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Subsection 2.2.2 Component 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 \(a \text{.}\)

Remark 2.2.2.1.

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 \(a \text{.}\)

Homework 2.2.2.1.

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.

Answer

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.

Homework 2.2.2.2.

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.

Answer

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.

Homework 2.2.2.3.

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: \(\widehat b^H b^\perp = 0 \text{.}\)

Answer

ALWAYS.

Now prove it.

Solution
\begin{equation*} \begin{array}{l} \widehat b^H b^\perp \\ ~~~=~~~~ \lt \mbox{ substitute } \widehat b \mbox{ and } b^\perp \gt \\ \left( \frac{a a^H}{a^H a} b \right)^H ( b - \widehat b ) \\ ~~~=~~~~ \lt ( A x )^H = x^H A^H; \mbox{ substitute } b-\widehat b \gt \\ b^H \left(\frac{a a^H}{a^H a}\right)^H ( I - \frac{a a^H}{a^Ha}) b \\ ~~~=~~~~\lt ( ( ( x y^H ) / \alpha )^H = y x^H / \alpha \mbox{ if } \alpha \mbox{ is real } \gt \\ b^H \frac{a a^H}{a^H a}( I - \frac{a a^H}{a^Ha}) b \\ ~~~=~~~~ \lt \mbox{ last homework } \gt \\ b^H 0 b \\ ~~~=~~~~ \lt 0 x = 0 ; y^H 0 = 0 \gt \\ 0. \end{array} \end{equation*}
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