Let $s\in \mathcal{L}(F,E)$

$$\displaystyle F \overset{s}{\longrightarrow} E\overset{^ts}{\longrightarrow} F$$ I spent two days to show that :$$\text{Im}(s)\cap \ker (\,^ts)=\{0_E\}\qquad \tag{1}$$

I'm not sure that is right, I tried with specific matrix (3x3), and it works.

But when I want to provide a general proof, I struggle.

I used the Annihilator but no way to find a solution, may be this statement is wrong...?

I add more information about $E$ and $F$ regarding the comments

$F=\mathbb{R}^p$ and $E=\mathbb{R}^n$ with $p\le n$

Or $F=(\mathbb{R}^p,\langle\cdot,\cdot\rangle)$ and $E=(\mathbb{R}^n,\langle\cdot,\cdot\rangle)$

The matrix associated at $u$ is $M$ and $M\in \mathcal{M}_{n,p}(\mathbb{R})$

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    $\begingroup$ I think the problem is not clear. What is the base field here? Presumably, you have nondegenerate bilinear forms on $E$ and $F$. However, not all bilinear forms work. If the base field is of a positive characteristic, then this is false. Even when the characteristic of the base field is $0$, you can come up with a nondegenerate symmetric bilinear form that contradicts the claim. $\endgroup$ – Batominovski Nov 16 '18 at 17:03
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    $\begingroup$ So, I would like the OP to confirm whether: (1) the base field is $\mathbb{R}$, (2) $E$ and $F$ are finite-dimensional vector spaces, and (3) the vector spaces are equipped with positive-definite symmetric bilinear forms. $\endgroup$ – Batominovski Nov 16 '18 at 17:05
  • $\begingroup$ I added more informations, but I didn't want to use the inner product at first...but it seems I must do. $\endgroup$ – Stu Nov 16 '18 at 17:23

It is sufficient to show that $$ \ker s^t\subset(\text{Im } s)^\perp $$ because we can then conclude by: $$ \ker s^t \cap \text{Im}\ s \subset (\text{Im}\ s)^\perp \cap \text{Im}\ s =\{0_E\} $$

The claim $\ker s^t\subset(\text{Im } s)^\perp$ can be proven as follows:

\begin{align*} e\in\ker s^t &\Rightarrow s^t(e)=0_F \\ &\Rightarrow \forall f\in F,\ \langle f,s^t(e) \rangle_F=0 \\ &\Rightarrow \forall f\in F,\ \langle s(f),e\rangle_E=0 \\ &\Rightarrow \forall f\in F,\ s(f) \perp e \\ &\Rightarrow e \in (\text{Im}\ s)^\perp \end{align*}

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    $\begingroup$ I would like to note that the equality $\ker (s^\top)=\big(\text{im}(s)\big)^\perp$ is true for any nondegenerate symmetric bilinear form $\langle\_,\_\rangle_E$. However, the last equality $\big(\text{im}(s)\big)^\perp\cap\text{im}(s)=\{0_E\}$ must rely on some strong conditions on $\langle\_,\_\rangle_E$ such as positive-definiteness. $\endgroup$ – Batominovski Nov 16 '18 at 18:27
  • $\begingroup$ thanks a lot I need some time to understand everything, and I'll be back $\endgroup$ – Stu Nov 16 '18 at 18:28
  • $\begingroup$ @Batominovski you are right, and that is the reason why I simply wrote positive definite right at the beginning. Also a more general result would also hold only using duality bracket $<. ,.>_{E\times E^*}$ and $<. ,.>_{F\times F^*}$ (without introducing scalar products $<. ,.>_E$ and $<. ,.>_F$ $\endgroup$ – Picaud Vincent Nov 16 '18 at 18:30
  • $\begingroup$ Shorter proof (sorry it is nearly a complete rewrite, but the idea is the same) $\endgroup$ – Picaud Vincent Nov 16 '18 at 21:39

In order that the transpose is a map $F\to E$ you need some assumptions: both vector spaces need to be equipped with a nondegenerate bilinear form, so that we can define isomorphisms $E\to E^*$ and $F\to F^*$ (the dual spaces), assuming finite dimensionality.

In particular, if the base field is $\mathbb{R}$, the two space might be inner product spaces.

If $\langle{\cdot},{\cdot}\rangle_E$ is the form on $E$, then for each $v\in E$, the map $$ e_v\colon E\to K,\qquad e_v(x)=\langle v,x\rangle_E $$ is an element of $E^*$. If $v\ne0$, then nondegeneracy implies there exists $x\in E$ with $\langle v,x\rangle_E\ne0$, so $e\colon E\to E^*$, $v\mapsto e_v$ is an isomorphism (it is clear from bilinearity that $e_v$ is linear, for every $v\in E$, and $e$ is linear as well).

Then the transpose ${}^{t\!}s$ is the dual map composed with these isomorphisms: $$ {}^{t\!}s=e^{-1}\circ s^*\circ e $$ (I denote by $e$ both maps $E\to E^*$ and $F\to F^*$, no confusion should arise).

If $w=s(v)$ and ${}^{t\!}s(w)=0$, then also $s^*\circ e_w=0$, which means $e_w\circ s=0$, that is, $$ e_w(s(x))=0\quad\text{for all $x\in V$} $$ hence $$ \langle w,s(x)\rangle_F=0 $$ In particular, $\langle s(v),s(v)\rangle_F=0$, so by nondegeneracy, $s(v)=0$ and $w=0$.


