1. ## Eigen valves question

Plz help me to find the answer for following question;

A matrix

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

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

the result gain $( \lambda -16)( \lambda -4)( \lambda -9)$

and

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

x
$\begin{pmatrix}
a \\
b \\
c
\end{pmatrix} = 16
$

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

$$x $\begin{pmatrix} a \\ b \\ c \end{pmatrix} = 4$ also I did for $\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 $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 $P^{-1} AP$ is a diagonal matrix (d) If $S=P^{-1} AP$, state the special property of each of $S$ and $S^{-1}$ (e) Using the above result reduce the quadratic from Q(x) to form Q(y) where $Q(x)=16x_1^2 + \frac{29}{5} x_2^2 + \frac{36}{5} x_3^2 + \frac{24}{5} x_2 x_3$ $Q(y)=16y_1^2 + +9y_2^2 + 4y_3^2$ (f) obtain the relationship between $x^t = (x_1 ,x_2 ,x_3)^t$ and $y^t (y_1 ,y_2 ,y_3)^t$ Now I solved part (a) as follows $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 $( \lambda -16)( \lambda -4)( \lambda -9)$ and part (b) solved as follows $ \begin{pmatrix} 16 & 0 & 0 \\ 0 & \frac{29}{5} & \frac{12}{5} \\ 0 & \frac{12}{5} & \frac{36}{5} \end{pmatrix} $ x $\begin{pmatrix} a \\ b \\ c \end{pmatrix} = 16 $ then if $\lambda =4$ $\begin{pmatrix} 16 & 0 & 0 \\ 0 & \frac{29}{5} & \frac{12}{5} \\ 0 & \frac{12}{5} & \frac{36}{5} \end{pmatrix} $ [tex]x $\begin{pmatrix} a \\ b \\ c \end{pmatrix} = 4$ also I did for $\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 $ 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 $\lambda =16$, then we have the vector of [a,b,c] is it true? likewise $\lambda =4$, and $\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. $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 $( \lambda -16)( \lambda -4)( \lambda -9)$ Assume ur eigenvalues are right, so $\lambda=16,4,9$ For b, to find eigenvectors, sub each eigenvalues into the $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 $\lambda =16$, the the mat will be $ $ $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 $\frac{51}{5}y+ \frac{12}{5}z=0$ and $\frac{12}{5}y+ \frac{44}{5}z=0$, the results shold be x=0, y=0, z=o then the vector $v_1=(0,0,0)$ If $\lambda=9$, then the matrix will be$$ $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
$-7x=0$, $\frac{16}{5}y+ \frac{12}{5}z=0$ and $\frac{12}{5}y+\frac{9}{5}z=0$, results shouldbe x=0, y=3, z=-4, the vector is $v_2(0,3,4)$

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