Question

If Ais any matrix, show that the linear transformation $\mathit{L}\mathbf{(}\overrightarrow{\mathbf{x}}\mathbf{}\mathbf{)}\mathbf{=}\mathit{A}\overrightarrow{\mathbf{x}}$from $\mathbf{im}\mathbf{\left(}{\mathbf{A}}^{\mathbf{T}}\mathbf{\right)}$to $\mathbf{im}\mathbf{\left(}\mathbf{A}\mathbf{\right)}$is an isomorphism. This provides yet another proof of the formula $\mathbf{rank}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{=}\mathbf{rank}\mathbf{\left(}{\mathbf{A}}^{\mathbf{T}}\mathbf{\right)}$.

Short answer

L is an isomorphism.

Step-by-step solution

L is an isomorphism.

Let $v\in im\left(A^T\right)$such that $v\in k\mathrm{e}r\left(L\right)$. Then it has $A^{T}w=v$ for some w. Also $L\left(v\right)=0$implies Av=0.

Thus, it has $A\left(A^{T}w\right)=0$.

If it multiplies both side of this equation by $w^T$from the left, then it gets

$w^{T}A\left(A^T\mathrm{w}\right)=0$

Which implies that $\left(A^T\mathrm{W}\right)-\left(A^T\right(W)=0$.

Then it gets ${||A^{T}W||}^2=0$.

Thus, kernel is Ker(L)=0.

So, L is injective. now it knows that${\left(im\left(A\right)\right)}^{\perp }=ker\left(A^I\right)$for any matrix A. so if it applies this formula to$A^T$then,$(im{\left(A\right))}^{\perp }=\mathrm{ke}r(A)$.

Also, it knows that $R^n=im\left(A^T\right)\oplus \left(im\right(A){)}^{\perp }$.

Which implies that $R^n=im\left(A^T\right)\oplus ker\left(A\right)$.

From the above terms it gets $W=W_1+W_2$.

Where$w_1\in im\left(A^T\right),w_2\in ker\left(A\right)$

Then, it is written as,

$\begin{array}{l}V=Aw=A{w}_1+A{w}_2\\ =A{w}_1+0\\ =A{w}_1\end{array}$

Thus, it gets $v=A{W}_1$, where $w_1\in im\left(A^T\right)$. So,$A{w}_1=L\left(w_1\right)$ and $v\in im\left(L\right)$.

Hence, L is subjective and an isomorphism.