1. ## Eigen valves question

Plz help me to find the answer for following question;

A matrix

$\displaystyle A= \begin{bmatrix} 0 & 0 & 0 \\ 0 & \frac{29}{5} & \frac{12}{5} \\ 0 & \frac{12}{5} & \frac{36}{5} \end{bmatrix}$

Find
(a) Find Eigen valves A
(b) Egen vectors A
(c) obtain a matrix P such that $\displaystyle P^{-1} AP$ is a diagonal matrix
(d) If $\displaystyle S=P^{-1} AP$, state the special property of each of $\displaystyle S$ and $\displaystyle S^{-1}$
(e) Using the above result reduce the quadratic from Q(x) to form Q(y) where
$\displaystyle Q(x)=16x_1^2 + \frac{29}{5} x_2^2 + \frac{36}{5} x_3^2 + \frac{24}{5} x_2 x_3$
$\displaystyle Q(y)=16y_1^2 + +9y_2^2 + 4y_3^2$
(f) obtain the relationship between $\displaystyle x^t = (x_1 ,x_2 ,x_3)^t$ and $\displaystyle y^t (y_1 ,y_2 ,y_3)^t$

Now I solved part (a) as follows
$\displaystyle A- \lambda I=\begin{bmatrix} 16- \lambda & 0 & 0 \\ 0 & \frac{29}{5} - \lambda & \frac{12}{5} \\ 0 & \frac{12}{5} & \frac{36}{5} - \lambda \end{bmatrix}$
the result gain $\displaystyle ( \lambda -16)( \lambda -4)( \lambda -9)$

and

part (b) solved as follows
$\displaystyle \begin{pmatrix} 16 & 0 & 0 \\ 0 & \frac{29}{5} & \frac{12}{5} \\ 0 & \frac{12}{5} & \frac{36}{5} \end{pmatrix}$
x
$\displaystyle \begin{pmatrix} a \\ b \\ c \end{pmatrix} = 16$

then if $\displaystyle \lambda =4$
$\displaystyle \begin{pmatrix} 16 & 0 & 0 \\ 0 & \frac{29}{5} & \frac{12}{5} \\ 0 & \frac{12}{5} & \frac{36}{5} \end{pmatrix}$

