Question

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

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 columns 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

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

Theorem: Dimension of the image.

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

By the theorem, the value and .

Substitute $rank\left(A{A}^T\right)$for role="math" localid="1660132784152" $dim\left\{im\left(A{A}^T\right)\right\}=rank\left(A{A}^T\right)$in the $dim\left\{im\left(A^{T}A\right)\right\}=rank\left(A^{T}A\right)$as follows.

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

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

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

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

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

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