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 $dim X=n<infty$ 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.

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

2 Answers

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

$ker T ne {0}, tag 6$


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

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

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