0
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Both the zero matrix and the identity matrix are diagonal, upper/lower triangular, and symmetric. Check the definitions. $\endgroup$ – Ethan Alwaise Dec 9 '16 at 4:55
  • 2
    $\begingroup$ Just check the hypothesis $\endgroup$ – Learnmore Dec 9 '16 at 4:56
1
$\begingroup$

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 $i,j$.

Similar reasoning applies to the identity matrix.

$\endgroup$
1
$\begingroup$

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.

The same idea applies to the diagonal, lower triangular and symmetric matrices.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.