If you represent a linear map by a matrix then the image of the map is the sub-space spanned of the columns of the matrix. And the kernel of the map is the sub-space that is perpendicular to the rows of the matrix (because the inner product of each row with any vector in the kernel must be zero).

So the image of $s$ is the sub-space that is spanned by the columns of the matrix representing $s$ (relative to specific bases in $E$ and $F$).

And the kernel of $^ts$ is the sub-space that is perpendicular to the sub-space spanned by the rows of the matrix representing $^ts$.

But the columns of the first matrix are the rows of the second matrix. The result follows.


Let $s:F\to E$ be a linear transformation from a vector space $F$ to another vector space $E$ over the base field $\mathbb{R}$. Both $E$ and $F$ are equipped with nondegenerate symmetric bilinear forms $\langle\_,\_\rangle_E$ and $\langle\_,\_\rangle_F$, respectively. Suppose further that $\langle\_,\_\rangle_E$ is positive-definite. Write $s^\top:E\to F$ for the transpose of $s$. We claim that $$\text{im}(s)\cap\ker(s^\top)=\{0_E\}\,.$$

Recall that $s^\top$ is the unique linear transformation from $E$ to $F$ such that $$\big\langle s(x),y\big\rangle_E=\big\langle x,s^\top(y)\rangle_F$$ for all $x\in F$ and $y\in E$. Suppose that $y\in\text{im}(s)\cap\ker(s^\top)$. Then, $y\in\text{im}(s)$, so $y=s(x)$ for some $x\in F$. Sinc $y\in\ker(s^\top)$, we get $$s^\top\big(s(x)\big)=s^\top(y)=0_F\,.$$ Therefore, $$\Big\langle x,s^\top\big(s(x)\big)\Big\rangle_F=\big\langle x,0_F\big\rangle_F=0\,.$$ By the definition of $s^\top$, we have $$\langle y,y\rangle_E=\big\langle s(x),s(x)\big\rangle_E=\Big\langle x,s^\top\big(s(x)\big)\Big\rangle_F=0\,.$$ Since $\langle\_,\_\rangle_E$ is positive-definite, we have $y=0_E$. This proves the claim.

P.S.: If the ground field is $\mathbb{C}$, we instead assume that $\langle\_,\_\rangle_E$ and $\langle\_,\_\rangle_F$ are nondegenerate Hermitian forms. The transpose map has to be replaced by the Hermitian conjugate. For the claim to hold, we need to assume further that $\langle\_,\_\rangle_E$ is positive-definite.

If the ground field $\mathbb{K}$ is a field of characteristic $0$, but there is no positive-definiteness assumption on $\langle\_,\_\rangle_E$ (which makes no sense when $\mathbb{K}$ is not a subfied of $\mathbb{R}$ anyhow), then the claim is not true. For example, let $\mathbb{K}:=\mathbb{R}$, $E:=\mathbb{R}^2$, and $F:=\mathbb{R}^2$. Then we set $e_1:=(1,0)$, $e_2:=(0,1)$. Let $\langle\_,\_\rangle$ be the nondegenerate symmetric bilinear form on both $E$ and $F$ given by $$\langle a_1e_1+a_2e_2,b_1e_1+b_2e_2\rangle:=a_1b_1-a_2b_2$$ for all $a_1,a_2,b_1,b_2\in\mathbb{R}$. Take $s:F\to E$ to be the linear map sending both $e_1$ and $e_2$ to $e_1+e_2$. That is, $s$ is given by the matrix $$\begin{bmatrix}1&1\\1&1\end{bmatrix}$$ with respect to the ordered basis $(e_1,e_2)$ of both $F$ and $E$. Then, $s^\top:E\to F$ is given by the matrix $$\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$$ with respect to the ordered basis $(e_1,e_2)$ of both $E$ and $F$. This shows that $$\text{im}(s)=\text{span}_\mathbb{R}\{e_1+e_2\}=\ker(s^\top)\,.$$

The claim fails in positive characteristics. If the ground field $\mathbb{K}$ is of a positive characteristric $p$, then take $E$ and $F$ to be both $\mathbb{K}^p$. Fix a basis $\{e_1,e_2,\ldots,e_p\}$ of both $E$ and $F$. Equip $E$ and $F$ with the nondegenerate symmetric bilinear form $\langle\_,\_\rangle$ given by $$\left\langle\sum_{i=1}^p\,a_ie_i,\sum_{i=1}^p\,b_ie_i\right\rangle:=\sum_{i=1}^p\,a_ib_i$$ for all $a_1,a_2,\ldots,a_p,b_1,b_2,\ldots,b_p\in\mathbb{K}$. If the map $s:F\to E$ is the $\mathbb{K}$-linear map sending $e_i\mapsto e_1+e_2+\ldots+e_p$ for all $i=1,2,\ldots,p$, then its transpose $s^\top:E\to F$ also sends $e_i\mapsto e_1+e_2+\ldots+e_p$ for all $i=1,2,\ldots,p$. In particular, this shows that $$\text{span}_\mathbb{K}\{e_1+e_2+\ldots+e_p\}=\text{im}(s)\cap\ker(s^\top)\,.$$


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