$$x \displaystyle \begin{pmatrix} a \\ b \\ c \end{pmatrix} = 4 also I did for \displaystyle \lambda =9 now I need help to find rest of parts in this question, also I need to know the above calculations are in correct or not!! if some are in incorrect plz. let me know the place and how I correct it 2. Originally Posted by dhammikai Plz help me to find the answer for following question; A matrix \displaystyle A= \begin{bmatrix} 0 & 0 & 0 \\ 0 & \frac{29}{5} & \frac{12}{5} \\ 0 & \frac{12}{5} & \frac{36}{5} \end{bmatrix}  Find (a) Find Eigen valves A (b) Egen vectors A (c) obtain a matrix P such that \displaystyle P^{-1} AP is a diagonal matrix (d) If \displaystyle S=P^{-1} AP, state the special property of each of \displaystyle S  and \displaystyle S^{-1} (e) Using the above result reduce the quadratic from Q(x) to form Q(y) where \displaystyle Q(x)=16x_1^2 + \frac{29}{5} x_2^2 + \frac{36}{5} x_3^2 + \frac{24}{5} x_2 x_3 \displaystyle Q(y)=16y_1^2 + +9y_2^2 + 4y_3^2 (f) obtain the relationship between \displaystyle x^t = (x_1 ,x_2 ,x_3)^t and \displaystyle y^t (y_1 ,y_2 ,y_3)^t Now I solved part (a) as follows \displaystyle A- \lambda I=\begin{bmatrix} 16- \lambda & 0 & 0 \\ 0 & \frac{29}{5} - \lambda & \frac{12}{5} \\ 0 & \frac{12}{5} & \frac{36}{5} - \lambda \end{bmatrix} the result gain \displaystyle ( \lambda -16)( \lambda -4)( \lambda -9) and part (b) solved as follows \displaystyle \begin{pmatrix} 16 & 0 & 0 \\ 0 & \frac{29}{5} & \frac{12}{5} \\ 0 & \frac{12}{5} & \frac{36}{5} \end{pmatrix}  x \displaystyle \begin{pmatrix} a \\ b \\ c \end{pmatrix} = 16  then if \displaystyle \lambda =4 \displaystyle \begin{pmatrix} 16 & 0 & 0 \\ 0 & \frac{29}{5} & \frac{12}{5} \\ 0 & \frac{12}{5} & \frac{36}{5} \end{pmatrix}  [tex]x \displaystyle \begin{pmatrix} a \\ b \\ c \end{pmatrix} = 4 also I did for \displaystyle \lambda =9 now I need help to find rest of parts in this question, also I need to know the above calculations are in correct or not!! if some are in incorrect plz. let me know the place and how I correct it For matrix A, is row 1 column 1 suppose to be 16 or 0? for a, i didnt check, but it looks right? b. Eigenvectors should be a vector... not a number c. the diagonal values of S is the eigenvalues of A 3. Hi thanks for the reply and comments. Yes I got an error the matrix A should be \displaystyle A= \begin{bmatrix} 16 & 0 & 0 \\ 0 & \frac{29}{5} & \frac{12}{5} \\ 0 & \frac{12}{5} & \frac{36}{5} \end{bmatrix}  In part (b) the vector means if \displaystyle \lambda =16, then we have the vector of [a,b,c] is it true? likewise \displaystyle \lambda =4, and \displaystyle \lambda =6 also have vector [a,b,c] is it true? in part (c) you mean the eigen values of A is insert to the diagonal matrix? Plz explain me... 4. \displaystyle A- \lambda I=\begin{bmatrix} 16- \lambda & 0 & 0 \\ 0 & \frac{29}{5} - \lambda & \frac{12}{5} \\ 0 & \frac{12}{5} & \frac{36}{5} - \lambda \end{bmatrix} the result gain \displaystyle ( \lambda -16)( \lambda -4)( \lambda -9) Assume ur eigenvalues are right, so \displaystyle \lambda=16,4,9 For b, to find eigenvectors, sub each eigenvalues into the \displaystyle A-\lambda I matrix and find the nullspace of the matrix, it should yield exactly 1 nullspace vector for each eigenvalue (for this example) c. yes, the diagonal entries of the diagonal matrix is the eigenvalues of A 5. Hi thnks for the reply and help.So now we complete part (a), I approch the part (b) as follows \displaystyle \lambda =16, the the mat will be \displaystyle  \displaystyle math]A= \begin{bmatrix} 0 & 0 & 0 \\ 0 & \frac{51}{5} & \frac{12}{5} \\ 0 & \frac{12}{5} & \frac{44}{5} \end{bmatrix}  then according to equations \displaystyle \frac{51}{5}y+ \frac{12}{5}z=0 and \displaystyle \frac{12}{5}y+ \frac{44}{5}z=0, the results shold be x=0, y=0, z=o then the vector \displaystyle v_1=(0,0,0) If \displaystyle \lambda=9, then the matrix will be$$$\displaystyle A= \begin{bmatrix} -70 & 0 & 0 \\ 0 & \frac{16}{5} & \frac{12}{5} \\ 0 & \frac{12}{5} & \frac{9}{5} \end{bmatrix}$
Then according to equations
$\displaystyle -7x=0$, $\displaystyle \frac{16}{5}y+ \frac{12}{5}z=0$ and $\displaystyle \frac{12}{5}y+\frac{9}{5}z=0$, results shouldbe x=0, y=3, z=-4, the vector is $\displaystyle v_2(0,3,4)$

again $\displaystyle \lambda=4$, then the equations should be $\displaystyle -12x=0$, $\displaystyle \frac{-9}{5}y+\frac{12}{5}z=0$ and $\displaystyle \frac{12}{5}y-\frac{16}{5}z=0$. results are x=0, y=4, z=3

plz help me the above procedure is correct or wrong..and if wrong let me know the right answer/way..