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Consider a $m \times n$ matrix $A$ where $m \gg n$. Can $AA^T$ be a diagonal matrix (meaning all the diagonal entries are non-zero and the off-diagonal entries are 0).

Arthur
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applied_math
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    That's not what "diagonal matrix" means. Any matrix whose off-diagonal entries are zero is called a diagonal matrix. There is no requirement for the diagonal entries to be nonzero. –  May 24 '19 at 14:12
  • Case $vv^T$ where $v$ is a vector is the simplest to consider.. – Widawensen Jul 11 '19 at 12:23

4 Answers4

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No. Since $m \gg n$, there are many more rows than columns. We can let each row be a a vector in $n$ dimensional space. A diagonal matrix would only be produced if any distinct pair of vectors has a zero dot product, implying all vectors are perpendicular to each other, but since there are more vectors than the dimension this is impossible.

Arthur
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auscrypt
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$AA^T$ would be an $m\times m$ matrix of rank at most $n$, so with the requirement that

all the diagonal entries are non-zero

the answer is no, as this requirement would give the matrix rank $m$. If you remove this requirement from your definition of "diagonal", however (which is customary), then the answer is yes. For instance the $0$ matrix.

One can even make it so that there are $n$ non-zero entries. In particular, if the top $n$ rows of $A$ makes an $n\times n$ identity matrix, and all entries below that are $0$, then $AA^T$ comes out to a diagonal matrix where the first $n$ diagonal entries are $1$, and the rest are $0$.

This can easily be tweaked to give any (positive) values you want along the diagonal. Since $AA^T$ is positive semi-definite, you can't get negative values along the diagonal.

Arthur
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No, because $AA^T$ cannot have full rank.

Matt Samuel
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This is not true only because we chose $m>n$. Consider that any diagonal matrix is invertible (it will clearly have a non-zero determinant), and hence represents a sort of (linear) bijective map between two vector spaces. But we know that if the composition of two maps is bijective, then we know that the first map must be injective. We know from the rank-nullity theorem that the linear map represented by matrix multiplication is injective (i.e., has a null kernel) if the rank of the matrix is equal to its dimension, but clearly an $m\times n$ matrix where $m>n$ has rank less than its dimension and is not injective, so $AA^T$ cannot be invertible.

As a bit of an aside, to see why this would be possible if $m<n$, consider the relatively trivial case of $1\times 2$ and its transpose $2\times 1$:

$\begin{bmatrix}1 & 0\end{bmatrix}\begin{bmatrix} 1 \\ 0\end{bmatrix} = \begin{bmatrix}1 \end{bmatrix}$

Clearly $AA^T = \begin{bmatrix}1\end{bmatrix}$ is trivially diagonal, whereas the matrix $A=\begin{bmatrix}1&0\end{bmatrix}$ is rectangular.

Jack Crawford
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