Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\det A =0$?

share|improve this question
4  
The matrix formed by the outer product of two vectors has rank 1... –  J. M. Oct 1 '11 at 12:30
12  
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
1  
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

1 Answer 1

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 $\sigma\to\tau\sigma$).

share|improve this answer
3  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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