Find Basis for diagonal matrix

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A diagonal matrix contains non-zero elements only along its main diagonal, and a basis for these matrices consists of a set of nxn matrices that can express all nxn diagonal matrices as linear combinations. For a 2x2 diagonal matrix, the basis includes matrices like [1 0; 0 0] and [0 0; 0 1]. For a 3x3 diagonal matrix, the basis expands to include matrices such as [1 0 0; 0 0 0; 0 0 0]. The notation eii refers to matrices with a 1 in the (i, i) position and 0 elsewhere, indicating that the basis for an nxn diagonal matrix contains n matrices. Thus, the basis is finite, consisting of n distinct matrices for an nxn diagonal matrix.
Clandry
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I'm not sure how to start this problem.
All i know is a diagonal matrix consists of all 0 elements except along the main diagonal.

But how do I even find a basis for this?
 

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What would a basis look like? It would be set of nxn matrices such that... you can do what with them?
 
haruspex said:
What would a basis look like? It would be set of nxn matrices such that... you can do what with them?

For this case, a basis consists of all matrices such that all nxn diagonal matrices can be written as a linear combination of them?
 
Clandry said:
For this case, a basis consists of all matrices such that all nxn diagonal matrices can be written as a linear combination of them?
Yes. What's the simplest matrix you can think of that might be useful in creating such a basis?
 
haruspex said:
Yes. What's the simplest matrix you can think of that might be useful in creating such a basis?

This is where I get stuck. I've only been taught and done problems where the basis is a set of "vectors."

I saw somewhere that the basis for a 2x2matrix is
1 0
0 0

0 1
0 0

0 0
1 0

0 0
0 1


if it were a 2x2 diagonal would it be
1 0
0 0

and
0 0
0 1
?
 
Ok. Now try 3x3.
 
haruspex said:
Ok. Now try 3x3.

1 0 0
0 0 0
0 0 0

0 0 0
0 1 0
0 0 0

0 0 0
0 0 0
0 0 1if it's an nxn matrix, wouldn't that give an infinite amount of matrices for the bases?The answer in the back of the book is
37. B = {eii | 1 ≤ i ≤ n} the "ii" part is supposed to be subscripts for e. I'm bad at interpreting these kind of answers, what is it saying?
 
Clandry said:
if it's an nxn matrix, wouldn't that give an infinite amount of matrices for the bases?
You had 2 for 2x2 and 3 for 3x3. Why would you get infinitely many for nxn?
 
haruspex said:
You had 2 for 2x2 and 3 for 3x3. Why would you get infinitely many for nxn?

oops i mean n amount.
 
  • #10
So you have the answer.
The eii notation used in the book apparently means the nxn matrix that has 1 at the (i, i) position and 0 everywhere else. I don't know how standard that is. Should be defined in the book somewhere.
 

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