Can a Nontrivial Subspace Have the Same Dimension as Its Original Space?

AI Thread Summary
A nontrivial subspace can indeed have the same dimension as its original space. For instance, in a three-dimensional space, the set of all linear combinations of two linearly independent vectors forms a subspace with dimension 2, which is less than the original space. However, in infinite-dimensional vector spaces, such as the space of all polynomials, a nontrivial subspace like the space of all even polynomials can have the same infinite dimension. Additionally, the set of symmetric matrices within square matrices can also demonstrate this property. Understanding these examples can aid in grasping concepts for linear algebra exams.
dan0
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Hi,
I'm just learning for my linear algebra exam and I wonder if somebody could give me an example of a nontrivial subspace which has as many dimensions as the original space.
Thanks a lot
 
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What do you mean by "nontrivial". I would call a subspace nontrivial as long as it was not just the 0 vector. If that's the case then just take any vectors space V with V as the subspace.

Another possibility, a lot more non-trivial would be an infinite dimensional vector space with an infinite dimensional subspace.
The vector space of all polynomials with subspace the space of all even polynomials comes to mind.
 
!

Sure, no problem! One example of a nontrivial subspace with the same dimension as the original space is the set of all linear combinations of two linearly independent vectors in a three-dimensional space. This subspace would have a dimension of 2, the same as the original three-dimensional space. Another example could be the set of all symmetric matrices in a space of square matrices, which would have a dimension equal to the number of variables in the original space. I hope this helps with your exam preparation! Good luck!
 
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