Diagonalize Matrix, Given an Eigenvalue and Eigenvector

[BraedenP]

Homework Statement

\begin{bmatrix}<br /> -7 &amp;&amp; -16 &amp;&amp; 4\\<br /> 6 &amp;&amp; 13 &amp;&amp; -2\\<br /> 12 &amp;&amp; 16 &amp;&amp; 1<br /> \end{bmatrix}

Diagonalize the matrix (if possible), given that one eigenvalue is 5, and that one eigenvector is {-2, 1, 2}

Homework Equations

A=PDP^{-1}

The Attempt at a Solution

If I were allowed to simply calculate the eigenvalues and corresponding eigenvectors, I'd be able to determine if it's diagonalizable and if so, to diagonalize it. The problem here is that I have to go only on the provided information.

I'm stuck regarding how to proceed with this question. Where do I start?

Thanks!

 

[vela]

Start by finding the eigenvalue associated with the given eigenvector and determining what system of equations the eigenvector for λ=5 has to satisfy.

 

[BraedenP]

Oh! I just read the question wrong. I assumed that the eigenvector provided was in the eigenspace of the provided eigenvalue. That probably means there will be two eigenvalues, and one of them will have an algebraic multiplicity of 2.

I'll give it a shot and see if that works. Thanks. :)

Edit:

Well that was easy! Perhaps I should spend more time reading these questions :P

Thanks!

 

[HallsofIvy]

Oh! I just read the question wrong. I assumed that the eigenvector provided was in the eigenspace of the provided eigenvalue.

Well, that obviously is NOT true because (-7)(-2)+ (-16)(1)+ (4)(2)= 14- 16+ 8= 6, not -10.

That probably means there will be two eigenvalues, and one of them will have an algebraic multiplicity of 2.

I'll give it a shot and see if that works. Thanks. :)

Edit:

Well that was easy! Perhaps I should spend more time reading these questions :P

Thanks!

Why did you say "If I were allowed to simply calculate the eigenvalues and corresponding eigenvectors"? There is nothing in the statement of the problem that prohibits that. The characteristic equation is cubic and knowing one solution allows you to reduce it to a quadratic.

 

[BraedenP]

Well, that obviously is NOT true because (-7)(-2)+ (-16)(1)+ (4)(2)= 14- 16+ 8= 6, not -10.


Why did you say "If I were allowed to simply calculate the eigenvalues and corresponding eigenvectors"? There is nothing in the statement of the problem that prohibits that. The characteristic equation is cubic and knowing one solution allows you to reduce it to a quadratic.

Regarding your first point, I didn't actually calculate anything; I just took it by assumption (yes, bad.. I know) that the eigenvector and value were associated.

Regarding the second point, we were explicitly told not to calculate eigenvalues using the characteristic equation.

But it's all good now. Thanks :)