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Diagonalizing matrix problem with parameters

Hi,

I have this problem I'm stuck on:

A is a 3x3 matrix

A= Attachment 24871

a,b,c are real numbers

1) determine for what values of a,b,c the matrix A is diagnoizable. (advice: distinguish between the cases a=c and a=\=c)

My attempt at a solution has been finding det(tI-A)=0 Attachment 24874, the characteristic polynomial is (t-c)(t-a)^2 and therefore the eigenvalues are **a** and **c.**

for t=c we get the matrix Attachment 24872 let it be C. for C__x__=__0__ we get the system of linear equations:

(c-a)x_{1}-x_{3}=0

(c-a)x_{2} -bx_{3}=0

likewise, for t=a we get

-1x_{3}=0

-bx_{3}=0

(a-c)x_{3}=0

Now, when I tried to find the eigenspaces that correspond to a and c to find out their dimensions I got pretty lost.

I have no idea what to do. help would be very welcomed!

Re: Diagonalizing matrix problem with parameters

*Case 1.* $\displaystyle a\neq c.$ Then, $\displaystyle \lambda=a$ (double), $\displaystyle \lambda=c$ (simple). By a well known theorem, $\displaystyle \dim V_c=1=\mbox{ multiplic }(c).$ On the other hand:

$\displaystyle \dim V_a=3-\mbox{rank}\begin{bmatrix}{0}&{0}&{1}\\{0}&{0}&{b} \\{0}&{0}&{c-a}\end{bmatrix}=3-1=2=\mbox{ multiplic }(a).$ That is, $\displaystyle A$ is diagonalizable.

*Case 2.* $\displaystyle a=c.$ Then, $\displaystyle \lambda=a$ (triple). So,

$\displaystyle \dim V_a=3-\mbox{rank}\begin{bmatrix}{0}&{0}&{1}\\{0}&{0}&{b} \\{0}&{0}&{0}\end{bmatrix}=3-1=2\neq \mbox{ multiplic }(a).$ In this case, $\displaystyle A$ is not diagonalizable.

Re: Diagonalizing matrix problem with parameters

Re: Diagonalizing matrix problem with parameters

we can try to proceed with your original approach, and "see what happens". i often do this "just out of curiosity".

it's clear that you found the characteristic equation properly, it's (t - c)(t - a)^{2}.

now let's find some eigenvectors. first let's tackle c:

$\displaystyle (A - cI)x = 0 \iff \begin{bmatrix}a-c&0&1\\0&a-c&b\\0&0&0 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix}0\\0\\0 \end{bmatrix}$

it's clear that x_{3} is "free", and provided a ≠ c:

$\displaystyle x_2 = \frac{b}{c-a}x_3$

$\displaystyle x_1 = \frac{1}{c-a}x_3$

which means that any eigenvector (belonging to c) is a scalar multiple of (1,b,c-a) (we have a one-dimensional eigenspace).

(note that if we DID have a = c, we we get instead is: x_{3} = 0. more on this later).

next we look at a:

$\displaystyle \begin{bmatrix}0&0&1\\0&0&b\\0&0&c-a \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix}0\\0\\0 \end{bmatrix}$

gives:

x_{3} = 0

bx_{3} = 0 (and thus x_{3} = 0, this just duplicates the first equation)

(c-a)x_{3} = 0 (another duplication).

thus we are free to choose x_{1} and x_{2} independently, so we can pick (1,0,0) and (0,1,0) as linearly independent eigenvectors.

so if {(1,0,0),(0,1,0),(1,b,c-a)} are linearly independent, A is diagonalizable (we have an eigenbasis). it should be clear that this will be true if and only if c -a ≠ 0, that is: if a ≠ c.

on the other hand, if a = c, we only get an eigenspace of dimension 2: namely {(x,y,0) in R^{3}}, so there is no eigenbasis, and A is not diagonalizable.