Question

If Ais any matrix with $\mathit{k}\mathit{e}\mathit{r}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{=}\mathbf{\left\{}\overrightarrow{\mathbf{0}}\mathbf{\right\}}$then the matrix $\mathit{A}{\mathit{A}}^{\mathbf{T}}$ represents the orthogonal projection onto the image of A.

Short answer

False

Step-by-step solution

Theorem 

The matrix of an orthogonal projection: The matrix P of the orthogonal projection onto $\mathbf{V}\mathbf{=}\mathbf{\text{span}}\mathbf{\{}{\overrightarrow{\mathbf{u}}}_{\mathbf{1}}{\overrightarrow{\mathbf{u}}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\overrightarrow{\mathbf{u}}}_{\mathbf{m}}\mathbf{\}}$is given by, where $\mathit{Q}\mathbf{=}\mathbf{\{}{\overrightarrow{\mathbf{u}}}_{\mathbf{1}}{\overrightarrow{\mathbf{u}}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\overrightarrow{\mathbf{u}}}_{\mathbf{m}}\mathbf{\}}$ and ${\mathit{Q}}^{\mathbf{T}}$is the transpose of $\mathit{Q}$and $\mathit{Q}$ is an orthonormal basis of V.

Determine whether the statement is true or false

Thus, according to the matrix of an orthogonal projection.

A cannot be any matrix with$\ker \left\{0\right\}=\left\{0\right\}$. So that, A must be orthonormal.

Hence, the statement is false.