Question

Describe the kernel and range of
(a) the orthogonal projection on the xz -plane
(b) the orthogonal projection on the yz-plane
(c) the orthogonal projection on the plane defined by the equation y=x

Answer 1

(a) The linear transformation is\[T(x,y,z) = (x, 0, z).\]\[\ker{T} = \left\{(x,y,z) \in \mathbb{R}^3\ \colon\ (x, 0, z) = (0, 0, 0)\right\} = \left\{(0, y, 0)\ \colon\ y\in \mathbb{R}\right\}\]which is the Y axis.
\[\mathop{\mathrm{im}}{T}=\left\{(x, 0, z) \in \mathbb{R}^3\ \colon\ x, y \in \mathbb{R}\right\}\]which is the XZ plane.
(b) The linear transformation is\[T(x,y,z) = (0, y, z).\]\[\ker{T} = \left\{(x,y,z) \in \mathbb{R}^3\ \colon\ (0, y, z) = (0, 0, 0)\right\} = \left\{(x, 0, 0)\ \colon\ x\in \mathbb{R}\right\}\]which is the X axis.
\[\mathop{\mathrm{im}}{T}=\left\{(0, y, z) \in \mathbb{R}^3\ \colon\ y, z \in \mathbb{R}\right\}\]which is the YZ plane.
(c) The linear tranformation is\[T(x, y, z) = \left(\frac{1}{2}(x+y),\frac{1}{2}(x+y),z\right).\]\begin{eqnarray*}\ker{T} &=& \left\{(x,y,z)\in\mathbb{R}^3\ \colon\ \left(\frac{1}{2}(x+y),\frac{1}{2}(x+y),z\right) = (0,0,0)\right\}\\&=&\{(x,y,0)\ \colon\ x = -y\}\\&=&\{(x, -x, 0)\ \colon\ x \in \mathbb{R}\}\end{eqnarray*}which is the subspace generated by a normal vector of the plane.\[\mathrm{\mathop{im}}{T} =\left\{\left(\frac{1}{2}(x+y),\frac{1}{2}(x+y),z\right) \in \mathbb{R}^3\ \colon\ x,y,z\in \mathbb{R}\right\} = \{(x,x,z)\ \colon\ x,z\in\mathbb{R}\}\]which is the plane defined by the equation\[y=x.\]