# Let ${^ts}:E\to F$ be the transpose of $s:F\to E$. Show that $\text{Im}(s)\cap \ker (\,^ts)=\{0_E\}$.

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 $$M\in \mathcal{M}_{n,p}(\mathbb{R})$$

• 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. – Batominovski Nov 16 '18 at 17:03
• 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. – Batominovski Nov 16 '18 at 17:05
• I added more informations, but I didn't want to use the inner product at first...but it seems I must do. – 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*}

• 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. – Batominovski Nov 16 '18 at 18:27
• thanks a lot I need some time to understand everything, and I'll be back – Stu Nov 16 '18 at 18:28
• @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 $<. ,.>_F$ – Picaud Vincent Nov 16 '18 at 18:30
• Shorter proof (sorry it is nearly a complete rewrite, but the idea is the same) – 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 $$\text{span}_\mathbb{K}\{e_1+e_2+\ldots+e_p\}=\text{im}(s)\cap\ker(s^\top)\,.$$