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I am trying to solve Problem 15 on page 107 of Brešar's Uvod v Algebro. The question translates into English as:

Explain why any endomorphism of the algebra $M_2(\mathbb{R})$ can't map the matrix $E_{11}$ to the matrix $E_{12}$.

In this question, what does the notation $E_{ij}$ mean?

Xander Henderson
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InsertNameHere
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    Maybe it could refer to the matrix that has a $1$ in row $i$, column $j$ and zero in everything else. – azif00 Jul 09 '19 at 13:13
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    $$E_{ij} = \begin{pmatrix} e_{11} & e_{12} & \dotsb & e_{1n} \ e_{21} & e_{22} & \dotsb & e_{2n} \ \vdots & \vdots && \vdots \ e_{n1} & e_{n2} & \dotsb & e_{nn} \end{pmatrix}, $$ where each $e_{ij}$ is a number. Possibly. It would be helpful to know what book you are reading out of. – Xander Henderson Jul 09 '19 at 13:13
  • It's not in English so I doubt that would help. I'm gonna write out the problem in my question for (hopefully) more clarity. – InsertNameHere Jul 09 '19 at 13:16
  • There are people here who speak languages other than English. Moreover, it is often possible to parse the mathematics without actually being able to read the surrounding text. Knowing what book you are using is, frankly, vital context, since the notation you are using is not universal. That being said, the context you added suggests that @Azif00's comment may be correct. – Xander Henderson Jul 09 '19 at 13:19
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    $$E_{11}=\begin{pmatrix} 1&0 \ 0&0 \end{pmatrix} \qquad E_{12}=\begin{pmatrix} 0&1 \ 0&0 \end{pmatrix} \qquad$$ – azif00 Jul 09 '19 at 13:20
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    If $e_k$ is the vector of zeros with one in the $k$th position, then $E_{ij} = e_i e_j^T$. – copper.hat Jul 09 '19 at 13:20
  • @XanderHenderson It's Brešar's Uvod v algebro – InsertNameHere Jul 09 '19 at 13:22
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    See page 35 of your book. – azif00 Jul 09 '19 at 13:24
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    Thanks, Azif, it's all there. And for anyone else wondering, Azif's first comment is correct. – InsertNameHere Jul 09 '19 at 13:27

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The notation is explained at the top of page 35 of the same text. In this text, $E_{ij} \in M_{2\times 2}(\mathbb{R})$ is the standard basis matrix with a $1$ in the $j$-th column of the $i$-th row, and $0$s everywhere else. There are four such matrices: $$ E_{11} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},\quad E_{12} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},\quad E_{21} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix},\quad\text{and}\quad E_{22} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}. $$ In (linear) algebra and analysis, it is common to use the notation $e_i$ (or $e^{(i)}$) to denote the $i$-th standard basis vector in $\mathbb{R}^n$. That is, $e_i$ is a vector of length $n$ with a $1$ in the $i$-th position and $0$s everywhere else, i.e. $$ e_1 = (1, 0, \dotsc, 0),\qquad e_2 = (0, 1, \dotsc, 0),\qquad\dotsc,\qquad e_n = (0, 0, \dotsc, 1). $$ Every vector in $\mathbb{R}^n$ can be written as a linear combination of these vectors, so they give us a natural way of decomposing general vectors into simpler objects. Similarly, every matrix in $M_{2\times 2}(\mathbb{R})$ may be written as a linear combination of the $E_{ij}$s, hence the author's adoption of this notation—using a capital $E$ to distinguish matrices from vectors—is reasonable and not entirely unheard of.

It may also be worth noting, as per copper.hat's comment, that $$ E_{ij} = e_{i}^\intercal e_{j}, $$ where $e_{i}^\intercal$, a column vector, is thought of as an $m\times 1$ matrix; and $e_{j}$, a row vector, is thought of as a $1\times n$ matrix.

Xander Henderson
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