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I have a question as follows:

Let $$A = \begin{bmatrix}0&0\\0&1\end{bmatrix}.$$ Prove that the ideal $\left<A\right>$ of the ring $M_2(\mathbb{R})$ is the whole ring.

I took some advice from this question, and tried to find a matrix $X$ which would give me an identity matrix for $AX + XA$. However, it seems any matrix I would try to multiply A with results in the first element being zero because of the zero elements in $A$. My initial thought was that this has something to do with A already being a part of the identity matrix equation. Am I anywhere near the solution?

user26857
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TootsieRoll
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1 Answers1

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Note the following: $$ \begin{bmatrix} & 1\\ 1 & \end{bmatrix} \begin{bmatrix} & \\ & 1 \end{bmatrix} \begin{bmatrix} & 1\\ 1 & \end{bmatrix} = \begin{bmatrix} 1 & \\ & \end{bmatrix} =: \boldsymbol B, $$ and $\boldsymbol A + \boldsymbol B = \boldsymbol I_2$. By the definition of an ideal, $\boldsymbol B \in I, \boldsymbol A + \boldsymbol B \in I$, hence $I = \mathrm M_2(\Bbb R)$.

Another thing to correct:

It should be

Prove that the ideal containing $\boldsymbol A$ is the whole ring.

user26857
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xbh
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