For which values of k the matrix is diagonalized ?

[Yankel]

Hello

I need some help with this question...

I have a mtrix A:

$$\begin{pmatrix} k &-2 &1 \\ 4 &-k &2 \\ 0 &0 &1 \end{pmatrix}$$and I need to find for which values of k, the matrix is diagonalized. I know the way to check if a matrix is diagonalized is to check the algebraic and geometric multipliers, but technically, how to do it with a parameter is beyond my skill, I need some help with the solution. I found the characteristic polynomial, it is:

$$-\lambda ^{3}+\lambda ^{2}+k ^{2}\lambda-k ^{2}+8-8\lambda$$

and the final answer should be a^2>8 AND a!=3 (not equal 3)

Thanks !

 

[Fernando Revilla]

The characteristic polynomial is

$\begin{vmatrix} k-\lambda &-2 &1 \\ 4 &-k-\lambda &2 \\ 0 &0 &1-\lambda \end{vmatrix}=(1-\lambda)\begin{vmatrix} k-\lambda &-2 \\ 4 &-k-\lambda \end{vmatrix}=(1-\lambda)(\lambda^2-k^2+8)$

So, the eigenvalues are $\lambda=1$ and $\lambda=\pm \sqrt{k^2-8}$. If $k^2<8$ then, there are no real roots so, $A$ is not diagonalizable in $\mathbb{R}$. Now, consider that if $k^2\geq 8$ there are multiple roots iff $k=\pm 3$ (Edited: or $k^2=8$).

 

[I like Serena]

there are multiple roots iff $k=\pm 3$.

What if $k^2=8$?

 

[Fernando Revilla]

What if $k^2=8$?

You know the answer (I suppose). :)

 

[I like Serena]

The OP suggested that A is only diagonizable iff $k^2 > 8$ and $k \ne 3$.

That does not seem right.
Either way, each of the values $k=\pm 3$ and $k=\pm \sqrt 8$ require extra investigation.
$A$ turns out to be diagonizable with exactly one of those, but it is not $k=-3$.

 

[Deveno]

i suppose then, there is naught to do but investigate the eigenspaces. if $k = 3$ then solving $A - I = 0$ we get the eigenvectors (1,1,0) and (0,1,2) so the eigenspace belonging to 1 has dimension 2, so in this case $A$ is diagonalizable.

if $k = -3$ then solving $A - I = 0$ leads to the single eigenvector (1,-2,0) so in this case $A$ is not diagonalizable.

if $k = \sqrt{8}$ we get the single eigenvector (1,√2,0)

if $k = -\sqrt{8}$ we get the single eigenvector (1,-√2,0). so in both these cases $A$ is not diagonalizable. the computations of these facts is dreary, and i omit them. the relevant point is:

$\dim(E_{\lambda}) = \dim(\ker(A - \lambda I))$

therefore the $k$ for which the matrix $A$ is diagonalizable is:

$k \in (-\infty,-3) \cup (-3,-\sqrt{8}) \cup (\sqrt{8}, \infty)$

that is $k^2 > 8$ and $k \neq -3$

(the "answer" in the OP is incorrect).

 

[Fernando Revilla]

Also: in the cases where the given matrix $A$ has only two distinct eigenvalues $\lambda_1$ and $\lambda_2$, $A$ is diagonalizable $\Leftrightarrow\;(A-\lambda_1I)(A-\lambda_2I)=0$.

 

[Yankel]

sorry about the delayed response.

I don't understand this:

if k2≥8 there are multiple roots iff k=±3

how did we get ±3 ?

 

[Yankel]

$k \in (-\infty,-3) \cup (-3,-\sqrt{8}) \cup (\sqrt{8}, \infty)$

that is $k^2 > 8$ and $k \neq -3$

(the "answer" in the OP is incorrect).

but this is the answer I wrote, so it is correct, I am confused here...

 

[I like Serena]

sorry about the delayed response.

I don't understand this:

if k2≥8 there are multiple roots iff k=±3

how did we get ±3 ?

As Fernando showed, the characteristic polynomial factorizes to $(1-\lambda)(\lambda^2-k^2+8) = 0$.

What do you get if you substitute for instance k=-3?
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[I like Serena]

but this is the answer I wrote, so it is correct, I am confused here...

You wrote a^2>8 AND a!=3 (not equal 3).

We're assuming you intended k instead of a.
The second part should be that k is not equal to -3 instead.

 

[Yankel]

As Fernando showed, the characteristic polynomial factorizes to $(1-\lambda)(\lambda^2-k^2+8) = 0$.

What do you get if you substitute for instance k=-3?
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I get that the eigenvalues are 1,1,-1 ?

 

[I like Serena]

I get that the eigenvalues are 1,1,-1 ?

Yes. That means that 1 is a multiple root.
1 will be a multiple root when k is either -3 or +3.
Furthermore, 0 will be a multiple root when k is either -√8 or +√8.

Your matrix is only diagonizable if it has 3 real eigenvalues, and furthermore if there are 3 independent eigenvectors.

If the eigenvalues are distinct, the eigenvectors are guaranteed to be independent.
If they are not, we have to check if we can find independent eigenvectors for the duplicated eigenvalues.
This is what deveno did.

Can you find the eigenvectors for the eigenvalue 1 if k=-3?