Orthogonal Diagonalization

**demode** · May 1st 2010, 02:07 AM

So if there exists a P such that $P^{-1}AP$ is diagonal, then A is diagonalizable. $P=[P_1 P_2 P_3]$ where $P_1,P_2,P_3$ 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 $P^TAP$ is diagonal.

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

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

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

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

**tonio** · May 1st 2010, 02:42 AM

Originally Posted by demode

So if there exists a P such that $P^{-1}AP$ is diagonal, then A is diagonalizable. $P=[P_1 P_2 P_3]$ where $P_1,P_2,P_3$ 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 $P^TAP$ 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:

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

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

This should be $x^3-qx^2-prx+pqr=(x-q)(x^2-pr)$ , 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 $x^2-pr$ 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 $q$ , 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?

.

**demode** · May 1st 2010, 08:45 PM

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 $(x-q)(x^2-pr)$ are $\pm \sqrt{pr} , q$ . 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 $A=A^T$ ? But

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

$<br /> A^{T}=\begin{pmatrix}0 & 0 & r \\ 0 & q & 0 \\ p & 0 & 0 \end{pmatrix}<br />$

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 $\exists P$ such that $P^{-1}AP$ is a diagonal matrix. If there is such P, and $P^{-1}=P^T$ , then A is orthogonally diagonalizable.

**tonio** · May 2nd 2010, 04:25 AM

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 $(x-q)(x^2-pr)$ are $\pm \sqrt{pr} , q$ . So each of these could correspond to an eigenvector. Then we need to check if the 3 eigenvectors are linearly independent?

When $A=A^T$ ? But

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

$<br /> A^{T}=\begin{pmatrix}0 & 0 & r \\ 0 & q & 0 \\ p & 0 & 0 \end{pmatrix}<br />$

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 $\exists P$ such that $P^{-1}AP$ is a diagonal matrix. If there is such P, and $P^{-1}=P^T$ , then A is orthogonally diagonalizable.

.

**demode** · May 2nd 2010, 08:29 PM

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 $\pm \sqrt{pr} , q$ ).

**tonio** · May 3rd 2010, 01:25 AM

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.

The answer to all your three questions is already in all these posts.

Tonio

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

.

**demode** · May 3rd 2010, 10:05 PM

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 $\lambda^2-pr$ from $\lambda^3-q\lambda^2-pr\lambda+pqr=(\lambda-q)(\lambda^2-pr)$ ?

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

**tonio** · May 4th 2010, 02:46 AM

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 $\lambda^2-pr$ from $\lambda^3-q\lambda^2-pr\lambda+pqr=(\lambda-q)(\lambda^2-pr)$ ?

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 $q\neq \pm\sqrt{pr}$ 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 $pr=0\Longrightarrow$ then the characteristic pol. is $p_A(x)=(x-q)x^2$ , so:

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

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

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

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

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

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

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

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

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

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

Tonio

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