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[SOLVED] Compute the ideal

Fermat

Active member
Nov 3, 2013
188
Compute the ideal generated by $\begin{bmatrix}1&0&0\\0&0&0\\0&0&0 \end{bmatrix}$

My answer:
$\begin{bmatrix}a&b&c\\d&e&f\\g&h&i \end{bmatrix}$$\begin{bmatrix}1&0&0\\0&0&0\\0&0&0 \end{bmatrix}$$\begin{bmatrix}a'&b'&c'\\d'&e'&f'\\g'&h'&i' \end{bmatrix}$=$\begin{bmatrix}aa'&ab'&ac'\\da'&db'&dc'\\ga'&gb'&gc' \end{bmatrix}$.

Is this correct, and is there a nicer way to express the matrices in the ideal?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
Compute the ideal generated by $\begin{bmatrix}1&0&0\\0&0&0\\0&0&0 \end{bmatrix}$

My answer:
$\begin{bmatrix}a&b&c\\d&e&f\\g&h&i \end{bmatrix}$$\begin{bmatrix}1&0&0\\0&0&0\\0&0&0 \end{bmatrix}$$\begin{bmatrix}a'&b'&c'\\d'&e'&f'\\g'&h'&i' \end{bmatrix}$=$\begin{bmatrix}aa'&ab'&ac'\\da'&db'&dc'\\ga'&gb'&gc' \end{bmatrix}$.

Is this correct, and is there a nicer way to express the matrices in the ideal?
That is a correct, but not very informative, way of saying that the ideal is the whole algebra of $3\times3$ matrices. In fact, that algebra has no nontrivial ideals, so the ideal generated by any nonzero element is the whole algebra.

Edit. As later comments point out, you only get the rank 1 matrices this way. These do not form an ideal: you then need to take sums of them to get the full ideal.
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Wouldn't the rank of the matrix always be 1?
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
The product given in the first post is indeed of rank one, yes.

However, any elementary matrix $E_{ij}$ can be produced by such a product by multiplying by $E_{1j}$ on the right, and $E_{i1}$ on the left.

Since any possible matrix can be obtained by $R$-linear combinations of the $E_{ij}$ and all such $R$-linear combinations of elements of the ideal are in the ideal, we obtain all 3x3 matrices.

Perhaps more elegantly, we have (writing $J$ for the ideal):

$E_{11},E_{22},E_{33} \in J \implies I = E_{11} + E_{22} + E_{33} \in J \implies J = \text{Mat}_3(R)$

where $R$ is the commutative ring we are taking our matrix entries from (which may, or may not be, a field, the original poster does not say).
 

Fermat

Active member
Nov 3, 2013
188
That is a correct, but not very informative, way of saying that the ideal is the whole algebra of $3\times3$ matrices. In fact, that algebra has no nontrivial ideals, so the ideal generated by any nonzero element is the whole algebra.

Edit. As later comments point out, you only get the rank 1 matrices this way. These do not form an ideal: you then need to take sums of them to get the full ideal.
So whats the generated ideal then? I thought by definition the generated ideal is the set of things of the form axa'
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
So whats the generated ideal then? I thought by definition the generated ideal is the set of things of the form axa'
Ideals must be closed under addition.....