# Identity Matrices

Definition: A square $n \times n$ matrix $I$ is considered an Identity Matrix if all entries along the main diagonal are 1 and all other entries are 0. Alternatively we can define identity matrices such that if $(I)_{ij} = 1$ if $i =j$ and $(I)_{ij} = 0$ otherwise. |

We note that all identity matrices are square. For example, the identity matrix $I_{3 \times 3}$ (commonly denoted $I_{3}$ for brevity) looks like:

(1)

\begin{align} I_{3 \times 3} = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \end{align}

Theorem 1: Let $A$ be an $m \times n$ matrix and $I$ be an identity matrix and assume all operations are defined. Then:a) $AI_n = A$ and $I_mA = A$.b) $\mathrm{tr}(I_m) = m$. |

**Proof of (a):**Recall that any entry of a matrix product can be determined by the formula $(AB)_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + … + a_{in}b_{nj}$. Let $B = I_n$. Recall that $b_{nj} = 1$ if $n = j$ (and $b_{nj} = 0$ otherwise). Let $n = j$, then the entry $(AI)_{in} = a_{i1}b_{1n} + a_{i2}b_{2n} + … + a_{in}b_{nn}$. Since all other $b$ entries are $0$, it follows that $(AI)_{in} = a_{in} \cdot 1 = (A)_{in}$. Therefore $AI_n = A$. $\blacksquare$

**Proof of (b):**$I_m$ has only $1$s on the main diagonal. There will be exactly $m$ entries on the main diagonal, so $1 \cdot m = m$, so $tr(I_m) = m$. $\blacksquare$