MAT223: 5_Projections

Projections and Vector Components §

Projection §

• Projection - Let $X$ be a set, the projection of the vector $\vec{v}$ onto $X$, written ${\text{proj}}_X\vec{v}$ is the closest point in $X$ to $\vec{v}$

  • Intuitively, ${\text{proj}}_{xy}\vec{v}$ is the “shadow” that $\vec{v}$ would on ${\text{proj}}_{xy}$ if the sun were perpendicular to $X$

• If there are two points other as the closest point to $\vec{v}$, OR there are no closest point to $\vec{v}$, then ${\text{proj}}_X\vec{v}$ is undefined

• For a fixed set $X$, $P(\vec{v})={\text{proj}}_X\vec{v}$ is a function that inputs and outputs vectors. The domain of this function is exactly the vectors is which $P(\vec{v})$ is defined.

  • If $X$ is a line or a plane in ${\mathbb{R}}^n$, then the domain of $P(\vec{v})$ is all of ${\mathbb{R}}^n$ (if projection exist)

  Let $\vec{v}=(1,2,3)$, ${\text{proj}}_{xy}\subseteq {\mathbb{R}}^3$ be the xy-plane, ${\ell }_y\subseteq {\mathbb{R}}^3$ by the y-axis §

  Intuitively, or by the picture, it can be concluded that ${\text{proj}}_{P_{xy}}\vec{v}=\left[\begin{array}{c}1\\ 2\\ 0\end{array}\right]$

  By definition, every vector in ${\ell }_y$ takes the form ${\vec{u}}_t=(0,t,0)$ for some $t\in \mathbb{R}$. The distance between ${\vec{u}}_t$ and $\vec{v}$ is

  $$\Vert {\vec{u}}_t-\vec{v}\Vert =\Vert \left[\begin{array}{c}0\\ t\\ 0\end{array}\right]-\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\Vert =\sqrt{1^2+(t-2{)}^2+3^2}$$

  Since $(t-2{)}^2$ would be always position, this whole expression would be minimized when $(t-2{)}^2=0$, thus when $t=2$. Therefore ${\vec{u}}_2$ is the closest vector in ${\ell }_y$ to $\vec{v}$.

  $${\text{proj}}_{{\ell }_y}\vec{v}={\vec{u}}_2=\left[\begin{array}{c}0\\ 2\\ 0\end{array}\right]$$

• If $X$ is a line or plane and $\vec{v}\notin X$ is a vector, then $\vec{v}-{\text{proj}}_X\vec{v}$ is a normal vector for $X$

  • When projecting onto lines and planes, right angles appear in key places

  Let $\ell \subseteq {\mathbb{R}}^2$ be the line given in vector form by $\vec{x}=t\left[\begin{array}{c}1\\ 1\end{array}\right]+\left[\begin{array}{c}3\\ -2\end{array}\right]$, and let $\vec{v}=\left[\begin{array}{c}-1\\ -1\end{array}\right]$. Use the fact that $\vec{v}-{\text{proj}}_{\ell }\vec{v}$ is a normal vector to $\ell$ to find ${\text{proj}}_{\ell }\vec{v}$ §

  Since $\vec{v}-{\text{proj}}_{\ell }\vec{v}$ is a normal vector to $\ell$, then its orthogonal to $\vec{d}=\left[\begin{array}{c}1\\ 1\end{array}\right]$. Let $\left[\begin{array}{c}x\\ y\end{array}\right]={\text{proj}}_{\ell }\vec{v}$ for some unknown $x,y\in \mathbb{R}$, get:

  $$(\vec{v}-{\text{proj}}_{\ell }\vec{v})\cdot \vec{d}=\left(\left[\begin{array}{c}-1\\ -1\end{array}\right]-\left[\begin{array}{c}x\\ y\end{array}\right]\right)\cdot \left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}-1-x\\ -1-y\end{array}\right]\cdot \left[\begin{array}{c}1\\ 1\end{array}\right]=-2-x-y=0$$

  That is $x+y=-2$ (*1)

  Since ${\text{proj}}_{\ell }\vec{v}\in \ell$ then this means:

  $${\text{proj}}_{\ell }\vec{v}=\left[\begin{array}{c}x\\ y\end{array}\right]=t\left[\begin{array}{c}1\\ 1\end{array}\right]+\left[\begin{array}{c}3\\ -2\end{array}\right]=\left[\begin{array}{c}t+3\\ t-2\end{array}\right]$$

