Chapter 6.1, Problem 32PS

To determine

(a) To determine:

Give a basis for the null space and a basis for the column space.

Eigenvector $u$ makes the basis for null space.

Eigenvectors $v\text{ and }w$ makes the basis for the column space of the matrix.

Given:

4 has eigenvalues $O,3,5$ with independent eigenvectors $u,v,w$.

Concept used:

A vector which is represented by a matrix $M$ such that when it is multiplied with matrix $A$ then the direction of the resultant matrix would remain the same as vector $M$ is called Eigen vector.

Calculation:

The equation relating matrix $A$, eigenvalue $\lambda$ and eigenvector $x$ is given as

   $Ax=\lambda x\mathrm{.....}\left(1\right)$.

The eigenvalues of matrix $A$ given in the question are $0,3\text{ and }5$ and the corresponding

Eigenvectors are $u.\text{ v and }w$ respectively.

By substituting for $\lambda =0\text{ and }x=u$ we get

   $Au=0$

Multiplying above equation by a constant $c\in \square$ we get.

   $\begin{array}{l}c\left(Au\right)=0\\ A\left(cu\right)=0\end{array}$

So from above equation we get that eigenvector $u$ makes the basis for null space

Also as only eigenvectors of nonzero eigenvalues lie in the column space of the matrix, so eigenvectors $\text{v and }w$ makes the basis for the column space of the matrix

Conclusion:

Eigenvector $u$ makes the basis for null space.

Eigenvectors $v\text{ and }w$ makes the basis for the column space of the matrix.

To determine

(b) To determine:

Find a particular solution to $Ax=\upsilon +w$. Find all solution.

A particular solution as $\left(\begin{array}{ccc}0& \frac{1}{3}& \frac{1}{5}\end{array}\right)$

General solution as $\left(\begin{array}{ccc}0& \frac{1}{3}& \frac{1}{5}\end{array}\right)+cu$

Given:

A has eigenvalues $0,3,5$ with independent eigenvectors $u,v,w$.

Concept used:

Particular solution is free of arbitrary parameters while general solution has.

Calculation:

The equation to be solved is given as

   $Ax=u+v\cdots \cdots \left(2\right)$

As the eigenvectors are au linearly independent, they form a basis of

So for any vector $x\in {\square }^3$, there exists constant $c_1,c_2,c_3\in \square$ such that

   $x=c_{1}u+c_{2}v+c_{3}w\cdots \cdots \left(3\right)$

So on substituting equation $\left(3\right)\text{ in }\left(2\right)$ and using equation $\left(1\right)$ we get,

   $\begin{array}{l}A\left(c_{1}u+c_{2}v+c_{3}w\right)=v+w\\ c_{1}Au+c_{2}Av+c_{3}Av=v+w\\ 3c_{2}v+5c_{3}w=v+w\end{array}$

So from above equation we get

   $c_1=0,c_2=\frac{1}{3},c_3=\frac{1}{5}$

So we get a particular solution as $\left(0\frac{1}{3}\frac{1}{5}\right)$.

To get a general solution, we will add $cu$ to the particular solution where $c\in \square$. So we get the general solution as

   $\left(\begin{array}{ccc}0& \frac{1}{3}& \frac{1}{5}\end{array}\right)+cu$

Conclusion:

A particular solution as $\left(\begin{array}{ccc}0& \frac{1}{3}& \frac{1}{5}\end{array}\right)$

General solution as $\left(\begin{array}{ccc}0& \frac{1}{3}& \frac{1}{5}\end{array}\right)+cu$

To determine

(c) To determine:

   $Ax=u$ Has no solution. If it did then $\mathrm{.........}$ would he in the column space.

   $Ax=u$ Has no solution. If it did the $u$ would be in the column space.

Given:

   $A$ Has eigenvalues $0,3,5$ with independent eigenvectors $u,v,w$.

Concept used:

Solution of equation matrix

Calculation:

If $Ax=u$ solution, then eigenvalue $u$ will have a nonzero eigenvalue and as eigenvalue is nonzero then $u$ lie in the column space of the matrix.

So we get the answer as $u$.

Conclusion:

   $Ax=u$ Has no solution. If it did the $u$ would he in the column space.