This question aims to create an "abstract duplicate" of numerous questions that ask about determinants of specific matrices (I may have missed a few):

The general question of this type is

Let $A$ be a square matrix of rank$~1$, let $I$ the identity matrix of the same size, and $\lambda$ a scalar. What is the determinant of $A+\lambda I$?

A clearly very closely related question is

What is the characteristic polynomial of a matrix $A$ of rank$~1$?


The formulation in terms of the characteristic polynomial leads immediately to an easy answer. For once one uses knowledge about the eigenvalues to find the characteristic polynomial instead of the other way around. Since $A$ has rank$~1$, the kernel of the associated linear operator has dimension $n-1$ (where $n$ is the size of the matrix), so there is (unless $n=1$) an eigenvalue$~0$ with geometric multiplicity$~n-1$. The algebraic multiplicity of $0$ as eigenvalue is then at least $n-1$, so $X^{n-1}$ divides the characteristic polynomial$~\chi_A$, and $\chi_A=X^n-cX^{n-1}$ for some constant$~c$. In fact $c$ is the trace $\def\tr{\operatorname{tr}}\tr(A)$ of$~A$, since this holds for the coefficient of $X^{n-1}$ of any square matrix of size$~n$. So the answer to the second question is

The characteristic polynomial of an $n\times n$ matrix $A$ of rank$~1$ is $X^n-cX^{n-1}=X^{n-1}(X-c)$, where $c=\tr(A)$.

The nonzero vectors in the $1$-dimensional image of$~A$ are eigenvectors for the eigenvalue$~c$, in other words $A-cI$ is zero on the image of$~A$, which implies that $X(X-c)$ is an annihilating polynomial for$~A$. Therefore

The minimal polynomial of an $n\times n$ matrix $A$ of rank$~1$ with $n>1$ is $X(X-c)$, where $c=\tr(A)$. In particular a rank$~1$ square matrix $A$ of size $n>1$ is diagonalisable if and only if $\tr(A)\neq0$.

See also this question.

For the first question we get from this (replacing $A$ by $-A$, which is also of rank$~1$)

For a matrix $A$ of rank$~1$ one has $\det(A+\lambda I)=\lambda^{n-1}(\lambda+c)$, where $c=\tr(A)$.

In particular, for an $n\times n$ matrix with diagonal entries all equal to$~a$ and off-diagonal entries all equal to$~b$ (which is the most popular special case of a linear combination of a scalar and a rank-one matrix) one finds (using for $A$ the all-$b$ matrix, and $\lambda=a-b$) as determinant $(a-b)^{n-1}(a+(n-1)b)$.

  • $\begingroup$ Does this method always finds all the eigenvalues? $\endgroup$ – shinzou Feb 6 '16 at 12:26
  • $\begingroup$ @kuhaku: Yes, the set of roots of the characteristic polynomial gives you all the eigenvalues. $\endgroup$ – Marc van Leeuwen Feb 7 '16 at 20:20

Here’s an answer without using eigenvalues: the rank of $A$ is $1$ so its image is spanned by some nonzero vector $v$. Let $\mu$ be such that $$Av=\mu v.$$

We can extend this vector $v$ to a basis of $\mathbb{C}^n$. With respect to this basis now, we have that the matrix of $A$ has all rows except the first one equal to $0$. Since determinant and trace are basis-independent it follows by expanding the first column of $A$ wrt to this basis that $$\det(A-\lambda I)= (-1)^n(\lambda -\mu)\lambda^{n-1}.$$ Using this same basis as above we also see that $\text{Tr}(A) =\mu$, so the characteristic polynomial of $A$ turns out to be

$$(-1)^n(\lambda -\text{Tr}(A))\lambda^{n-1}.$$

  • $\begingroup$ But... $\mu$ is eigenvalue of matrix $A$ by definition...? $\endgroup$ – IPPK Jul 6 '20 at 13:04
  • $\begingroup$ Yea true I think what I meant to say is: this is a proof not using any theorems about the relation between eigenvectors and the characteristic polynomial $\endgroup$ – M. Van Jul 6 '20 at 13:05
  • $\begingroup$ Oh, okay. Anyway, your answer is very good! $\endgroup$ – IPPK Jul 6 '20 at 13:07
  • $\begingroup$ This is a nice answer (except that you use the wrong definition of characteristic polynomial, which is $\det(IX-A)$ <rant> no matter how many teachers/textbooks say otherwise; being a monic polynomial might not be relevant when one is just looking for roots, but it is relevant in many other contexts</rant>). Maybe you could say that all other basis vectors have their image a multiple of $v$, which is what makes those rows after the first all zero, just for ease of understanding. Alternatively extend a basis of the $n-1$ dimensional kernel to a full basis; all columns but the last will be zero. $\endgroup$ – Marc van Leeuwen Jul 6 '20 at 13:40

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