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Proving A subspace

  • Thread starter nautolian
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  • #1
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1. Let V={X [itex]\in[/itex] R2 : (1 2) X= (0)}
..........................(3 4) ..... (0)
Show that V is a subspace of R^2 with the usual operations. What is the dimension of V

2. Homework Equations

3. I am really kind of lost, the statement seems to make no sense. X is in R but it also = the matrix [1 2, 3 4] but also equals the matrix [0,0]?
Is v not a subspace of R^2 because it equals the zero vector? Thanks for the help.
 
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  • #2
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1. Let V={X [itex]\in[/itex] R2 : (1 2) X= (0)}
..........................(3 4) ..... (0)
Show that V is a subspace of R^2 with the usual operations. What is the dimension of V

2. Homework Equations

3. I am really kind of lost, the statement seems to make no sense. X is in R
No, X is a vector in R2.
but it also = the matrix [1 2, 3 4]
No, this is a 2 x 2 matrix, so it can't possibly be equal to a vector in R2.
but also equals the matrix [0,0]?
No again.

Here is your matrix:
$$\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix} $$
Let's call it A

Here is your vector x
$$\begin{bmatrix}x_1 \\ x_2\end{bmatrix} $$

V is all of the vectors in R2 such that Ax = 0, where 0 is the zero vector in R2.
Is v not a subspace of R^2 because it equals the zero vector? Thanks for the help.
 
  • #3
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Isn't the only solution to that equation [0,0]? Hence, it is not a subspace because the only solution is the zero vector? Thanks again
 
  • #4
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Isn't the only solution to that equation [0,0]?
Maybe. Why do you think so?
Hence, it is not a subspace because the only solution is the zero vector?
A set with only the zero vector in it is a subspace of whatever space it's in.
Thanks again
 
  • #5
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So essentially the dimension is then 1 because the basis is the zero vector? Is that possible and I think so because reducing the matrix [1 2, 3 4] gives the identity which set to 0 means that the only solution is at x=[0,0]
 
  • #6
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So essentially the dimension is then 1 because the basis is the zero vector? Is that possible and I think so because reducing the matrix [1 2, 3 4] gives the identity which set to 0 means that the only solution is at x=[0,0]
You're mostly on track here. Since the basis for V is just the zero vector, the dimension of V is zero, not one. Your matrix A maps every nonzero vector in R to some other nonzero vector.
 
  • #7
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Thanks! Okay, so how would I go about proving V is a subspace of R^2 though? Like in terms of addition and multiplication being closed? Do I just pretend that two vectors in V are [a,b] and [c,d]??
 
  • #8
Dick
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Thanks! Okay, so how would I go about proving V is a subspace of R^2 though? Like in terms of addition and multiplication being closed? Do I just pretend that two vectors in V are [a,b] and [c,d]??
The only element of V is [0,0]. Doesn't that make it pretty easy to show it's a subspace? It's a trivial subspace.
 

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