GATE DS&AI 2024 | Question: 37

Question

Select all choices that are subspaces of $\mathbb{R}^{3}$.

Note: $\mathbb{R}$ denotes the set of real numbers.

• A. $\left\{\mathbf{x}=\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right] \in \mathbb{R}^{3}: \mathbf{x}=\alpha\left[\begin{array}{l}1 \\ 1 \\ 0\end{array}\right]+\beta\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right], \alpha, \beta \in \mathbb{R}\right\}$
   
• B. $\left\{\mathbf{x}=\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right] \in \mathbb{R}^{3}: \mathbf{x}=\alpha^{2}\left[\begin{array}{l}1 \\ 2 \\ 0\end{array}\right]+\beta^{2}\left[\begin{array}{l}1 \\ 0 \\ 1\end{array}\right], \alpha, \beta \in \mathbb{R}\right\}$
   
• C. $\left\{\mathbf{x}=\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right] \in \mathbb{R}^{3}: 5 x_{1}+2 x_{3}=0,4 x_{1}-2 x_{2}+3 x_{3}=0\right\}$
   
• D. $\left\{\mathbf{x}=\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right] \in \mathbb{R}^{3}: 5 x_{1}+2 x_{3}+4=0\right\}$

Answer 1

The answer is A, C.

A: given space is a spanning set of two Linearly independent vectors of R³, so it will be a subspace of R³ of dimension 2.

B: It is only the non-negative linear combination. so not a subspace.

C: given space is a subset of R³ under two Linear homogeneous restrictions it will make a subspace.

D: given restriction is not homogenous, so it will not be a subspace.

NOTE: Any subset of Rⁿ under the Linear homogeneous restrictions forms a subspace of R^n.

Comments

$\alpha^2$ and $\beta^2$ can't be negative so I don't think that scalar multiplication with $-1$ is closed in option $B$ for it to be a subspace.

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What can be if in option C one homogenous and other non-homogenous equation of planes were given, will it be considered in subspaces or not ?

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No that does not form a subspace. Lets assume first equation equals to 4 instead of 0 when we do elimination we get  vector X =  Xp+Xn here Xp is Particular solution (4, -10, 0) and Xn free solution c(-2, 3/2, 1) where c belongs to R. So X= (4, -10, 0) + c(-2, 3/2, 1) we can put c=0 but still won't get zero vector.