Question

If Ais an$\mathbf{n}\mathbf{\times }\mathbf{m}$matrix, is the formula$\mathbf{im}\left(A\right)\mathbf{=}\mathbf{im}\left({\mathrm{AA}}^T\right)$necessarily true? Explain.

Short answer

Yes.

Step-by-step solution

Determine the image property.

Consider a$n\times m$matrix A.

Theorem: For any matrix B, $\mathrm{im}\left(\mathrm{B}\right)={\mathrm{kerB}}^{\perp }$.

By the theorem,$im\left(A\right)={\left(kerA\right)}^{\perp }$.

Theorem: If B is an $n\times m$matrix then, $\ker \left(\mathrm{B}\right)=\ker \left({\mathrm{B}}^{\mathrm{T}}\mathrm{B}\right)$.

By the theorem, $ker\left(A\right)=ker\left(A^{T}A\right)$.

Take $\perp$both side in the equation $ker\left(A\right)=ker\left(A^{T}A\right)$as follows.

$$\begin{gathered} ker\left(A\right)=ker\left(A^{T}A\right) \\ {\left(ker\left(A\right)\right)}^{\perp }={\left(ker\left(A^{T}A\right)\right)}^{\perp } \end{gathered}$$

Theorem: For a subset V of ${\mathrm{ℝ}}^{\mathrm{n}}$then${\left({\mathrm{V}}^{\perp }\right)}^{\perp }=\mathrm{V}$.

By the theorem,localid="1659501115571" ${\left(ker\left(A^{T}A\right)\right)}^{\perp }=im\left(A^{T}A\right)$.

Compare the equations ${\left(ker\left(A^{T}A\right)\right)}^{\perp }=im\left(A^{T}A\right),{\left(ker\left(A\right)\right)}^{\perp }={\left(ker\left(A^{T}A\right)\right)}^{\perp }andim\left(A\right)={\left(kerA\right)}^{\perp }$,and as follows.

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

Hence, for any $n\times m$matrix A the equation $im\left(A^{T}A\right)=im\left(A\right)$holds true.