# Let $T:X to Y$ be a linear operator and $dim X=dim Y<infty$. Show $Y=mathscr{R}(T)$ if and only if $T^{-1}$ exists, without dimension theorem.

Mathematics Asked by André Armatowski on December 2, 2020

The problem I am trying to solve is:

Let $$T:X to Y$$ be a linear operator and $$dim X = dim Y = n < infty$$. Show that $${scr{R}}(T)=Y$$ if and only if $$T^{-1}$$ exists.

Here $${scr{R}}(T) := text{Range} T$$.

This question asks the same thing but the answer uses the dimension theorem which has yet to be presented and so I am interested in if another proof exists.

My current progress:

The proof that the existence of $$T^{-1}$$ implies $${scr{R}}(T)=Y$$ follows by the following theorem:

Let $$T:{scr{D}}(T)to Y$$ be a linear operator whose inverse exists. If $${x_{1},dots,x_{n}}$$ is a linearly independet set in $${scr{D}}(T)$$ then $${T x_{1},dots,T x_{n}}$$ is linearly independent in Y.

The argument is: Since $$dim X=n there exists a linearly independent set of $$n$$ vectors $${x_{1},dots,x_{n}}$$ in $$X$$ and since $$T^{-1}$$ exists we get that $${Tx_{1},dots,Tx_{n}}$$ is a linearly independent set in $$Y$$ by the above theorem. Since $$dim Y=n$$ the set $${T x_{1},dots,T x_{n}}$$ forms a basis for $$Y$$. So, for any $$yin Y$$ there exists scalars $$alpha_{1},dots,alpha_{n}$$ such that by the linearity of $$T$$:$$y=alpha_{1}Tx_{1}+dotsalpha_{n}Tx_{n}=T(alpha_{1} x_{1}+dots+alpha_{n}x_{n}).$$
Therefore $$yin {scr{R}}(T)$$ and because $$yin Y$$ was chosen arbitrarily $${scr{R}}(T)=Y$$.

Suppose now instead that $${scr{R}}(T)=Y$$. Then to prove that $$T^{-1}$$ exists it suffices to show that $$T$$ is injective. First let us pick a basis $${y_{1},dots,y_{n}}$$ for $$Y$$. Since $${scr{R}}(T)=Y$$ there exists vectors $$x_{1},dots,x_{n}in X$$ such that $$T x_{1}=y_{1},dots,T x_{n} = y_{n}$$. Then it is immediate that if $$Ta = Tb$$ and we write $$Ta$$ and $$Tb$$ in terms of linear combinations of $$Tx_{1},dots,Tx_{n}$$, that is
$$Ta = alpha_{1}Tx_{1}+dots+alpha_{n}Tx_{n}, quad Tb=beta_{1}Tx_{1}+dots+beta_{n}Tx_{n}$$

it will result in $$alpha_{j}=beta_{j}$$ (since $${Tx_{1},dots,Tx_{n}}$$ is a basis).

Here I am stuck. I think the claim would follow either if $$x_{j} mapsto Tx_{j}$$ was unique (but this is sort of what we want to prove). Or, if we can show the set $${x_{1},dots,x_{n}}$$ to be linearly independent.

Question:
Am I missing something obvious in this last argument and is the first part correct?

If $$T^{-1}$$ exists, that is, if $$T$$ is invertible, then for any

$$y in Y tag 1$$

we have

$$T(T^{-1}(y)) = y, tag 2$$

which shows that

$${scr R}(T) = Y, tag 3$$

that is, $$T$$ is surjective; likewise, if (3) holds, then

$$X/ker T cong Y, tag 4$$

i.e., $$X/ker T$$ and $$Y$$ are isomorphic as vector spaces; from this,

$$dim(X/ker T) = dim(Y); tag 5$$

but if

$$ker T ne {0}, tag 6$$

then

$$dim(X/ker T) < dim X, tag 7$$

and (5) and (7) in concert yield

$$dim(Y) < dim(X), tag 8$$

contradicting the given hypothesis $$dim(X) = dim(Y)$$; therefore

$$ker(T) = {0}, tag 9$$

and thus $$T$$ is injective; since $$T$$ is both surjective and injective, it is invertible, or in other words, $$T^{-1}$$ exists.

Correct answer by Robert Lewis on December 2, 2020

Let $$x_1,...,x_n$$ be a basis of $$X$$. Then $$T(x_1),...,T(x_n)$$ span $$T(X)=Y$$. Since $$Y$$ is $$n$$-dimensional, $$y_1=T(x_1),...,y_n=T(x_n)$$ is a basis of $$Y$$. Consider the linear map $$S:Yto X$$ which takes every linear combination $$sum alpha_iy_1$$ to $$sumalpha_ix_i$$. Then $$Tcirc S$$ ($$T$$ acts first is identity on the basis $$x_1,...,x_n$$. So $$Tcirc S$$ is the identity map. Hence $$S=T^{-1}$$. Conversely, if for some $$S: Yto X$$, $$Tcirc S=mathrm{identity}$$. Then $$T$$ is surjective.

Answered by JCAA on December 2, 2020