Question

1 $V_1 = \{(x, y) \in \mathbb{R}^2 : 11x + 7y = 0\}$ \\ 2 $V_2 = \{(x, y) \in \mathbb{R}^2 : x^2 + y^2 = 7\}$ \\ 3 $V_3 = \{(x, y) \in \mathbb{R}^2 : x = 7\}$ \\ 4 $V_4 = \{(x, y) \in \mathbb{R}^2 : y = 0\}$ and \\ 5 $V_5 = \{(x, y) \in \mathbb{R}^2 : y = 9x^3\}$ \\ Then the ONLY subspaces of $\mathbb{R}^2$ are:

          1 $V_1 = \{(x, y) \in \mathbb{R}^2 : 11x + 7y = 0\}$ \\
2 $V_2 = \{(x, y) \in \mathbb{R}^2 : x^2 + y^2 = 7\}$ \\
3 $V_3 = \{(x, y) \in \mathbb{R}^2 : x = 7\}$ \\
4 $V_4 = \{(x, y) \in \mathbb{R}^2 : y = 0\}$ and \\
5 $V_5 = \{(x, y) \in \mathbb{R}^2 : y = 9x^3\}$ \\
Then the ONLY subspaces of $\mathbb{R}^2$ are:

$1 V1 = {(x, y) ∈ℝ^2 : 11x + 7y = 0} 2 V2 = {(x, y) ∈ℝ^2 : x^2 + y^2 = 7} 3 V3 = {(x, y) ∈ℝ^2 : x = 7} 4 V4 = {(x, y) ∈ℝ^2 : y = 0} and 5 V5 = {(x, y) ∈ℝ^2 : y = 9x^3} Then the ONLY subspaces of ℝ^2 are:$

Added by Natalie V.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

04/24/2023

Step 1

In other words, it must contain the zero vector, be closed under addition, and be closed under scalar multiplication. Step 2: Let's analyze each set given: Show more…

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1V_(1)={(x,y)inR^(2):11x+7y=0} 2V_(2)={(x,y)inR^(2):x^(2)+y^(2)=7} 3:V_(3)={(x,y)inR^(2):x=7} 4V_(4)={(x,y)inR^(2):y=0} and 5V_(5)={(x,y)inR^(2):y=9x^(3)} Then the ONLY subspaces of R^(2) are: 2 4 3 1 5 5 1 0 3 4 2 Then the ONLY subspaces of IR2 are: