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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The characteristic polynomial, defined for a matrix $A$:

$ c(x; A) = \det (A-I x ) = 0 $

has nice properties related to the eigenvalues $\lambda$, of the matrix:

$ c(x; A) = (x-\lambda_1)(x-\lambda_2)(x-\lambda_3) \ldots $

What, if any, is the connection to the eigenvalues of a matrix to this function:

$ p(x; A) = \det (x A-I ) $

share|cite|improve this question
Do you mean $Ax-I$? – Alex Becker Mar 17 '12 at 1:26
@AlexBecker does it matter? Isn't $Ax=xA$ for any scalar, or does the notation imply something different? – Hooked Mar 17 '12 at 1:40
I suppose its entirely a question of notation, that is just one I rarely if ever see. – Alex Becker Mar 17 '12 at 1:42
up vote 3 down vote accepted

If $x$ is a root of $p(x;A)$, then $Ax-I$ is not invertible, and therefore has a nonzero null space. So there exists a nonzero vector $\vec{v}$ such that $$(Ax-I)\vec{v}=\vec{0}$$

So $Ax\vec{v}=\vec{v}$. Note that $x$ cannot be zero for two reasons - we have specified $\vec{v}\neq\vec{0}$ and $\det(A\cdot0-I)\neq0$. We then have that $A\vec{v}=x^{-1}\vec{v}$, and $x^{-1}$ is an eigenvalue for $A$.

So the roots of $p(x;A)$ are inverses of the nonzero eigenvalues of $A$. (If $A$ has zero as an eigenvalue, then $p(x;A)$ has degree less than $n$.)

A more complete picture added later:

If the Jordan canonical form of $A$ (over $\mathbb{C}$) is given by $A=PJP^{-1}$, with $J$ a composite of Jordan blocks, then $$ \begin{align} \det(Ax-I) & =\det(PJP^{-1}x-I)\\ &=\det(PJP^{-1}x-PIP^{-1})\\ &=\det(P)\det(Jx-I)\det(P^{-1})\\ &=\det(Jx-I) \end{align} $$ so let's assume that $A$ is already in its Jordan canonical form. Some of the Jordan blocks of $A$ have eigenvalue $0$ and some do not. Write $$A=\begin{bmatrix} Z & 0\\ 0 & Y \end{bmatrix} $$ where $Z$ has the Jordan blocks with eigenvalue $0$, and $Y$ has the other (nonzero eigenvalued) Jordan blocks. It's important to understand that $Z$ has $0$'s everywhere except for some $1$s at selected places along the $+1$-off-diagonal. Then $$ \begin{align} \det(Ax-I)&=\det\left(\begin{bmatrix}Zx & 0\\0 & Yx\end{bmatrix}-I\right)\\ &=\det(Zx-I_{z\times z})\det(Yx-I_{y\times y})\\ &=(-1)^z\det(Yx-I_{y\times y})\\ &=(-1)^zp(x;Y) \end{align} $$ where $z$ is the multiplicity of $0$ as an eigenvalue of $A$ and $y$ is the complement: $n-z$.

Since $Y$ is invertible, Alex Becker's answer can be applied. In summary: $$\begin{align}p(x;A)&=(-1)^zp(x;Y)\\&=(-1)^z(-1)^y\det(Y)c(x;Y^{-1})\\&=(-1)^n\det(Y)c(x;Y^{-1})\end{align}$$ That is, $p(x;A)$ is a certain multiple of the characteristic polynomial of $Y^{-1}$, where $Y$ is the composite of $A$'s invertible Jordan blocks. Said one more way, $p$ is a polynomial whose roots are the inverses of $A$'s nonzero eigenvalues, and the multiplicities are respected.

share|cite|improve this answer
Great answer(s) Alex(s), I learned a great deal. Thanks! – Hooked Mar 19 '12 at 5:24

If $A$ is an $n\times n$ invertible matrix, then $$\begin{eqnarray} p(x;A)&=&\det(Ax-I)\\ &=&\det(A)\det(Ix-A^{-1})\\ &=&(-1)^n\det(A)\det(A^{-1}-Ix)\\ &=&(-1)^n\det(A)c(x;A^{-1})\end{eqnarray}$$ and so the roots are the eigenvalues of $A^{-1}$, which are the inverses of the eigenvalues of $A$.

share|cite|improve this answer
Also, I think $p(x; A)$ is the polynomial reverse of $c(x; A)$ up to sign difference. i.e. $$c(x; A) = (-1)^n x^n p(\frac{1}{x}; A)$$ but I have to verify it. – user2468 Mar 17 '12 at 1:40
What if $A$ is not invertible? – alex.jordan Mar 17 '12 at 5:29

Your Answer


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.