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Find the dimension of the subspace of all vectors in set of real numbers r superscript 4ℝ4 whose firstfirst and thirdthird entries are equal.

User EvilDuck
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1 Answer

1 vote

The subspace has dimension 3.


Intuitively, you can think that you're removing one degree of freedom from
\mathbb{R}^4. In fact, this space is generated by the following span:



\lambda_1 (1,0,0,0) + \lambda_2 (0,1,0,0) + \lambda_3 (0,0,1,0) + \lambda_4 (0,0,0,1)


Since the first and third coordinate must be the same, we have
\lambda_1 = \lambda_3. So, the span becomes




\lambda_1 (1,0,0,0) + \lambda_2 (0,1,0,0) + \lambda_1 (0,0,1,0) + \lambda_4 (0,0,0,1)


which is the same as



\lambda_1 (1,0,1,0) + \lambda_2 (0,1,0,0) + \lambda_4 (0,0,0,1)


And so you can see that the space is generated by three vectors.

User Mleykamp
by
7.7k points
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