# Normal matrix?

1. Sep 9, 2006

### pivoxa15

If A*A=AA*

than A is a normal matrix. But does A must also have an orthonormal basis?

2. Sep 9, 2006

### matt grime

Matrices don't have orthonormal bases. Vector spaces have bases, and any basis can be defined to be orthonormal with respect to some inner product.

3. Sep 9, 2006

### pivoxa15

Maybe I mean does A must also have columns which form an orthonormal set? If not than would they have to form an orthogonal set?

4. Sep 9, 2006

### matt grime

A matrix whose rows/columns are an orthonormal basis is one that satisfies XX^t =Id, the orthogonal matrices. If Normal were the same as orthogonal why would we define normal? Using this heuristic, we conclude they are different, so try to find an example. (These heuristics are not foolproof, but not unreasonable.)

5. Sep 9, 2006

### pivoxa15

I was a bit confused last night maybe what I was really trying to ask is, if A is a normal matrix than A can be written as a matrix T (with respect to B where B is an orthogonormal basis in R^n). The reason would be because of the spectral theorem. In fact from this theorem, T with respect to the standard basis in R^n is a diagonal matrix.
An aside question on wording, would 'for' be more appropriate to replace the word in bold?

Going back to what I was asking before and consider the second case. If a normal matrix form an orthogonal set than this would imply AA*=A*A=D where D is any diagonal matrix. But a normal matrix dosen't always have to have AA*=diagonal matrix. So the columns dosen't have to be orthogonal to each other either.

Last edited: Sep 9, 2006
6. Sep 10, 2006

### matt grime

You can use either 'in' or 'for'. You cannot however use:

"If a normal matrix form an orthogonal set"

and you shouldn't say:

"AA*=A*A=D where D is any diagonal matrix."

It is 'a' diagonal matrix, not any diagonal matrix.

A matrix is something written with respect to a basis, by the way. The linear map it represents is what you ought to be talking about, or the matrices conjugate to A.

7. Sep 10, 2006

### pivoxa15

The basis can be a set in R^n or do you mean a set of matrices in a (matrix) vector space?

The linear map depends on the basis in which the matrix is written in so the basis is very important.

8. Sep 11, 2006

### matt grime

When you write a matrix it comes with a basis attached: a matrix is written with respect to a basis. It appears your book is teaching you things backwards.

9. Sep 28, 2006

### pivoxa15

I have to admit my fundalmentals are poor. I just like to clear this matter up

People usually just write a matrix like

$$A=\left( \begin{array}{cc} 2 & 3\\ 3 & 2 \end{array} \right)$$

without giving any bases. But would you assume they are stating it with respect to the standard basis?

However, if they wrote the same matrix and said it was with respect to {(1,0), (1,1)} than it you could work out what that matrix is with respect to the standard basis in R^2 which turns out to be

$$A=\left( \begin{array}{cc} -1 & 0\\ 3 & 5 \end{array} \right)$$

Is that would you are getting at?

Last edited: Sep 28, 2006
10. Sep 28, 2006

### matt grime

You have just said that A=/=A. A matrix is a matrix is a matrix. Just a set of numbers. You can't change the numbers in matrix and say it is the same matrix. It may be the same linear map but it is a different matrix.

The vector (1,0) means e_1 irrespective of what your choice of e_1 is, i.e. it just means 'your first basis vector'.

11. Sep 28, 2006

### pivoxa15

With the point you raised which was "When you write a matrix it comes with a basis attached: a matrix is written with respect to a basis." I assume you mean always?

What do you mean here?

I have try to formalise my previous post so as to make it more clear and to hopefully clear the confusions.

Assuming $$B=\{(1,0), (1,1)\}$$ and $$B_{o}=\{(1,0), (0,1)\}$$

If
$$A=\left( \begin{array}{cc} 2 & 3\\ 3 & 2 \end{array} \right)$$

then
$$A_{B_{o}}=\left( \begin{array}{cc} 2 & 3\\ 3 & 2 \end{array} \right)$$
----------------------------------

If
$$A_{B}=\left( \begin{array}{cc} 2 & 3\\ 3 & 2 \end{array} \right)$$

then
$$A_{B_{o}}=\left( \begin{array}{cc} -1 & 0\\ 3 & 5 \end{array} \right)$$

Last edited: Sep 28, 2006
12. Sep 29, 2006

### matt grime

I know what you're attempting to say.

If I write down a matrix and say it is a linear map on a vector space I have implicitly assumed some basis. I don't necessarily have to say what one basis looks like relative to another. The matrix A above sends the basis vector e_1 to 2e_1 +3e_2. See, I've not said what e_1 is. Now, if I want to figure out how the linear map this is is written with respect to a different matrix I can. It would be really helpful if you just remembered that the vector (1,0) just means e_1, the first basis vector and didn't think it meant the unit vector in the x direction.

13. Sep 29, 2006

### pivoxa15

So the statement "When you write a matrix it comes with a basis attached: a matrix is written with respect to a basis." Corresponds to "If I write down a matrix and say it is a linear map on a vector space I have implicitly assumed some basis." only.

In other words
If
A matrix is a linear map on a vector space
then
It automatically comes with a basis.

What about the reverse
If
A matrix has a basis
then
It is a linear map on a vector space

But one does not always define a matrix to be a linear map on a vector space? So in those cases, matrices don't have to come with a basis.

Last edited: Sep 29, 2006
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