1. ## Orthogonal Diagonalization

So if there exists a P such that $\displaystyle P^{-1}AP$ is diagonal, then A is diagonalizable. $\displaystyle P=[P_1 P_2 P_3]$ where $\displaystyle P_1,P_2,P_3$ are eigenvectors of A. So, A is diagonalizable if it has 3 distinct eigenvalues.
A is orthogonally diagonalizable if there exists an orthonormal set of 3 eigenvectirs if A. Also, it is orthogonally diagonalizable if there exists P such that $\displaystyle P^TAP$ is diagonal.

So for (a), I need to show if there is a matrix P that diagonalizes A. First finding the eigenvalues of A:

$\displaystyle det(\lambda I -A)=0$

$\displaystyle \lambda^3-q \lambda-rp \lambda -rpq = 0$

And I'm stuck here, because I don't know the values of r,p,q. Can anyone help?

2. Originally Posted by demode

So if there exists a P such that $\displaystyle P^{-1}AP$ is diagonal, then A is diagonalizable. $\displaystyle P=[P_1 P_2 P_3]$ where $\displaystyle P_1,P_2,P_3$ are eigenvectors of A. So, A is diagonalizable if it has 3 distinct eigenvalues.

No. It is diagonalizable iff it has three linearly independent eigenvectors.

A is orthogonally diagonalizable if there exists an orthonormal set of 3 eigenvectirs if A. Also, it is orthogonally diagonalizable if there exists P such that $\displaystyle P^TAP$ is diagonal.

It should be " is orthog. diag. if there exists orthogonal P such that..."

So for (a), I need to show if there is a matrix P that diagonalizes A. First finding the eigenvalues of A:

$\displaystyle det(\lambda I -A)=0$

$\displaystyle \lambda^3-q \lambda-rp \lambda -rpq = 0$

This should be $\displaystyle x^3-qx^2-prx+pqr=(x-q)(x^2-pr)$ , and then we can use the theorem that says that a matrix is diagonalizable iff its

minima polynomial is a product of different linear factors, so it must be that $\displaystyle x^2-pr$ has a two different roots in the definition field of the vector space we're

talking about, and since we're talking , apparently, of the complex field, then we must have that none of these two different roots equals $\displaystyle q$ , which

already is a root of the first factor...

Finally, remember (or prove) that a matrix orthogonally diagonalizable iff it is symmetric, and in our case that happens when...

Tonio

And I'm stuck here, because I don't know the values of r,p,q. Can anyone help?
.

3. Hi!
I don't understand your theorem very well! Why did you say that none of these two different roots should equal q?
The roots of the polynomial $\displaystyle (x-q)(x^2-pr)$ are $\displaystyle \pm \sqrt{pr} , q$. So each of these could correspond to an eigenvector. Then we need to check if the 3 eigenvectors are linearly independent?

Originally Posted by tonio
Finally, remember (or prove) that a matrix orthogonally diagonalizable iff it is symmetric, and in our case that happens when...
When $\displaystyle A=A^T$? But

$\displaystyle A= \begin{pmatrix}0 & 0 & p \\ 0 & q & 0 \\ r & 0 & 0 \end{pmatrix}$

$\displaystyle A^{T}=\begin{pmatrix}0 & 0 & r \\ 0 & q & 0 \\ p & 0 & 0 \end{pmatrix}$

A could be symmetric if p=r. But we don't know the values of p and r...

It should be " is orthog. diag. if there exists orthogonal P such that..."
Yes, here's the definition we use in our course: A is diagonalizable if $\displaystyle \exists P$ such that $\displaystyle P^{-1}AP$ is a diagonal matrix. If there is such P, and $\displaystyle P^{-1}=P^T$ , then A is orthogonally diagonalizable.

4. Originally Posted by demode
Hi!
I don't understand your theorem very well! Why did you say that none of these two different roots should equal q?
The roots of the polynomial $\displaystyle (x-q)(x^2-pr)$ are $\displaystyle \pm \sqrt{pr} , q$. So each of these could correspond to an eigenvector. Then we need to check if the 3 eigenvectors are linearly independent?

When $\displaystyle A=A^T$? But

$\displaystyle A= \begin{pmatrix}0 & 0 & p \\ 0 & q & 0 \\ r & 0 & 0 \end{pmatrix}$

$\displaystyle A^{T}=\begin{pmatrix}0 & 0 & r \\ 0 & q & 0 \\ p & 0 & 0 \end{pmatrix}$

A could be symmetric if p=r. But we don't know the values of p and r...

I think you haven't yet understood the question at all (although perhaps it's me who hasn't...): I think the question asks you to check what the values of p, q, r must be so that the matrix a will be diagonalizable, or orthogonally diagonaizable, or etc.
You have to decide what the values pq,r must be so that condition 1 will be satisfied, and then which ones will satistfy condition 2 and etc...

Tonio

Yes, here's the definition we use in our course: A is diagonalizable if $\displaystyle \exists P$ such that $\displaystyle P^{-1}AP$ is a diagonal matrix. If there is such P, and $\displaystyle P^{-1}=P^T$ , then A is orthogonally diagonalizable.
.

5. Originally Posted by tonio
I think you haven't yet understood the question at all (although perhaps it's me who hasn't...): I think the question asks you to check what the values of p, q, r must be so that the matrix a will be diagonalizable, or orthogonally diagonaizable, or etc.
You have to decide what the values pq,r must be so that condition 1 will be satisfied, and then which ones will satistfy condition 2 and etc...
So, if p=r we know that the matrix A is symmetric, otherwise it is not. But I don't see any theorems in my textbook which says "a matrix is diagonalizable if it is symmetric".

There is a theorem which says: "A matrix is orthogonally diagonalizable if and only if it is symmetric". Again this doesn't mean every symmetric matrix is orthogonally diagonalizable, right?

