7
$\begingroup$

I have a matrix $A$ and I introduce $(I+A)^{m},$ where $I$ is the identity matrix of same order with $A$ and $m$ is a positive integer. I want to show that if $(1+ \lambda )^m$ is a simple eigenvalue of $(I+A)^m$, that is, it is not a repeated root of the characteristic polynomial of $(I+A)^m,$ then $ \lambda$ is a simple eigenvalue of $A.$ Any help with this will be highly appreciated.

Edit: what I mean by a simple eigenvalue.

For an arbitrary matrix $A$ of order $n \times n,$ let $ \lbrace \lambda_1,\lambda_2,\lambda_3, \dots ,\lambda_k \rbrace $ be the set of distinct eigenvalues of $A.$ The characteristic polynomial of $A,$ denoted $p(\lambda),$ can be written in the factorized form \begin{equation} p(\lambda) = (\lambda_1-\lambda)^{n_1}(\lambda_2-\lambda)^{n_2} \dotsm (\lambda_k-\lambda)^{n_k}\end{equation} with $n_1+n_2+ \dotsm +n_k = n.$ The exponent $n_i,$ corresponding to each eigenvalue $\lambda_i,$ is called the algebraic multiplicity of $ \lambda_i$ and the dimension of the null space of $ A - \lambda_i I, \dim(N(A - \lambda_i I )),$ is called the geometric multiplicity of $ \lambda_i.$ If $n_i=1$ for some $i \in \lbrace 1, 2, \dots, k \rbrace,$ then $ \lambda_i$ is said to be a simple eigenvalue of $A.$

$\endgroup$
1
  • 1
    $\begingroup$ The $I+A$ is purely ornamental here ; your question can be rephrased equivalently (and more simply) as : show that if $\mu^m$ is a simple eigenvalue of $B^m$, then $\mu$ must be a simple eigenvalue of $B$. $\endgroup$ Commented Jan 5, 2013 at 12:59

3 Answers 3

Reset to default
2
$\begingroup$

Here is the “algebraic” version of Ofir’s answer.

In that case also, it is easier to show the contraposite : if $\lambda$ is a repeated ALGEBRAIC eigenvalue of $A$, then $(1+\lambda) ^m$ is a repeated ALGEBRAIC eigenvalue for $(I+A)^m$.

So suppose that $\lambda$ is a repeated algebraic eigenvalue of $A$. The matrix $B=A-\lambda I$ has determinant zero, so its kernel is nontrivial : there is a nonzero $v_1$ such that $Bv_1=0$. Let $C$ be the lower rightmost $(n-1)\times(n-1)$ square submatrix of $A$ defined by $C=(c_{ij})_{1 \leq i ,j \leq n}$ with $c_{ij}=a_{i+1,j+1}$. Then $\chi_A(t)=(t-\lambda)\chi_C(t)$ where $\chi$ denotes the characteristic polynomial. We deduce that $\lambda$ is also an eigenvalue of $C$, so there is a $v_2$ independent from $v_1$ such that $Cv_2=0$. Therefore $Bv_2$ is of the form $cv_1$, for some constant $c$ : in other words $Av_1=\lambda v_1$ and $Av_2=\lambda v_2+cv_1$.

So $$ A^2v_2=A(\lambda v_2+cv_1)=\lambda Av_2+cAv_1=\lambda^2v_2+(2c\lambda)v_1 \tag{1} $$ More generally, we have by induction that $A^kv_2=\lambda^kv_2+(ck\lambda^{k-1})v_1$ for any $k\geq 0$. It follows that

$$P(A)v_2=P(\lambda)v_2+c\frac{\partial P}{\partial \lambda}(\lambda)v_1 \tag{2} $$ for any polynomial $P$.

Completing the pair $(v_1,v_2)$ in a basis $(v_1,v_2, \ldots ,v_n)$ of ${\mathbb K}^n$, and working in this new basis, we see that $P(A)$ is of the form

$$ P(A)=\left( \begin{array}{ccl} P(\lambda) & c\frac{\partial P}{\partial \lambda}(\lambda) & \ldots \\ 0 & P(\lambda) & \ldots \\ \mathbf O & \mathbf O & \ldots \\ \end{array} \right) $$

So $P(\lambda)$ is always a repeated algebraic eigenvalue of $P(A)$, for any polynomial $P$.

