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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 ) $

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Do you mean $Ax-I$? –  Alex Becker Mar 17 '12 at 1:26
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@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

2 Answers 2

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.

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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 $A$.

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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
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What if $A$ is not invertible? –  alex.jordan Mar 17 '12 at 5:29

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