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For which values of k the matrix is diagonalized ?

Yankel

Active member
Jan 27, 2012
398
Hello

I need some help with this question...

I have a mtrix A:

[tex]\begin{pmatrix} k &-2 &1 \\ 4 &-k &2 \\ 0 &0 &1 \end{pmatrix}[/tex]


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:

[tex]-\lambda ^{3}+\lambda ^{2}+k ^{2}\lambda-k ^{2}+8-8\lambda[/tex]

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

Thanks !!
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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 $k^2=8$).
 
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Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
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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 $k=-3$.
 
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Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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

Well-known member
MHB Math Helper
Jan 29, 2012
661
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

Active member
Jan 27, 2012
398
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

Active member
Jan 27, 2012
398
$k \in (-\infty,-3) \cup (-3,-\sqrt{8}) \cup (\sqrt{8}, \infty)$

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...
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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?
[FONT=MathJax_Main]

[/FONT]
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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

Active member
Jan 27, 2012
398
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?
[FONT=MathJax_Main]

[/FONT]
I get that the eigenvalues are 1,1,-1 ?

:confused:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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?
 
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