Question

Use the formula${\left(imA\right)}^{\mathbf{\perp }}\mathbf{=}\mathit{k}\mathit{e}\mathit{r}\left(A^T\right)$ prove the equation$\mathit{r}\mathit{a}\mathit{n}\mathit{k}\left(A\right)\mathbf{=}\mathit{r}\mathit{a}\mathit{n}\mathit{k}\left(A^T\right)$.

Short answer

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

Step-by-step solution

Determine the value of rank{A} .

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

Theorem: Property of the orthogonal compliment.

Consider a subspace V of${\mathbf{ℝ}}^{\mathbf{n}}{\mathbf{ℝ}}^{\mathbf{n}}$ .

1. The orthogonal complement V- of V is a subspace of ${\mathbf{ℝ}}^{\mathbf{n}}{\mathbf{ℝ}}^{\mathbf{n}}{\mathbf{ℝ}}^{\mathbf{n}}{\mathbf{ℝ}}^{\mathbf{n}}$.
2. The intersection of V and role="math" localid="1660129666065" $V^{\perp }$consist of the zero vector: $V\cap V^{\perp }=\left\{\overrightarrow{0}\right\}$.
3. $dim\left(V\right)+dim\left(V^{\perp }\right)=n$
4. ${\left(V^{\perp }\right)}^T=V$
   

As$lm\left(A\right)$ is subs space of${\mathrm{ℝ}}^m{\mathrm{ℝ}}^m$ , by the theorem$dim\left\{imA\right\}+dim{\left\{im\left(A\right)\right\}}^{\perp }=m$ .

Theorem: Dimension of the image.

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

By the theorem, the value$dim\left\{imA\right\}=rank\left\{A\right\}$ .

Substitute$rank\left\{A\right\}$ for$dim\left\{imA\right\}$ in the$dim\left\{imA\right\}+dim\left\{imA\right\}=m$ as follows.

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

Determine the value of AT .

By the Rank-Nullity theorem, the value of the equation is$dim\left\{ker\left(A^T\right)\right\}+dim\left\{im\left(A^T\right)\right\}=m$.

Theorem: Dimension of the image.

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

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

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

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

As${\left(imA\right)}^{\perp }=ker\left(A^T\right)$ implies$dim\left\{{\left(imA\right)}^{\perp }\right\}=dim\left\{ker\left(A^T\right)\right\}$ , substitute the value$dim\left\{ker\left(A^T\right)\right\}$ for$dim\left\{{\left(imA\right)}^{\perp }\right\}$ in the equation$rank\left\{A\right\}=m-dim{\left\{im\left(A\right)\right\}}^{\perp }$ as follows.

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

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

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

Hence, the equation if$rank\left\{A^T\right\}=rank\left\{A\right\}if{\left(imA\right)}^{\perp }=ker\left(A^T\right)$ .