I am trying to prove that suppose that a linear transformation $T$ is invertible, then its transpose $T^t$ is also invertible.

Is the following proof correct?


Let $T$ be an invertible linear transformation that maps elements from the vector space $V$ to $W$. Let $V^\ast$ and $W^\ast$ be the dual spaces respectively.

Suppose that $T^t(g_1)=T^t(g_2)$ where $g_i$ is an element in $W^\ast$. By definition, $T^t(g)=gT$. Therefore, $g_1T=g_2T$. Adding the additive inverse of $g_2T$ to both sides of the equation, we get $g_1T-g_2T=0$ and therefore $(g_1-g_2)T=0$.

Since $T$ is invertible, $T^{-1}$ exist. Applying both sides of $(g_1-g_2)T=0$ to $T^{-1}$, we get $(g_1-g_2)TT^{-1}=0T^{-1}$ and $(g_1-g_2)I=0$. Hence $g_1-g_2=0$.

Finally, this means that $T^t(g_1)=T^t(g_2)$ implies $g_1=g_2$, proving that $T^t$ is injective.

To show that $T^t$ is surjective, we need to show that for all $f$ in $V^\ast$, there exist some $g$ in $W^\ast$ such that $T^t(g)=f$, or $gT=f$. Since $T$ is invertible, given any such $f$, a $g$ can be found and is given by $fT^{-1}$. Hence, $T^t$ is surjective.

Since $T^t$ is bijective, it is invertible.

  • $\begingroup$ Apparently you defined the transpose of a lin. trans. on the dual of the original vector space. I'm not sure this is standard usage (though I could easily be wrong). Have you already studied determinants (are you working in finite dimensional spaces at all, or in general ones?) $\endgroup$ – DonAntonio Nov 27 '13 at 5:34
  • $\begingroup$ I learnt this definition from Hoffman and Kunze's book. I have not done a study of determinants, although I have taken a brief look at it. I am trying to define transpose of a lin.trans. in general and not just on matrices, so how will the study of determinants be relevant? (I know that a non-zero determinant of a square matrix A implies the existence of a non-zero determinant for the transpose of matrix A, therefore implying that A transpose is invertible. What other ways of defining transpose of lin.trans. are there? $\endgroup$ – Student Nov 27 '13 at 6:05

I think we can prove a little stronger result that

$$ T \text{ is surjective } \implies T^{t} \text{ is injective}. $$

But let's do it by putting all the definitions first.

Let $ V $ and $ W $ be linear spaces over a field $ F $.

$T : V \mapsto W $ be an isomorphism i.e a linear transformation that is injective and surjective. Of course, as a bijective map, there is a map $ T^{-1} : W \mapsto V $ and one can prove that this map is itself a linear transformation.

Let $ l : V \mapsto F $ be a "linear functional" i.e. a linear transformation from $ V $ to the one-dimensional vector space $ F $.

The set $ V^{*}= \left \{ l| l : V \mapsto F \text{ is a linear functional } \right \} $ forms a vector space over $ F $, with addition and scalar multiplication defined in the obvious way. We call it the dual of $ V $. Similarly, let $ W^{*} $ denote the dual of $ W $.

If $ l \in W^{*} $, then the composition $ l \circ T : V \mapsto F $ is a linear functional over $ V$, and so we get an assignment, $ T^{t} $, which sends $ l \in W^{*} $ to $ m = l \circ T $ in $ V^{*} $. We can prove that $ T^{t} : W ^{ * } \mapsto V ^ { * } $ is a linear transformation, such that, for $ v \in V $, we have

$$ (T^{t}( l)) ( v ) = l ( T ( v ) ) = m(v). $$

We now prove the following:

(1) If $ T $ is surjective, then $ T^{t} $ is injective.

(2) If $ T $ is invertible, then $ T^{t} $ is surjective.

$ \textit{Proof:} $ We prove (1) first. Suppose that $ T^{t}(g_{1}) = T^{t} ( g_{2} ) $ where $ g_{1}, g_{2} \in W^{*}$. Then, $ g_{1}T = T^{t}(g_{1} ) = T^{t}(g_{2}) = g_{2}T $. Thus, for every $ v \in V $, we have that

$$ g_{1}(T(v)) = g_{2}(T(v)). $$

As $ T $ is surjective, every $ w \in W $ can be written as $ T(v) $ for some $ v \in V $, and so, the last equality says that $ g_{1} $ and $ g_{2} $ agree on every $ w \in W $, and so $ g_{1} = g_{2} $.

