Question

Does the equation$\mathit{r}\mathit{a}\mathit{n}\mathit{k}\left(A\right)\mathbf{=}\mathit{r}\mathit{a}\mathit{n}\mathit{k}\left(A^{T}A\right)$ holds for all$\mathit{n}\mathbf{\times }\mathit{m}$ matrices A? Explain.

Short answer

The matrix B is${\left(A^{T}A\right)}^{-1}{A}^T$ .

Step-by-step solution

Determine the value of rank{A} and rank{ATA} .

Consider a $n\times m$matrix A where $ker\left(A\right)=\left\{\overrightarrow{v}\in \overline{){\mathrm{ℝ}}^n}A\overrightarrow{v}=0\right\}andIm\left(A\right)=\left\{\overrightarrow{w}\in \overline{){\mathrm{ℝ}}^m}A\overrightarrow{v}=\overrightarrow{w}\right\}$.

By the Rank-Nullity theorem, the sum of the dimensions of the kernel and the image of a matrix is equal to the number of column of that matrix.

As the matrix A is$n\times m$ , then$A^{T}A$ is an$n\times n$ matrix, and therefore the number of their column is the same means.

$$\begin{gathered} dim\left\{ker\left(A\right)\right\}+dim\left\{lm\left(A\right)\right\}=m \\ dim\left\{ker\left(A^{T}A\right)\right\}+dim\left\{lm\left(A^{T}A\right)\right\}=m \end{gathered}$$

Theorem: Dimension of the image.

For a matrix A ,$dim\left(imA\right)=rank\left(A\right)$ .

By the theorem, the value$dim\left\{imA\right\}=rank\left\{A\right\}$ and$dim\left\{im\left(A^{T}A\right)\right\}=rank\left(A^{T}A\right)$ .

Substitute$rank\left\{A\right\}$ for$dim\left\{lmA\right\}$ in the$dim\left\{ker\left(A\right)\right\}+dim\left\{im\left(A\right)\right\}=m$ as follows.

$$\begin{gathered} dim\left\{ker\left(A\right)\right\}+dim\left\{im\left(A\right)\right\}=m \\ rank\left(A\right)+dim\left\{ker\left(A\right)\right\}=m \\ rank\left(A\right)=-dim\left\{ker\left(A\right)\right\} \end{gathered}$$

Substitute$rank\left(A^{T}A\right)$ for$dim\left\{im\left(A^{T}A\right)\right\}$ in the$dim\left\{ker\left(A^{T}A\right)\right\}+dim\left\{im\left(A^{T}A\right)\right\}=m$ as follows.

$$\begin{gathered} dim\left\{ker\left(A^{T}A\right)\right\}+dim\left\{im\left(A^{T}A\right)\right\}=m \\ rank\left(A^{T}A\right)+dim\left\{ker\left(A^{T}A\right)\right\}=m \\ rank\left(A^{T}A\right)=-dim\left\{ker\left(A^{T}A\right)\right\} \end{gathered}$$

Compare the equations$rank\left\{A\right\}=m-dim\left\{ker\left(A\right)\right\}$ and$rank\left(A^{T}A\right)=m-dim\left\{ker\left(A^{T}A\right)\right\}$ as follows.

$rank\left(A\right)=rank\left(A^{T}A\right)$

Hence, the equation$rank\left(A\right)=rank\left(A^{T}A\right)$ for all $n\times m$matrices.