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- Thread starter ArcanaNoir
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I like Serena

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Hey Arcana!

A diagonal matrix has only non-zero entries on its main diagonal.

The main diagonal has 3 entries, each of which can either be 0 or 1.

That makes 8 possible diagonal matrices.

If the ordering of eigenvalues is not relevant, that leaves 4 diagonal matrices.

Note that the eigenvalues cannot be distinct, since you have 3 entries on the diagonal but only 2 possible eigenvalues.

According to the spectral theorem, a real matrix is diagonalizable if and only if it is symmetric.

I think it also holds for matrices over the field GF(2).

That means that you have a free choice for the upper triangular matrix for 2^6=64 choices. After that the remaining part of the lower triangle is fixed.

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No, this is false. There are plenty of matrices which are diagonalizable but not symmetric.According to the spectral theorem, a real matrix is diagonalizable if and only if it is symmetric.

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I like Serena

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From the link I provided:No, this is false. There are plenty of matrices which are diagonalizable but not symmetric.

"Another way to phrase the spectral theorem is that a real n×n matrix A is symmetric if and only if there is an orthonormal basis of consisting of eigenvectors for A."

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A diagonalizable matrix need not have an orthonormal basis of eigenvectors.From the link I provided:

"Another way to phrase the spectral theorem is that a real n×n matrix A is symmetric if and only if there is an orthonormal basis of consisting of eigenvectors for A."

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Good point. I'll take the time to digest that.A diagonalizable matrix need not have an orthonormal basis of eigenvectors.

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Office_Shredder

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I Like Serena: All you need is a basis of eigenvectors, doesn't matter if they're orthogonal to each other.

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WannabeNewton

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Also, a matrix doesn't have to have distinct eigenvalues to be diagonalizable. The converse is true of course but the original statement is false (take the identity matrix for example). All you need is for the geometric multiplicities of the eigenspaces to add up to the dimension of the original space.

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Office_Shredder

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I'm not seeing the connection between orthogonal and having determinant 1.Over GF(2) all 3x3 matrices that are invertible have determinant 1.

So all invertible matrices are orthogonal.

How would you define "orthogonal" in ##\mathbb{F}_2## anyway. There is no inner product on that space. At most, there is a symmetric bilinear form.Therefore each basis of independent eigenvectors has an associated matrix that is orthogonal.

[tex]\left(\begin{array} 1 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 1\end{array}\right)[/tex]

Counterexample:It means that in this particular case, each diagonalizable matrixissymmetric.

[tex]\left(\begin{array} 1 & 0 & 1\\ 0 & 1 & 0\\ 0 & 0 & 0\end{array}\right)[/tex]

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Still, you make good points.

So I wrote a program to count the invertible matrices, and to count the diagonalizable matrices.

I got the surprising result that there are 168 invertible matrices and 58 diagonalizable matrices.

It turns out that for instance the following symmetric matrix is not diagonalizable:

$$\begin{bmatrix}0&1&0 \\ 1&0&0 \\0&0&0\end{bmatrix}$$

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They need not be invertible.

I'm confused about this, first you say the main diagonal can only have non-zero entries, but then you say the main diagonal can have 0's or 1's. Also, diagonalizable is not the same as diagonal, right?Hey Arcana!

A diagonal matrix has only non-zero entries on its main diagonal.

The main diagonal has 3 entries, each of which can either be 0 or 1.

That makes 8 possible diagonal matrices.

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58? Crap! They must have started at n=0... :(:(:( Well in that case there is no way I'm going to do this sum over 58 matrices.

Still, you make good points.

So I wrote a program to count the invertible matrices, and to count the diagonalizable matrices.

I got the surprising result that there are 168 invertible matrices and 58 diagonalizable matrices.

It turns out that for instance the following symmetric matrix is not diagonalizable:

$$\begin{bmatrix}0&1&0 \\ 1&0&0 \\0&0&0\end{bmatrix}$$

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I like Serena

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Sorry for the confusion.I'm confused about this, first you say the main diagonal can only have non-zero entries, but then you say the main diagonal can have 0's or 1's. Also, diagonalizable is not the same as diagonal, right?

A diagonal matrix

It's just that all other entries have to be zero.

And yes, diagonalizable is not the same as diagonal.

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And thanks to everybody for all your input :)

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Ah well, just for the record, the sum of those 58 matrices is the identity matrix!58? Crap! They must have started at n=0... :(:(:( Well in that case there is no way I'm going to do this sum over 58 matrices.

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- #19

I like Serena

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Yeah, I saw it... but I do not consider it precalc.

With my first couple of attempts I couldn't prove associativity, although I could verify with a couple of examples that it does seem to be associative.

So for now I decided not to add to the zero-count of the thread.

Perhaps someone else is more creative than I am.

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Do you mean I'm allowed to post in the big-people sections now?Yeah, I saw it... but I do not consider it precalc.

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I would say the post is precalculus, since it's just an algebraic manipulation.

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I like Serena

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Dunno.Do you mean I'm allowed to post in the big-people sections now?

Until now no one has corrected me for posting in the big-people section.

So I guess that makes it okay.

As for your associativity problem, it seems to me that mere algebraic manipulation is not going to cut it.

Moreover, I'd consider associativity problems part of abstract algebra.

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It will. But it's tedious.As for your associativity problem, it seems to me that mere algebraic manipulation is not going to cut it.

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That's what I thought too.I would say the post is precalculus, since it's just an algebraic manipulation.

It is, it's actually a group theory thing I'm working on.Moreover, I'd consider associativity problems part of abstract algebra.

It will? Are you sure? Any hints would be greatly appreciated.It will. But it's tedious.

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It appears to turn into a higher order polynomial expression.It will. But it's tedious.

It may not be possible to solve that.

At any rate, Wolfram doesn't.

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