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Let $a_1,\ldots,a_n$ be complex numbers and let $b_1,\ldots,b_n$ be complex numbers.

Let $A$ be the matrix whose $(i,j)$-th entry is $A_{ij} = a_i b_j$. Then I think $\det A = 0$ when $n>1$. This is easy to compute when $n=2$.

Question. Why is $\det A =0$?

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The matrix formed by the outer product of two vectors has rank 1... – J. M. Oct 1 '11 at 12:30
Isn't it obvious that the rows and columns are linearly dependent? They are multiples of $(a_1,\ldots,a_n)^T$ and $(b_1,\ldots,b_n)$ respectively. – t.b. Oct 1 '11 at 12:31
Use the multi-linearity of the determinant: $\det A=b_1\ldots b_n\det A'$ where $A'_{ij}=a_i$, and for $n>1$ $A'$ is a matrix whose columns are equal hence... – Davide Giraudo Oct 1 '11 at 12:33
Aside: This is a good way to see that the image of the Segre embedding is closed. – Dylan Moreland Oct 1 '11 at 13:02
Gooz/shaye: may I suggest registering your account? :) Then all your questions are in one place, and you don't have the trouble of multiple accounts... – J. M. Oct 1 '11 at 14:57
up vote 9 down vote accepted

Algebraically (mentioned by t.b.): The columns of the matrix are $b_1\mathbf{a}$, $b_2\mathbf{a}$, $\dots, b_n\mathbf{a}$. If all of $\mathbf{b}$'s components are zero then the result is trivial, while if at least one of the components is nonzero, say $b_1$ without loss of generality, then (for $n>1$)

$$\color{Blue}{-(b_2+b_3+\cdots+b_n)}(b_1\mathbf{a})+\color{Blue}{b_1}(b_2\mathbf{a})+\color{Blue}{b_1}(b_3\mathbf{a})+\cdots+\color{Blue}{b_1}(b_n\mathbf{a})$$ $$=\big((\color{Blue}{-b_2}b_1+\color{Blue}{b_1}b_2)+(\color{Blue}{-b_3}b_1+\color{Blue}{b_1}b_3)\cdots+(\color{Blue}{-b_n}b_1+\color{Blue}{b_1}b_n)\big)\mathbf{a}$$ $$=(0+0+\cdots+0)\mathbf{a}=\mathbf{0}.$$ is a nontrivial linear combination of the matrix columns that evaluates to zero, hence the columns are linearly dependent and thus the matrix's determinant is zero.

Shortcut (via David): Use the multlinearity of the determinant to reduce $$\det\begin{pmatrix}b_1\mathbf{a}&b_2\mathbf{a}&\cdots&b_n\mathbf{a}\end{pmatrix}=b_1b_2\cdots b_n \det\begin{pmatrix}\mathbf{a}&\mathbf{a}&\cdots&\mathbf{a}\end{pmatrix}.$$ For $n>1$, it is impossible for $n$ copies of a vector $\mathbf{a}$ to be linearly independent, since e.g. $$1\mathbf{a}+(-1)\mathbf{a}+0\mathbf{a}+\cdots+0\mathbf{a}=\mathbf{0}.$$ Hence the original determinant must also be zero.

Geometrically: The parallelepiped formed by the matrix's columns are all contained in the one-dimensional subspace generated by $\mathbf{a}$: for $n>1$ this has zero $n$-dimensional content (hyper-volume), hence the determinant is zero.

Geometrically/Algebraically: As I posted in a different answer (paraphrased here),

We're assuming $\mathbf{a},\mathbf{b}\ne\mathbf{0}$. Then $\mathbf{b}^\perp$, the orthogonal complement of the linear subspace generated by $\mathbf{b}$ (i.e. the set of all vectors orthogonal to $\mathbf{b}$) is therefore $(n-1)$-dimensional. Let $\mathbf{c}_1,\dots,\mathbf{c}_{n-1}$ be a basis for this space. Then they are linearly independent and $$(\mathbf{a}\mathbf{b}^T)\mathbf{c}_i =\mathbf{a}(\mathbf{b}^T\mathbf{c}_i)= (\mathbf{b}\cdot\mathbf{c}_i)\mathbf{a}=\mathbf{0}.$$ Thus the eigenvalue $0$ has geometric multiplicity $n-1\qquad$ [...]

The determinant is the product of the matrix's eigenvalues, so if one of those is $0$ the product is necessarily zero as well.

Analytically/Combinatorially: Via Leibniz formula we have $$\det(\mathbf{a}\mathbf{b}^T)=\sum_{\sigma\in S_n}(-1)^{\sigma}\prod_{k=1}^na_{\sigma(k)}b_k$$ $$=\left(\sum_{\sigma\in S_n}(-1)^\sigma\right)\prod_{k=1}^na_kb_k=0.$$ Above we observe that $\sum (-1)^{\sigma}$ is zero because the permutations of even and odd parity are in bijection with each other (e.g. take an arbitrary transposition $\tau$ and define the map $\sigma\to\tau\sigma$).

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Wow. Way to put that to bed. There is one more that might be mentioned using rank. $A=a^Tb$ and the rank of $a$ is $1$, so the rank of $A$ has to be $1$, so $\det(A)=0$ unless it is a $1\times1$ matrix. – robjohn Oct 1 '11 at 13:31

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