There is another theorem that: "if A is a symmetric matrix then eigenvectors form different eigenspaces are orthogonal". So can I here judge that A is orthogonally diagonalizable because if there are 3 eigenvalues then the 3 corresponding eigenvectors are linearly independent?? (I've already shown that the 3 eigenvalues of A are $\displaystyle \pm \sqrt{pr} , q$).

6. Originally Posted by demode
So, if p=r we know that the matrix A is symmetric, otherwise it is not. But I don't see any theorems in my textbook which says "a matrix is diagonalizable if it is symmetric".

Good you don't because if you did you should throw that book away: from where did you get the idea that this is a theorem? Please do read

carefully what I wrote in my prior posts. Mathematics requires a minimum of attention and care.

There is a theorem which says: "A matrix is orthogonally diagonalizable if and only if it is symmetric". Again this doesn't mean every symmetric matrix is orthogonally diagonalizable, right?

Of course it does: don't you know/understand the mathematical meaning of iff = if and ony if ?!?

There is another theorem that: "if A is a symmetric matrix then eigenvectors form different eigenspaces are orthogonal". So can I here judge that A is orthogonally diagonalizable because if there are 3 eigenvalues then the 3 corresponding eigenvectors are linearly independent??

No, you can't since this much is true always: eigenvectors belonging to different eigenvalues are lin. ind.

I see you have some serious problems understanding all this stuff, which is very important to attack the problem and eventually solve it. I propose

you sit down with your notes/books and re-read definitions, theorems, etc, and don't mix stuff so that'll you know exactly what to do.

Tonio

(I've already shown that the 3 eigenvalues of A are $\displaystyle \pm \sqrt{pr} , q$).
.

7. Originally Posted by tonio
Of course it does: don't you know/understand the mathematical meaning of iff = if and ony if ?!?
Sorry, I wasn't careful there...

This theorem answers part (c). If p=r, A is symmetric and hence orthogonally diagonalizable. This is because for a symmetric matrix, the eigenvectors from different eigenspaces are orthogonal.

Now to clarify part (a). I'm reading your first post but I'm still not sure if I understand it correctly...

A is diagonalizable whenever q is not a root of $\displaystyle \lambda^2-pr$ from $\displaystyle \lambda^3-q\lambda^2-pr\lambda+pqr=(\lambda-q)(\lambda^2-pr)$?

& for part (b), A is diagonalizable, but not orthogonally diagonalizable whenever q is not a root of $\displaystyle \lambda^2-pr$ and A is not symmetric. Is this correct?

8. Originally Posted by demode
Sorry, I wasn't careful there...

This theorem answers part (c). If p=r, A is symmetric and hence orthogonally diagonalizable. This is because for a symmetric matrix, the eigenvectors from different eigenspaces are orthogonal.

Now to clarify part (a). I'm reading your first post but I'm still not sure if I understand it correctly...

A is diagonalizable whenever q is not a root of $\displaystyle \lambda^2-pr$ from $\displaystyle \lambda^3-q\lambda^2-pr\lambda+pqr=(\lambda-q)(\lambda^2-pr)$?

Again, I didn't write that. The following is part of what I wrote in my first post: "... then we must have that none of these two different roots

equals , which already is a root of the first factor..." .This is a necessary condition, not a sufficient one, meaning: if the matrix is diagonalizable then it

must be that $\displaystyle q\neq \pm\sqrt{pr}$ but not necessarily the other way around ...**sigh**

Remember: the characteristic and the minimal pol's of a matrix have the very same irreducible factors, and the matrix is diag. iff its minimal pol. is

a product of different linear factors, and now you must check the different cases:

1) if $\displaystyle pr=0\Longrightarrow$ then the characteristic pol. is $\displaystyle p_A(x)=(x-q)x^2$ , so:

$\displaystyle {}\;\;\;{}\ldots$1a) if also $\displaystyle q=0$ then $\displaystyle p_A(x)=x^3\Longrightarrow A$ is nilpotent and thus it is diag. iff it is the zero matrix (i.e, iff also $\displaystyle p=r=0$) ;

$\displaystyle {}\;\;\;{}\ldots$1b) if $\displaystyle q\neq 0$ then it must be that the minimal pol. is $\displaystyle m_A(x)=x(x-q)\iff A(A-qI)=0=$ the zero matrix. But you

can easily check this is possible iff $\displaystyle p=r=0$

2) if $\displaystyle pr\neq 0\Longrightarrow \sqrt{pr}\neq -\sqrt{pr}$ so:

$\displaystyle \ldots$2a) if $\displaystyle q=\sqrt{pr}\,\,\,or\,\,\,q=-\sqrt{pr}$ , then we get that $\displaystyle p_A(x)=(x-q)^2(x\pm \sqrt{pr})$ , so that $\displaystyle A$ is diagonalizable iff

$\displaystyle m_A(x)=(x-q)(x\pm \sqrt{pr})\iff (A-qI)(A\pm \sqrt{pr}I)=0=$ the zero matrix. Check that in one case this happens but not in the other one (unless I erred. Check this carefully)

$\displaystyle \ldots$2b) if $\displaystyle q\neq \pm\sqrt{pr}$ then all the roots are different and thus $\displaystyle A$ is diag.

Pay attention to the fact that in the whole point (2) above we must have that $\displaystyle \sqrt{pr}$ is an element of the definition field, so if we work over the rationals this means $\displaystyle pr$ must be a (rational) square!

& for part (b), A is diagonalizable, but not orthogonally diagonalizable whenever q is not a root of $\displaystyle \lambda^2-pr$ and A is not symmetric. Is this correct?

Yes it is , but what they want to know is exactly for which values of $\displaystyle p,q,r$ this happens.

Tonio