$\endgroup$
1
$\begingroup$

EDIT: This solution is wrong, as shown in the comments.

In general, $Av=\lambda v \implies p(A)v=p(\lambda)v$ for any polynomial $p$. This implication shows that the multiplicity of $p(\lambda)$ is at least the multiplicity of $\lambda$, since any eigenvector of $\lambda$ translates to an eigenvector of $p(\lambda)$. One consequence is that if $p(\lambda)$ is simple, $\lambda$ must also be simple. (Assuming $\lambda$ is an eigenvector of $A$.)

In your case, $p(x)=(1+x)^m$.

Proof of $Av=\lambda v \implies p(A)v=p(\lambda)v$: By linearity, it is enough to prove for $p(x)=x^k$. This follows my applying $A$ to both sides of $Av=\lambda v$ $k-1$ times.

$\endgroup$
2
  • $\begingroup$ My understanding is that the implication, mentioned in the first line, shows that the geometric multiplicity of $p( \lambda)$ is at least the geometric multiplicity of $ \lambda.$ But simple eigenvalue is defined by the algebraic multiplicity, so how could we get the consequence that if $p( \lambda)$ is a simple, $\lambda$ must also be simple? $\endgroup$ Commented Jan 5, 2013 at 12:06
  • $\begingroup$ @Zizo : you would do well to edit your original post and explain those geometric/algebraic subtleties. I think most people understood your question the way Ofir did, which is not what you intended. $\endgroup$ Commented Jan 5, 2013 at 12:50
1
$\begingroup$

As someone (I forgot who) pointed out in a deleted answer, your statement, as it stands, is false. Although every eigenvalue $\lambda$ of $A$ gives rise to an eigenvalue $(1+\lambda)^m$ of $(I+A)^m$, the converse is not true because the mapping $\lambda\mapsto(1+\lambda)^m$ is not one-to-one. For instance let $A=-3,\ \lambda=1$ and $m=2$. Then $(1+\lambda)^2=4$ is an eigenvalue of $(I+A)^2=4$, but $\lambda=1$ is not an eigenvalue of $A=-3$.

So, assume that $\lambda$ is an eigenvalue of $A$. Let $v$ be a corresponding eigenvector. By a change of basis, we may assume that $A$ and $(I+A)^m$ are of the form $$ A=\begin{pmatrix}\lambda&u^T\\0&C\end{pmatrix}, \ (I+A)^m=\begin{pmatrix}(1+\lambda)^m&v^T\\0&(I+C)^m\end{pmatrix}, $$ where $C$ is an $(n-1)\times(n-1)$ matrix. Since $\det(xI-A)=(x-\lambda)\det(xI-C)$, if $\lambda$ is a repeated eigenvalue of $A$, it must be an eigenvalue of $C$ and hence $1+\lambda$ is a repeated eigenvalue of $(I+C)^m$, thus making $(1+\lambda)^m$ a repeated eigenvalue of $(I+A)^m$, which is a contradiction.

$\endgroup$
5
  • $\begingroup$ Great! could you explain a contradiction with what? $\endgroup$ Commented Jan 6, 2013 at 15:46
  • $\begingroup$ @Zizo A contradiction to the assumption that $(1+\lambda)^m$ is a simple eigenvalue of $(I+A)^m$. $\endgroup$ Commented Jan 6, 2013 at 16:39
  • $\begingroup$ I wanted to know that if $(1+ \lambda)^m$ is a simple eigenvalue of $(I+A)^m,$ then $ \lambda$ is a simple eigenvalue of $A$ which you did show using a contrapositive argument, that is, if $ \lambda$ is repeated eigenvalue of $A, (1+ \lambda)^m$ is repeated eigenvalue of $(I+A)^m.$ $\endgroup$ Commented Jan 7, 2013 at 2:23
  • $\begingroup$ @Zizo Exactly. So we have a proof for your statement, haven't we? $\endgroup$ Commented Jan 7, 2013 at 8:22
  • $\begingroup$ YES. \=D/ .. THANKS. $\endgroup$ Commented Jan 7, 2013 at 8:57

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.