## Subsection2.2.2Component in the direction of a vector

In a previous linear algebra course, you may have learned that if $a, b \in \Rm$ then

\begin{equation*} \widehat b = \frac{a^T b}{a^T a} a = \frac{a a^T}{a^T a} b \end{equation*}

equals the component of $b$ in the direction of $a$ and

\begin{equation*} b^\perp = b - \widehat b = ( I - \frac{a a^T}{a^T a}) b \end{equation*}

equals the component of $b$ orthogonal to $a \text{,}$ since $b = \widehat b + b^\perp$ and $\widehat b^T b^\perp = 0 \text{.}$ Similarly, if $a, b \in \Cm$ then

\begin{equation*} \widehat b= \frac{a^H b}{a^H a} a = \frac{a a^H}{a^H a} b \end{equation*}

equals the component of $b$ in the direction of $a$ and

\begin{equation*} b^\perp = b - \widehat b = ( I - \frac{a a^H}{a^H a}) b \end{equation*}

equals the component of $b$ orthogonal to $a \text{.}$

###### Remark2.2.2.1.

The matrix that (orthogonally) projects the vector to which it is applied onto the vector $a$ is given by

\begin{equation*} \frac{a a^H}{a^H a} \end{equation*}

while

\begin{equation*} I - \frac{a a^H}{a^H a} \end{equation*}

is the matrix that (orthogonally) projects the vector to which it is applied onto the space orthogonal to the vector $a \text{.}$

###### Homework2.2.2.1.

Let $a \in \mathbb C^m \text{.}$

ALWAYS/SOMETIMES/NEVER>:

\begin{equation*} \left( \frac{a a^H}{a^H a} \right) \left( \frac{a a^H}{a^H a} \right) = \frac{a a^H}{a^H a} . \end{equation*}

Interpret what thi s means about a matrix that projects onto a vector.

ALWAYS.

Now prove it.

Solution
\begin{equation*} \begin{array}{l} \left( \frac{a a^H}{a^H a} \right) \left( \frac{a a^H}{a^H a} \right) \\ ~~~=~~~~\lt \mbox{ multiply numerators and denominators } \gt \\ \frac{a a^H a a^H }{(a^H a) ( a^H a)} \\ ~~~=~~~~\lt \mbox{ associativity } \gt \\ \frac{a (a^H a) a^H }{(a^H a) ( a^H a)} \\ ~~~=~~~~ \lt a^H a \mbox{ is a scalar and hence commutes to front} \gt \\ \frac{a^H a a a^H }{(a^H a)( a^H a) } \\ ~~~=~~~~ \lt \mbox{ scalar division } \gt \\ \frac{a a^H }{a^H a}. \end{array} \end{equation*}

Interpretation: orthogonally projecting the orthogonal projection of a vector yields the orthogonal projection of the vector.

###### Homework2.2.2.2.

Let $a \in \mathbb C^m \text{.}$

ALWAYS/SOMETIMES/NEVER:

\begin{equation*} \left( \frac{a a^H}{a^H a} \right) \left( I - \frac{a a^H}{a^H a} \right) = 0 \end{equation*}

(the zero matrix). Interpret what this means.

ALWAYS.

Now prove it.

Solution
\begin{equation*} \begin{array}{l} \left( \frac{a a^H}{a^H a} \right) \left( I - \frac{a a^H}{a^H a} \right) \\ ~~~=~~~~\lt \mbox{ distribute } \gt \\ \left( \frac{a a^H}{a^H a} \right) - \left( \frac{a a^H}{a^H a} \right) \left( \frac{a a^H}{a^H a} \right) \\ ~~~=~~~~\lt \mbox{ last homework } \gt \\ \left( \frac{a a^H}{a^H a} \right) - \left( \frac{a a^H}{a^H a} \right) \\ ~~~ = \\ 0. \end{array} \end{equation*}

Interpretation: first orthogonally projecting onto the space orthogonal to vector $a$ and then orthogonally projecting the resulting vector onto that $a$ leaves you with the zero vector.

###### Homework2.2.2.3.

Let $a, b \in \Cn \text{,}$ $\widehat b = \frac{a a^H}{a^H a} b \text{,}$ and $b^\perp = b-\widehat b \text{.}$

ALWAYS/SOMETIMES/NEVER: $\widehat b^H b^\perp = 0 \text{.}$