Compute Ideal Generated by Matrix 3x3

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The ideal generated by the matrix $\begin{bmatrix}1&0&0\\0&0&0\\0&0&0 \end{bmatrix}$ in the algebra of $3 \times 3$ matrices is the entire algebra of $3 \times 3$ matrices, denoted as $\text{Mat}_3(R)$. This conclusion arises from the fact that any nonzero matrix generates the whole algebra, as there are no nontrivial ideals in this context. While the product of the given matrix with other matrices yields rank 1 matrices, these do not form an ideal by themselves; the ideal must include sums of such matrices to encompass all possible $3 \times 3$ matrices.

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Fermat1
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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?
 
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Fermat said:
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:
Wouldn't the rank of the matrix always be 1?
 
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).
 
Opalg said:
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 what's the generated ideal then? I thought by definition the generated ideal is the set of things of the form axa'
 
Fermat said:
So what's 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...
 

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