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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 (59.7k points)
edited by | 2.2k 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
+2
$R^3$ 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.

3 Answers

+14 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 (7.7k 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?

+6 votes

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

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

answered by Active (1.6k 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
+3 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 (3.6k 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,

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

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