I have been following the text Linear Algebra by Larry Smith and while finding kernels of linear transformation, I discovered the following statement which I have stated in a clean way:
Claim. Let $M$ be a matrix of size $m\times n$. Let $M^{T}$ denote the transpose of the matrix $M$. Let $T: \mathbb{R}^{n}\to\mathbb{R}^{m}$ be the linear transformation whose matrix with respect to the standard bases is $M$ and similarly let $\hat{T}$ be the linear transformation whose matrix with respect to the standard bases is $M^{T}$. Then $\ker T = \text{Span} \{ \hat{T} (\hat{e_{1}}) , \ldots , \hat{T} (\hat{e_{m}}) \} = \text{Im } \hat{T}$ where $\hat{e_1} , \hat{e_2}, \ldots, \hat{e_m}$ is the standard basis of $\mathbb{R}^{m}$.
I tried hard enough to prove it but I couldn't. I know the following things:
$T(x_1 , \ldots , x_n ) = \left( \sum_{j=1}^{n} M_{1j} x_{j} , \ldots , \sum_{j=1}^{n} M_{mj} x_{j} \right) $ and
$\hat{T} (x_1 , \ldots , x_m ) = \left( \sum_{j=1}^{m} M_{j1} x_{j} , \ldots , \sum_{j=1}^{m} M_{jn} x_{j} \right) $.
I tried to show it by double inclusion but it's bit of a huge mess. I need to solve system of linear equations and it doesn't look like the best way to do.
Can I get some hints/ideas on how to prove this?
You’ve likely had trouble proving this because the claim doesn’t hold. Let $$M=\begin{bmatrix}1&0\\0&0\end{bmatrix}.$$ The kernel of $T$ is spanned by $(0,1)^T$, but the image of $\hat T$ is spanned by $(1,0)^T$.
What is true, however, is that the kernel of $T$ is the orthogonal complement (with respect to the usual Euclidean scalar product) of the image of $\hat T$. Similarly, the image of $T$ is the orthogonal complement of the kernel of $\hat T$.
Put in terms of dual vector spaces, the image of $T^*:W^*\to V^*$, the adjoint of a linear map $T:V\to W$, annihilates the kernel of $T$. That is, for every linear functional $\mathbf\alpha\in\operatorname{im}(T^*)$ and every $\mathbf v\in\ker T$, $\mathbf\alpha[\mathbf v]=0$. Similarly, the kernel of $T^*$ annihilates the image of $T$.