Chapter 6.1, Problem 13PS

To determine

(a) To Show:

   $u$ Is an eigenvector with $\lambda =1$.

Given information:

   $u=\left(\frac{1}{6},\frac{1}{6},\frac{3}{6},\frac{5}{6}\right)$

Concept used:

Transpose of matrix means row and column of matrix gets interchanged.

Proof:

   $A$ Projection matrix $P$ with rank one is given as

   $\begin{array}{c}P=u{u}^T\\ =\left(\begin{array}{l}\frac{1}{3}\\ \frac{1}{6}\\ \frac{3}{6}\\ \frac{5}{6}\end{array}\right)\left(\frac{1}{3}\frac{1}{6}\frac{3}{6}\frac{5}{6}\right)\\ =\left(\begin{array}{cccc}\frac{1}{36}& \frac{1}{36}& \frac{3}{36}& \frac{5}{36}\\ \frac{1}{36}& \frac{1}{36}& \frac{3}{36}& \frac{5}{36}\\ \frac{1}{36}& \frac{1}{36}& \frac{9}{36}& \frac{15}{36}\\ \frac{5}{36}& \frac{5}{36}& \frac{15}{36}& \frac{25}{36}\end{array}\right)\end{array}$

So, we get the matrix $P$ as above.

We will show that $u$ is an eigenvector of matrix $P$.

We will simplify the vector $Pu$ as shown below.

   $\begin{array}{c}Pu=\left(u{u}^T\right)u\\ =u\left(u^{T}u\right)\left(\text{As }u\text{ is a unit vector, so }{u}^{T}u=1\right)\\ =u\end{array}$

So, we get $Pu=u$. From these we get that $u$ is an eigenvector of $P$ with eigenvalue as $1$.

Conclusion:

   $u$ is a eigen vector with $\lambda =1$

To determine

(b) To show:

If $\upsilon$ is perpendicular to $\upsilon$ show that $P\upsilon =0$. Then $\lambda =0$.

Given information:

   $u=\left(\frac{1}{6},\frac{1}{6},\frac{3}{6},\frac{5}{6}\right)$

Concept used:

Transpose of matrix means row and column of matrix gets interchanged.

Proof:

   $A$ Projection matrix $P$ with rank one is given as

   $\begin{array}{c}P=u{u}^T\\ =\left(\begin{array}{l}\frac{1}{6}\\ \frac{1}{6}\\ \frac{3}{6}\\ \frac{5}{6}\end{array}\right)\left(\frac{1}{6}\frac{1}{6}\frac{3}{6}\frac{5}{6}\right)\\ =\left(\begin{array}{cccc}\frac{1}{36}& \frac{1}{36}& \frac{3}{36}& \frac{5}{36}\\ \frac{1}{36}& \frac{1}{36}& \frac{3}{36}& \frac{5}{36}\\ \frac{3}{36}& \frac{3}{36}& \frac{9}{36}& \frac{15}{36}\\ \frac{5}{36}& \frac{5}{36}& \frac{15}{36}& \frac{25}{36}\end{array}\right)\end{array}$

So, we get the matrix $P$ as above.

We will simplify the vector $Pu$ as shown below.

   $\begin{array}{c}Pu=\left(u{u}^T\right)u\\ =u\left(u^{T}u\right)\left(\text{As }u\text{ is perpendicular to }u\text{, so }{u}^{T}v=1\right)\\ =0\end{array}$

So, we get $Pv=0$. From these we get that $v$ is an eigenvector of $P$ with eigenvalue as $0$.

Conclusion:

If $\upsilon$ is perpendicular to $\upsilon$ show that $P\upsilon =0$. Then $\lambda =0$.

To determine

(c) To determine:

Find three independent eigenvectors of $P$ all with eigenvalue $\lambda =0$.

Eigenvectors as ${\left(-1,1,0,0\right)}^T,{\left(-3,0,1,0\right)}^T\text{ and }{\left(-5,0,0,1\right)}^T$.

Given information:

   $u=\left(\frac{1}{6},\frac{1}{6},\frac{3}{6},\frac{5}{6}\right)$

Concept used:

Transpose of matrix means row and column of matrix gets interchanged.

Proof:

   $A$ Projection matrix $P$ with rank one is given as

   $\begin{array}{c}P=u{u}^T\\ =\left(\begin{array}{l}\frac{1}{6}\\ \frac{1}{6}\\ \frac{3}{6}\\ \frac{5}{6}\end{array}\right)\left(\frac{1}{6}\frac{1}{6}\frac{3}{6}\frac{5}{6}\right)\\ =\left(\begin{array}{cccc}\frac{1}{36}& \frac{1}{36}& \frac{3}{36}& \frac{5}{36}\\ \frac{1}{36}& \frac{1}{36}& \frac{3}{36}& \frac{5}{36}\\ \frac{3}{36}& \frac{3}{36}& \frac{9}{36}& \frac{15}{36}\\ \frac{5}{36}& \frac{5}{36}& \frac{15}{36}& \frac{25}{36}\end{array}\right)\end{array}$

So, we get the matrix $P$ as above.

To get eigenvalues, we will solve the equation $Pv=0$.

   $Pv=0$

   $=\left(\begin{array}{cccc}\frac{1}{36}& \frac{1}{36}& \frac{3}{36}& \frac{5}{36}\\ \frac{1}{36}& \frac{1}{36}& \frac{3}{36}& \frac{5}{36}\\ \frac{3}{36}& \frac{3}{36}& \frac{9}{36}& \frac{15}{36}\\ \frac{5}{36}& \frac{5}{36}& \frac{15}{36}& \frac{25}{36}\end{array}\right)\left(\begin{array}{l}{v}_1\\ v_2\\ v_3\\ v_4\end{array}\right)=\left[\begin{array}{l}0\\ 0\\ 0\\ 0\end{array}\right]$

The above condition gives the following equation

   $v_1+v_2+3v_3+5v_4\cdots \cdots \left(1\right)$

As equation $\left(1\right)$ has $4$ variables and $1$ equation we can choose three variables and will get the 4^th variable from equation $\left(1\right)$.

So by choosing different values of these variables we get the following solutions.

   $v={\left(-1,1,0,0\right)}^T,{\left(-3,0,1,0\right)}^T,{\left(-5,0,0,1\right)}^T$

So, we get the eigenvectors as ${\left(-1,1,0,0\right)}^T,{\left(-3,0,1,0\right)}^T\text{ and }\left(-5,0,0,1^T\right)$.

Conclusion:

Eigenvectors as ${\left(-1,1,0,0\right)}^T,{\left(-3,0,1,0\right)}^T\text{ and }\left(-5,0,0,1^T\right)$.