  Combining with equation (*1) get the system of equations:

  $$\{\begin{array}{cc}x+y& =-2\\ -t+x& =3\\ -t+y& =-2\end{array}$$

  Solving the system, to get $x,y$, the value of $t$ does not matter. Therefore ${\text{proj}}_{\ell }\vec{v}=\left[\begin{array}{c}3/2\\ -7/2\end{array}\right]$

Projection onto sets §

let $\mathcal{T}\subseteq {\mathbb{R}}^2$ be a filled in triangle with verticals $(0,0),(1,0),(0,2)$ and let $\vec{a}=(1/4,1/4),\vec{b}=(1,1),\vec{c}=(3,1/2)$ Find ${\text{proj}}_{\mathcal{T}}\vec{a},{\text{proj}}_{\mathcal{T}}\vec{b},{\text{proj}}_{\mathcal{T}}\vec{c}$ §

(Pic 1) (Pic2)

From the picture, $\vec{a}\in {\text{proj}}_{\mathcal{T}}$ so ${\text{proj}}_{\mathcal{T}}\vec{a}=\vec{a}$.

Then, $\vec{b}$ is closest to the hypotenuse of $\mathcal{T}$, so ${\text{proj}}_{\mathcal{T}}\vec{b}$ is the same as projection of $\vec{b}$ onto the line $y=-2x+2$, thus by doing calculations for the normal of this line, ${\text{proj}}_{\mathcal{T}}\vec{b}=\left[\begin{array}{c}3/5\\ 4/5\end{array}\right]$

Lastly, for $\vec{c}$, draw some concentric cycles (pic2), finds that it is closes to the lower-right corner, so ${\text{proj}}_{\mathcal{T}}\vec{c}=\left[\begin{array}{c}1\\ 0\end{array}\right]$

Vector Components §

• Let $\vec{u}$ and $\vec{v}\ne \vec{0}$ be vectors, the vector component of $\vec{u}$ in the $\vec{v}$ direction, written \text{vcomp}_\vec{v} \vec{u} is a vector in the direction of $\vec{v}$ so that \vec{u} - \text{vcomp}_\vec{v} \vec{u} is orthogonal to $\vec{v}$ \vec{u} = \text{vcomp}_\vec{v} \vec{u} + (\vec{u} - \text{vcomp}_\vec{v} \vec{u})

• Since by the definition, \text{vcomp}_\vec{v} \vec{u} is a vector in the direction of $\vec{v}$, then: \text{vcomp}_\vec{v} \vec{u} = k\vec{v}

• Since, \vec{u} - \text{vcomp}_\vec{v} \vec{u} is orthogonal to $\vec{v}$ then: \vec{v} \cdot (\vec{u} - \text{vcomp}_\vec{v} \vec{u}) = \vec{v} \cdot (\vec{u} - k\vec{v}) = \vec{v} \cdot \vec{u} - k\vec{v} \cdot \vec{v} = 0

  • Since $\vec{v}\ne \vec{0}$ then $\vec{v}\cdot \vec{v}\ne 0$, solve for $k=\frac{\vec{v}\cdot \vec{u}}{\vec{v}\cdot \vec{v}}$, put it tougher:
    \text{vcomp}_\vec{v} \vec{u} = \left(\frac{\vec{v} \cdot \vec{u}}{\vec{v} \cdot \vec{v}} \right) \vec{v}

For vectors $\vec{u}$ and $\vec{v}\ne \vec{0}$, then \text{proj}_{\text{span}\{\vec{v}\}} \vec{u} = \text{vcomp}_\vec{v} \vec{u} §

• When projection onto the span of a single vector, vector components can be a computational shortcut, but if the set isn’t a span, then this shortcut cannot work

Compute the projection of $\vec{a}=\left[\begin{array}{c}3\\ 7\end{array}\right]$ onto $\ell =\text{span}\left\{\left[\begin{array}{c}1\\ -4\end{array}\right]\right\}$ §

Let $\vec{b}=\left[\begin{array}{c}1\\ -4\end{array}\right]$ since $\ell =\text{span}\{\vec{b}\}$ and $\vec{b}\ne \vec{0}$, using vector components as a shortcut

\text{proj}_\ell \vec{a} = \text{vcomp}_\vec{b} \vec{a} = \left(\frac{\vec{b} \cdot \vec{a}}{\vec{b} \cdot \vec{b}} \right) \vec{b} = \frac{3-28}{1+16} \begin{bmatrix} 1 \\ -4 \end{bmatrix} = \begin{bmatrix} -25/17 \\ 100/17 \end{bmatrix}