A matrix with Repeated eigenvalues and its corresponding eigenvectors.

[pondzo]

Homework Statement

I am asked to find the diagonal matrix of eigenvalues, D, and the matrix of corresponding eigenvectors, P, of the following matrix:
$$\begin{pmatrix}<br /> 1 & 0 & 0\\<br /> 0 & 1 & -2\\<br /> 0 & 0 & -1<br /> \end{pmatrix}$$

Homework Equations

The Attempt at a Solution

We just started this topic, and so far i haven't had much trouble answering questions like these, but this question is different.

I found the eigenvalues to be λ = 1 (with algebraic multiplicity 2) and λ = -1

for λ = 1 the matrix is $$\begin{pmatrix}<br /> 0 & 0 & 0\\<br /> 0 & 0 & 0\\<br /> 0 & 0 & -2<br /> \end{pmatrix}$$ after putting in row echelon form. This is where my confusion begins, solving gives z=0, and since x nor y are described, they can take any value I please (to my understanding). so for the first eigenvector i let x=1 and y=0, then i let x=0,y=1 to give $$\begin{bmatrix}<br /> 1\\<br /> 0\\<br /> 0 <br /> \end{bmatrix}$$ and $$\begin{bmatrix}<br /> 0\\<br /> 1\\<br /> 0 <br /> \end{bmatrix}$$ respectively.
For λ = -1 the matrix is $$\begin{pmatrix}<br /> 2 & 0 & 0\\<br /> 0 & 2 & -2\\<br /> 0 & 0 & 1<br /> \end{pmatrix}$$ which gives x=y=z=0 which confuses me even more as this is $$\begin{bmatrix}<br /> 0\\<br /> 0\\<br /> 0 <br /> \end{bmatrix}$$

So after all that I said D = $$\begin{pmatrix}<br /> 1 & 0 & 0\\<br /> 0 & 1 & 0\\<br /> 0 & 0 & -1<br /> \end{pmatrix}$$ which I am fairly sure is correct
and P = $$\begin{pmatrix}<br /> 1 & 0 & 0\\<br /> 0 & 1 & 0\\<br /> 0 & 0 & 0<br /> \end{pmatrix}$$ which I am fairly sure isn't (Mathematica told me at least one of the parts was wrong and i think its this one).

Could someone please help me understand this example, and forgive me for the bad formatting, I am not sure how to keep everything in the same line.

 

[pondzo]

Dont worry i think i found my mistake. I got the matrix for λ=-1 wrong and hence confused the shiz out of myself.
Just checking the correct matrix for λ= -1 would be $$\begin{bmatrix}<br /> 0\\<br /> 1\\<br /> 1 <br /> \end{bmatrix}$$ right?

 

[HallsofIvy]

Yes, 1 is a "double" eigenvalue and -1 is an eigenvalue. I find it simplest to use the definition of "eigenvalue" to find the eigenvectors: $\lambda$ is an eigenvalue of linear transformation A if and only if there exist a non-zero vector, v, such that $Av= \lambda v$. (Of course, then, $(A-\lambda)v= 0$ which is what you used.)

For eigenvalue "1":
$$\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & -1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}= \begin{pmatrix}x \\ y- 2z \\ -z\end{pmatrix}= \begin{pmatrix}x \\ y \\ z \end{pmatrix}$$

so we have x= x, y- 2z= y, and -z= z. The last equation gives z= 0 and the first two equations, x= x and y= y, are always true. Any vector of the form (x, y, 0)= x(1, 0, 0)+ y(0, 1, 0) is an eigenvalue corresponding to eigenvalue 1. For eigenvalue "-1":
$$\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & -1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}= \begin{pmatrix}x \\ y- 2z \\ -z\end{pmatrix}= \begin{pmatrix}-x \\ -y \\ -z \end{pmatrix}$$

so that we must have x= -x, y- 2z= -z, and -z= -z. The first equation gives x= 0, the third is true for any z and the second gives y= z. Any eigenvector corresponding to eigenvalue -1 is of the form (0, z, z)= z(0, 1, 1).

It is easy to see that the matrix $$P= \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{pmatrix}$$
has inverse $$P^{-1}= \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1\end{pmatrix}$$and that
$$P^{-1}AP= \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & -1\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{pmatrix}= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{pmatrix}$$

 

[pasmith]

It is easy to see that the matrix $$P= \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{pmatrix}$$
has inverse $$P^{-1}= \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$$

You may want to reconsider the claim $P^{-1} = I$.

 

[HallsofIvy]

You may want to reconsider the claim $P^{-1} = I$.

Thanks, I copied the wrong matrix from my work! I went back and edited my post to correct that.

 

[pondzo]

Thank you for the alternate way of approaching these types of questions halls. When, for example, you say x=-x which leads to x=0 I assume this is because the only possible way for this to be true is if x=0 ?

 

[HallsofIvy]

Thank you for the alternate way of approaching these types of questions halls. When, for example, you say x=-x which leads to x=0 I assume this is because the only possible way for this to be true is if x=0 ?

Isn't that what "x satisfies the equation" means?

More specifically, I observed that adding x to both sides of "x= -x" gives "2x= 0" and then, dividing both sides by 2, "x= 0".

 

[pondzo]

Oh wow, please excuse that last comment. I have no Idea what I was thinking... lol.

 

[pondzo]

One more thing, I noticed that mathematica allowed any matrix, P, of the form
$\begin {pmatrix} a & c & 0 \\ b & d & 1 \\ 0 & 0 & 1 \end {pmatrix}$
As long as (a, b,0) =/= h (c, d,0) where h is a scalar multiple. Is this because the eigenvectors for the eigenvalue of 1 could be of the form (a, b,0) and (c, d,0) as long as they are not scalar multiples of each other?

 

[HallsofIvy]

I don't know what you mean by "Mathematica allowed". But, in general, 1 is NOT necessarily an eigenvalue for such a matrix.