bmatrix} 0\\1 \\-1 \end{bmatrix} \right \}.$

Hence, Answer is $(A)$

answered by Boss (13.2k points)
edited by
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