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By definition of a diagonal matrix, a square matrix is said to be diagonal if all its non-diagonal elements are zero. So, can a 1x1 matrix be considered diagonal by this definition?

Zwolf
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  • Yes: $1 \times 1$ matrices are diagonal, even according to this definition. – Crostul Mar 01 '17 at 08:31
  • @Crostul even though it does not have any diagonal elements? – Zwolf Mar 01 '17 at 08:32
  • Shouldn't that be 'if all non-diagonal elements are zero'? In this case it is a diagonal matrix, since there are no non-diagonal elements. – Student Mar 01 '17 at 08:33
  • In this case, "all non-diagonal elements are zero" is vacuously true, because there is no non-zero non-diagonal element. – Crostul Mar 01 '17 at 08:34
  • And by extension, a 1x1 matrix with element zero is a non-diagonal matrix? – Zwolf Mar 01 '17 at 08:39
  • @Zwolf: nope. In the definition, the diagonal elements play no role, so any $1\times1$ matrix is diagonal because all its non-diagonal elements are zero. –  Mar 01 '17 at 08:42
  • @Zwolf there is a trivial isomorphism between scalars and 1x1 matrices that preserves addition and multiplication, so in a way yes. But as objects they are different, but they behave the same. – Henno Brandsma Mar 01 '17 at 08:53

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If $A$ is $n \times n$, it's a diagonal matrix iff for all $1 \le i \neq j \le n: A_{i,j} = 0$, all its off-diagonal entries are $0$.

A $1 \times 1$ matrix has none, so the demand is voidly/trivially fulfilled. In other words: there can be no counterexample, so it's true.

Henno Brandsma
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By definition of a diagonal matrix, a square matrix is said to be diagonal if all its diagonal elements are zero.

Actually, a square matrix is diagonal if all its nondiagonal elements are zero.

And yes, under this definition, becaues a $1\times 1$ matrix has no nondiagonal elements, it is by definition diagonal. The claim

"every nondiagonal element of a $1\times 1$ matrix has property $X$"

is true for all properties you can think of.

5xum
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Yes, every $1 \times 1$ matrix is diagonal. The only entry is along the diagonal, so the matrix consists of real numbers along the diagonal (which is $1$ spot) and $0$s everywhere else (which is nowhere).

Caleb Stanford
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