Usbrth

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 $mtimes n$. Let $M^T$ denote the transpose of the matrix $M$. Let $T: mathbbR^ntomathbbR^m$ be the linear transformation whose matrix with respect to the standard bases is $M$ and similarly let $hatT$ be the linear transformation whose matrix with respect to the standard bases is $M^T$. Then $ker T = textSpan hatT (hate_1) , ldots , hatT (hate_m) = textIm hatT$ where $hate_1 , hate_2, ldots, hate_m$ is the standard basis of $mathbbR^m$.

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

$hatT (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 $$M=beginbmatrix1&0\0&0endbmatrix.$$ 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 $T:Vto W$, annihilates the kernel of $T$. That is, for every linear functional $mathbfalphainoperatornameim(T^*)$ and every $mathbf vinker T$, $mathbfalpha[mathbf v]=0$. Similarly, the kernel of $T^*$ annihilates the image of $T$.