Question

Show that A = [(3, 2, 4), (2, 0, 2), (4, 2, 3)] a 3x3 matrix, is diagnolizable even though one eigenvector has algebraic multiplicity 2. Do this by brute force computation. Why would you expect this to be true even without calculation?

Answer 1

@lgbasallote @UnkleRhaukus @Ishaan94
Please help here

Answer 2

I am sorry I don't know what eigenvector is

Answer 3

\[\textbf A=\left(\begin{array} {ccc} 3&2&4\\2&0&2\\4&2&3 \end{array}\right)\left(\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right) \]

Answer 4

i guess we should find the eigen vectors,

Answer 5

\[\textbf A\textbf x=\lambda \textbf x\]\[(\textbf A-\textbf I\lambda)\textbf x=0\]\[\left|\textbf A-\textbf I\lambda\right|=0\]\[\left|\begin{array} {ccc} 3-\lambda&2&4\\2&0-\lambda&2\\4&2&3-\lambda \end{array}\right|=0\]

Answer 6

\[(3-\lambda)\left[(-\lambda)\times(3-\lambda)-2\times2\right]-2[2\times(3-\lambda)-2\times4]+4[2\times2-(-\lambda\times4)]=0\]
\[(3-\lambda)\left[-3\lambda+\lambda^2-4\right]-2\left[6-2\lambda-8\right]+4\left[4+4\lambda\right]=0\]

Answer 7

\[-9\lambda+3\lambda^2-12-3\lambda^2-\lambda^3+4\lambda-12+4\lambda+16+16+16\lambda=0\]
\[-\lambda-\lambda^3+16\lambda+8=0\]

Answer 8

which dosent factor nicely, so i must have made a mistake somewhere ..

Answer 9

your \(-3\lambda^2\) should be \(+3\lambda^2\)

Answer 10

giving you \(6\lambda^2\)

Answer 11

\[-9\lambda+3\lambda^2-12+3\lambda^2+\lambda^3+4\lambda-12+4\lambda+16+16+16\lambda=0\]