Diagonal matrix is a type of square matrix in which all the diagonal elements are present and non-diagonal elements are zero. Diagonal elements are those elements for which row number is equal to the column number i.e. i = j.
So, A3x3 = $\begin{bmatrix}
a &0 &0 \\
0 &b &0 \\
0 &0 &c
\end{bmatrix}$
In the above matrix A, it is clear that all non-diagonal elements are 0 and diagonal elements are present. So, A is a diagonal matrix.

To know about Block Diagonal Matrix, first we have to know about Block Matrix.

Block Matrix: A block matrix is a matrix which is split into blocks. These blocks are nothing but the smaller matrices and formed by grouping the elements in adjacent rows or columns or if a matrix is define in terms of smaller matrices adjacent rows and columns.
Let E = $\begin{bmatrix}
2 &2 &1 &1 \\
2 &2 &1 &1 \\
4 &4 &3 &3 \\
4 &4 &3 &3
\end{bmatrix}$ be any matrix. It can be partitioned into 4 x 2 blocks as follows:

E11 = $\begin{bmatrix}
2 &2 \\
2 &2
\end{bmatrix}$

E12 = $\begin{bmatrix}
1 &1 \\
1 &1
\end{bmatrix}$

E21 = $\begin{bmatrix}
4 &4 \\
4 &4
\end{bmatrix}$

E22 = $\begin{bmatrix}
3 &3 \\
3 &3
\end{bmatrix}$

So, E = $\begin{bmatrix}
E_{11} &E_{12} \\
E_{21} &E_{22}
\end{bmatrix}$ is the block matrix.

So, we can say if A is i x j order matrix and k and l are numbers such that 1 < k < i and 1 < l < j, then we can form a block matrix with its elements C, D, E and F as follows:

A = $\begin{bmatrix}
C &D \\
E &F
\end{bmatrix}$

Block Diagonal Matrix: A block diagonal matrix is a type of square matrix such that if a block matrix is formed from this square matrix then, its diagonal is an arrangement of square matrices and the non-diagonal blocks or sub-matrices are zero matrices.
Diagonal Matrix
Here, Cn x n is a block diagonal matrix of order n x n. Thus, C is a block diagonal matrix if $C_{\alpha \beta }$ = 0, where $\alpha = \beta$.
Let A = $\begin{bmatrix}
a &0 &0 \\
0 &b &0 \\
0 &0 &c
\end{bmatrix}$ be any diagonal matrix. To find the inverse of A, we use

A-1 = $\frac{adj A}{\left | A \right |}$

Now, for determinant A, we select first row for expansion.

Hence I A I = a $\begin{vmatrix}
b&0 \\
0 &c
\end{vmatrix}$ + 0 $\begin{vmatrix}
0 &0 \\
0 &c
\end{vmatrix}$ + 0 $\begin{vmatrix}
0 &b \\
0 &0
\end{vmatrix}$

= abc

Now, adj A = $\begin{bmatrix}
bc &0 &0 \\
0 &ac &0 \\
0 &0 &ab
\end{bmatrix}$

So, A-1 = $\frac{1}{\left | A \right |}$ x $\begin{bmatrix}
bc &0 &0 \\
0 &ac &0 \\
0 &0 &ab
\end{bmatrix}$

= $\begin{bmatrix}
1/a &0 &0 \\
0 &1/b&0 \\
0 &0 &1/c
\end{bmatrix}$

Hence, inverse of a diagonal matrix is defined only if the determinant of a diagonal matrix is non zero and the reciprocal of the diagonal elements are present in inverse matrix instead of actual diagonal elements.
Listed below are some of the diagonal matrix properties:

Property 1: Any diagonal matrix is also a symmetric matrix. If A is any diagonal matrix then, A = $A^T$Proof:

Let A3x3 = $\begin{bmatrix}
-2 &0 &0 \\
0 & 4 &0 \\
0 &0 &7
\end{bmatrix}$ is a diagonal matrix. Now, lets calculate AT

AT = $\begin{bmatrix}
-2 &0 &0 \\
0 & 4 &0 \\
0 &0 &7
\end{bmatrix}$ = A

Hence, A = AT.

Property 2: If two diagonal matrices are of same order, then their addition and multiplication are again a diagonal matrix. So, if C and D are diagonal matrices, then C + D and CD are again diagonal matrices.
Proof:

Let C3x3 = $\begin{bmatrix}
4 &0 &0 \\
0 &-7 &0 \\
0 &0 &9
\end{bmatrix}$ and D3x3 = $\begin{bmatrix}
-1 &0 &0 \\
0 & 3 &0 \\
0 &0 &2
\end{bmatrix}$ are two matrices of same order.

Then, C + D = $\begin{bmatrix}
4 &0 &0 \\
0 &-7 &0 \\
0 &0 &9
\end{bmatrix}$ + $\begin{bmatrix}
-1 &0 &0 \\
0 & 3 &0 \\
0 &0 &2
\end{bmatrix}$

= $\begin{bmatrix}
3 &0 &0 \\
0 & -4 &0 \\
0 &0 &11
\end{bmatrix}$ ..........................................(a)

Again, CD = $\begin{bmatrix}
4 &0 &0 \\
0 &-7 &0 \\
0 &0 &9
\end{bmatrix}$ . $\begin{bmatrix}
-1 &0 &0 \\
0 & 3 &0 \\
0 &0 &2
\end{bmatrix}$

= $\begin{bmatrix}
-4 &0 &0 \\
0 &-21 &0 \\
0 &0 &18
\end{bmatrix}$ .................................................(b)

From (a) and (b), it is clear that C + D and CD are diagonal matrices.

Property 3: If A and B are diagonal matrices, then AB = BA.
Proof:

A = $\begin{bmatrix}
9 &0 \\
0 &-1
\end{bmatrix}$ and B = $\begin{bmatrix}
1 &0 \\
0 &7
\end{bmatrix}$ are two matrices.

Then, AB = $\begin{bmatrix}
9 &0 \\
0 &-1
\end{bmatrix}$. $\begin{bmatrix}
1 &0 \\
0 &7
\end{bmatrix}$

= $\begin{bmatrix}
9 &0 \\
0 &-7
\end{bmatrix}$ ...................................... (i)

Now, BA = $\begin{bmatrix}
1 &0 \\
0 &7
\end{bmatrix}$ . $\begin{bmatrix}
9 &0 \\
0 &-1
\end{bmatrix}$

= $\begin{bmatrix}
9 &0 \\
0 &-7
\end{bmatrix}$ ........................................(ii)

Hence, from (i) and (ii), AB = BA which means matrix multiplication is commutative.