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I have a vector $x$ of dimension $N \times 1$ and let's say I create a matrix $S = x x'$ which a matrix of dimension $N \times N$.

If I calculate the determinant of $S$, I get it as $0$. Is this a property / known theorem ? Does it have name? Is there a proof?

uday
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  • Geometrically, your matrix acts as a projection onto the line spanned by $x$. So obviously the determinant is zero (it sends the orthogonal complement of your vector to zero and so isn't injective), unless you're in one dimension. – symplectomorphic Aug 31 '15 at 17:38

2 Answers2

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The rank of the matrix $xx'$ is $1$ because all of its columns are scalar multiples of just one column, $x$, so just one column spans the column space. As with columns, in this case so also with rows: they are all scalar multiples of just one row vector, $x'$. If the rank of an $n\times n$ matrix is less than $n$, then its determinant is $0$.

Michael Hardy
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This is the determinant of the matrix with $x_i x_j$ as element in row $i$, column $j$: \begin{align} \mbox{det } x x^t &= \mbox{det }(x_i x_j) \\ &= \left(\prod x_i^n\right) \underbrace{\mbox{det } (x_j)}_0 \\ &= \left(\prod x_i^n\right)^2 \underbrace{\mbox{det } (1)}_0 \\ &= 0 \end{align}

The determinant is a multilinear form, so we can pull $n$ times row multiples $x_i$ (leaving $n$ equal rows) and then column multiples, leaving a matrix with all elements being 1.

mvw
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