Question

For which n×mmatrices Adoes the equation

$\mathit{d}\mathit{i}\mathit{m}\left(ker\left(A\right)\right)\mathbf{=}\mathit{d}\mathit{i}\mathit{m}\left(ker\left(A^T\right)\right)$hold? Explain.

Short answer

$dimker\left(A\right)=dimker\left(A^T\right)\iff m=n$, i.e., the matrix A is square matrix.

Step-by-step solution

A is an n × m matrices.

It wants to find the matrices of size n× m such that

$dimker\left(A\right)=dimker\left(A^T\right)$

Observe that for any n× m matrices A it has

$dim\left(imA\right)=rank\left(A\right)$ ..…. (1)

According to the rank nullity theorem

$dim\left(kerA\right)=dim\left(im\left(A\right)\right)=n$ …… (2)

And ${\left(AB\right)}^T=B^T{A}^T$ for all m× pmatrices Aand for all p× nmatrices B.

$rank\left(A\right)=rank\left(A^T\right)$ …… (3)

Since, A is an n× mmatrix, so $A^T$is m x n ,

From the equation (2) and (3) it gives,

$$\begin{gathered} dim\left(ker{A}^T\right)+rank\left(A^T\right)=n \\ \Rightarrow dim\left(ker{A}^T\right)=n-rank\left(A^T\right) \\ \Rightarrow dim\left(kerA\right)=m-rank\left(A\right) \\ \mathrm{Further}\mathrm{solve}\mathrm{above}\mathrm{expression}, \\ \dim \ker \left(\mathrm{A}\right)=\mathrm{n}-\mathrm{rank}\left({\mathrm{A}}^{\mathrm{T}}\right) \\ \Rightarrow \mathrm{m}-\mathrm{rank}\left(\mathrm{A}\right)=\mathrm{n}-\mathrm{rank}\left({\mathrm{A}}^{\mathrm{T}}\right) \\ \Rightarrow \mathrm{m}-\mathrm{rank}\left(\mathrm{A}\right)=\mathrm{n}-\mathrm{rank}\left(\mathrm{A}\right) \\ \Rightarrow \mathrm{m}=\mathrm{n} \end{gathered}$$

Hence, $dimker\left(A\right)=dimker\left(A^T\right)\iff m=n$, i.e., the matrix A will be square matrix.