To prove (2), let $ m \in V^{*} $, then $ m \circ T^{-1} \in W ^ { * } $, and we have

$$ T^{t} ( m T^{ - 1} ) = (m T^{-1} ) T = m( T^{-1} T) = m ( I ) = m. $$

$ \textit{ Remark : } $ The only thing that I don't like about your proof is when you "apply" the linear functional $ ( g_{1} - g_{2})T $ to $ T^{t} $. You are composing two functions. Don't say apply. The functional $ ( g_{1} - g_{2} ) T $ can only be applied to vectors $ v \in V $. Otherwise, the proof seems fine.


Your proof looks fine to me, but it's possible to give a simpler argument.

Suppose $T\colon V\to W$ is invertible with inverse $S\colon W\to V$. I claim that $T^t\colon W^*\to V^*$ is invertible with inverse $S^t\colon V^*\to W^*$. Indeed, for any $f\in V^*$, $T^t(S^t(f)) = T^t(fS) = fST = f$ since $ST = \text{id}_V$, and for any $G\in W^*$, $S^t(T^t(g)) = S^t(gT) = gTS = g$, since $TS = \text{id}_W$.


I think there is a point that must be clarified: does the dimension of $V$ equals the dimension of $W$, or not? If the answer is not then you have to distinguish between left and right inverse.

Let's consider the matrix case for a moment. In this setting the question is: is $T$ a square matrix?

If $T$ is a square matrix then the following are equivalent

  • $T$ is injective
  • $T$ is surjective
  • $T$ is bijective
  • $\det(T)\neq 0$
  • $T$ admits a left inverse, i.e. $\exists \; L$ such that $LT=\mathbb{1}$
  • $T$ admits a right inverse, i.e. $\exists \; R$ such that $TR=\mathbb{1}$

On the other hand, if $T$ is not a square matrix, then $T$ could have a right inverse and not a left one or viceversa. If I'm not wrong we can say

  • $T$ is injective $\iff$ $T$ admits a left inverse
  • $T$ is surjective $\iff$ $T$ admits a right inverse

All of this preamble to say that, in your argument, you are assuming that $T$ has a right inverse, precisely when you say

T is invertible so we have a $T^{-1}$ such that $ \;(g_1-g_2)TT^{-1} = g_1-g_2 $.

Therefore I believe your argument is correct in the case $\dim(V)=\dim(W)$, but it is wrong in the general case. I think in the general setting what holds is the following

$$T \text{ is injective} \iff T^t \text{ is surjective} $$

  • $\begingroup$ Correct me if I am wrong, I believe that a linear transformation $T$ is invertible if and only if it is 1:1 (injective) and onto (surjective). The rank-nullity theorem states that for finite dimensional spaces $V$ and $W$, $rank(T) + nullity(T) = dim(V)$. Since $T$ is injective, $nullity(T)=0$. Therefore $rank(T)= dim(V)$. Since $T$ is onto, $rank(T)= dim(W)$. Therefore, $dim(W)= dim(V)$. This means that by supposing that $T$ is invertible, I am already assuming that $dim(W)= dim(V)$. $\endgroup$ – Student Nov 28 '13 at 9:53
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    $\begingroup$ Yes, I agree with you on the last point! If any map is not injective it cannot be invertible. What I'm saying is that you can have an injective map which is NOT surjective! In your first comment you say "since T is onto, ..." that is the point! $T$ doesn't need to be onto in general $\endgroup$ – Abramo Nov 28 '13 at 16:19
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    $\begingroup$ If by your definition of "invertible" you require $T$ to have both a left and right inverse, and further you require them to be the same matrix $T^{-1}$, then this implies that $T$ is an isomorphism and consequently $V$ and $W$ have the same dimension. In this setting your prove is correct, but it works only in the case $V$ and $W$ have the same dimension. $\endgroup$ – Abramo Nov 29 '13 at 19:34
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    $\begingroup$ But notice, for example, that the linear map $T:R\to R^2$ sending $e_1$ to $e_1$ is also invertible (or injective) but it not an isomorphism. Indeed $T$ is not surjective and of course there exists a left inverse $L$ such that $LT=1$, not a right inverse $R$ such that $TR=1$. I hope you get what I mean $\endgroup$ – Abramo Nov 29 '13 at 19:38
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    $\begingroup$ You are welcome! A +1 on the answer would be appreciated ;-) $\endgroup$ – Abramo Nov 30 '13 at 8:41

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