# Square Matrices

So I'm studying a few special families of square matrices, the diagonal matrices, upper triangular matrices, lower triangular matrices and symmetric matrices and I just had a few questions.

I know...

a diagonal matrix is if every nondiagonal entry is zero, $a_{ij}$=0 whenever $i$ doesn't equal $j$.

an upper triangular matrix is if all entries below the diagonal are zero, $a_{ij}=0$ whenever $i >j$.

a lower triangular matrix is if all entries above the diagonal are zero, $a_{ij}=0$ whenever $i < j$.

symmetric if $a_{ij}=a_{ji}$ for all $i$ and $j$.

But I was just wondering, can the diagonal matrices, upper triangular matrices, lower triangular matrices and symmetric matrices have the $0_{n\times n}$? Also I know the $I_{n\times n}$ matrix is in the diagonal matrices, but can it be in the other three types?

• Both the zero matrix and the identity matrix are diagonal, upper/lower triangular, and symmetric. Check the definitions. – Ethan Alwaise Dec 9 '16 at 4:55
• Just check the hypothesis – Learnmore Dec 9 '16 at 4:56

A diagonal matrix doesn't care what is on the diagonal; it only cares that the off-diagonal entries are $0$. So the zero matrix $\mathbf 0_{n \times n}$ is indeed diagonal.
Similarly, an upper triangular matrix only cares that the elements below the diagonal are $0$, so the zero matrix is upper triangular. For the same reason, it is also lower triangular.
It is also symmetric, because $a_{ij} = a_{ji} = 0$ for all $i,j$.
Take for example the definition of upper triangular matrix. It's a matrix which has $a_{ij}=0$ whenever $i>j$, but this doesn't mean that $a_{ij}\neq 0$ for $i\le j$, it's possible to have $a_{ij}=0$ if $i\le j$. Then both $O_{n\times n}$ and $I_{n\times n}$ are upper triangular matrices.