Linear transformaton and inverse

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Homework Statement



If T : V → W is an injective linear transformation, then T^-1: V →W is a linear transformation.

The Attempt at a Solution



Let w1, w2 in W. If w1=T(v1) and w2=T(v2), v1=/=v2 in V. Thus, T^-1: V →W is a function. Then, v1+v2=T^-1(w1) + T^-1(w2) and for a in F, T^-1(w1) = aT^-1(w1) = av1 for w in W.
 
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autre said:
If T : V → W is an injective linear transformation, then T^-1: V →W is a linear transformation.

This is not true. Injectivity is not enough. You need bijectivity.
 
Oh sorry, I left out a part. T:V -> W is a linear transformation, T(V) is a subspace of W.
 
micromass said:
This is not true. Injectivity is not enough. You need bijectivity.

but doesn't he have to use injectivity and surjectivity to show bijectivity?
 
autre said:
Oh sorry, I left out a part. T:V -> W is a linear transformation, T(V) is a subspace of W.

Not enough. You need T to be surjective.

mtayab1994 said:
but doesn't he have to use injectivity and surjectivity to show bijectivity?

Yes, but he didn't state that T is surjective.
 
what IS true, is that if T:V→W is an injective linear transformation, then

T-1:T(V)→V is a linear transformation.

now, prove this by explicitly defining what T-1 has to be.
 
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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