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$Im\left(A\right)=\left\{\stackrel{鈫�}{w}鈭�\overline{){\mathrm{鈩�}}^m}A\stackrel{鈫�}{v}=\stackrel{鈫�}{w}\right\}$Consider a $m脳n$matrix A where $ker\left(A\right)=\left\{\stackrel{鈫�}{v}鈭�\overline{){\mathrm{鈩�}}^n}A\stackrel{鈫�}{v}=0\right\}$and .

Theorem: Property of the orthogonal compliment.

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

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

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

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.

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

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

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

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$rank\left\{A^T\right\}=rank\left\{A\right\}if{\left(imA\right)}^{鈯�}=ker\left(A^T\right)$ .