"Span of a Subspace - Does it Equal x?

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Homework Statement



If x is a subspace of V so, span(x)=x

Homework Equations



span(x)=x


The Attempt at a Solution



If x is a subspace so, for any "a", "b" in x:
a+b∈x
and (c1)*a∈x

So a linear combination of x belongs to x but is equal to x?
 
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Your last sentence is badly worded. There is no such thing as "a linear combination of x". You mean "a linear combination of vectors in x".

What you have proved is only one direction- you have proved that the span of x is a subset of x. Now you need to prove that x is a subset of span of x. That is easy. Suppose a is vector in x. Then 1a is in the span of x.
 
Hi, thanks!
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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