Question

urgent please,thanks

Find all distinct (real or complex) eigenvalues of A. Then find a basis for the eigenspace of A corresponding to each eigenvalue. For each eigenvalue, specify the dimension of the eigenspace corresponding to that eigenvalue, then enter the eigenvalue followed by the basis of the eigenspace corresponding to that eigenvalue. 8 -1 9 A = -9 6 -15 |-6 4 -10 Number of distinct eigenvalues: 1 Dimension of Eigenspace: 1

Answer 1

A= 8 -9 -6 -1 6 4 9 -15 -10/

(A - x0) is,

s -9 {-6 -1 9} 100\ 6 -15)-10 10 4 -10/ lo 01/

(8 -2 -9 -6 6 -1 -7 4 9 -15 -10 - 1

R2 + R2+ 3 - . R1

R3 6 R3+ 3 - - R1

18- -1 12-14r +39 8-1 -4r+26 8-1 9 151-39 8-1 12+2r-26 8- 1 0

R2 HR

x2 – 14.1 +39 R3 + R3-4 - 4.c + 26 - R

1 -4-26 8-C -1 9 | r2-27-26 | 0 上 8-I [00 - r(r=-4x+13) 2012-13) ST

now matrix is in triangular form

hence determinant is equal to the multiplication of diagonal entry

(8 - x) - 4.0 + 26 8 - (x2 - 4x + 13) 2 (2.– 13)

1 (-4.0 +26) (- 0:22 - 4x + 13) 2 2.– 13)

(-1 (22 – 4.2 + 13)) = 0

(22 – 4.1 +13) = 0

I=0, 12 – 4.0 + 13 = 0

2-4x+13=0

-= V02 - 4ac 21, 2 = 20

-(-4) + X1,2 = (-4)2 – 4. 1. 13 2.1

4+ 16 52 T1, 2=-

4 = V-36 11,2 =

44 62 T1, 2=-

-21,2 = 2 +31

.

total 3 eigenvalues are

I=0, I = 2+3i, I = 2 – 31

.

.

for r=0 (A- rl) is,

s -1 9 \ 100) -9 6 -15 -0.10 10 {-6 4 -10/ to 01/ -1

78 -9 -6 -1 6 4 9 \ -15 -10/

R1 + R2

7-9 18 (-6 6 -1 4 -15} 9 -10/

R2 + R2+.. R

R3 + R3- R1

7-9 6 = 10 1 [ 0 0 -15) 13 0 /

R2 + R2

7-9 6 | 0 1 (0 0 -15) -1 0

R1 + R1 - 6. Ry

7-9 | 0 0 0 1 0 -9 -1 0/

R1 + RI

10 1 [01-1 (0 0 0/

reduced system is

10 1 [01-11 [0 0 0/

sz=fi 12- = 1 .. JO=2-RU 10=2+

2= 2.................... free

.

general solution is

11

take z=1

$\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}-1\\ 1\\ 1\end{pmatrix}$

eigenvector is

V1 = 1 dimension : 1

..

.

for I = 2 +31 (A- 1) is.

18 -1 -9 6 (-6 4 9 (100) -15 | - (2 + 3i) 1 0 1 0 -10/ 1001)

(6 – 3i - 1 - 94-3i -6 4 9 -15 -12 - 31/

R$ * *3-(3-3): R2 + R2 - 1

R3 R3 -

9 9 76 - 3i 0 -1 14 - i18 | 127 = - 21 + +

R2 HR

$=\begin{pmatrix}6-3i&-1&9\\ 0&\frac{16}{5}-i\frac{2}{5}&-\frac{24}{5}+i\frac{3}{5}\\ 0&\frac{14}{5}-i\frac{18}{5}&-\frac{21}{5}+i\frac{27}{5}\end{pmatrix}$

$R_3\:\leftarrow \:R_3-\left(1-i\right)\cdot \:R_2$

$\:R_2\:\leftarrow \left(\frac{4}{13}+i\frac{1}{26}\right)\cdot \:R_2$

$=\begin{pmatrix}6-3i&-1&9\\ 0&1&-\frac{3}{2}\\ 0&0&0\end{pmatrix}$

$R_1\:\leftarrow \:R_1+1\cdot \:R_2$

$=\begin{pmatrix}6-3i&0&\frac{15}{2}\\ 0&1&-\frac{3}{2}\\ 0&0&0\end{pmatrix}$

$R_1\:\leftarrow \left(\frac{2}{15}+i\frac{1}{15}\right)\cdot \:R_1$

$=\begin{pmatrix}1&0&1+i\frac{1}{2}\\ 0&1&-\frac{3}{2}\\ 0&0&0\end{pmatrix}$

reduced system is

$\begin{pmatrix}1&0&1+i\frac{1}{2}\\ 0&1&-\frac{3}{2}\\ 0&0&0\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}$

$\begin{Bmatrix}x+\left(1+i\frac{1}{2}\right)z=0\\ y-\frac{3}{2}z=0\end{Bmatrix}.....................\begin{Bmatrix}x=-\left(1+i\frac{1}{2}\right)z\\ y=\frac{3}{2}z\end{Bmatrix}$

2= 2.................... free

.

general solution is,

$\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}-\left(1+i\frac{1}{2}\right)z\\ \frac{3}{2}z\\ z\end{pmatrix}$

take z=2

– 2 أ \ /

eigenvector is

dimension : 1

.

.

now to find the eigenspace for x=2-3i,

just change the sign of i (complex part)

so third eigenspace is

-2+i 3 V3 = dimension : 1

.

.

.

final answer is

for dy = 0, 0,- () dimension for 11 = 0, 01 = dimension : 1

$for \:\:\:\:\lambda_2=2+3i, \:\:\:\:\:\:{\color{Red} v_2=\begin{pmatrix}-2-i\\ 3\\ 2\end{pmatrix}}\:\:\:\:\:dimension:\:1$

$for \:\:\:\:\lambda_3=2-3i, \:\:\:\:\:\:{\color{Red} v_3=\begin{pmatrix}-2+i\\ 3\\ 2\end{pmatrix}}\:\:\:\:\:dimension:\:1$