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 $A.$

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    $\begingroup$ 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 $B$. $\endgroup$ – Ewan Delanoy Jan 5 '13 at 12:59

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 $P$.


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.

  • $\begingroup$ 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? $\endgroup$ – Zizo Jan 5 '13 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. $\endgroup$ – Ewan Delanoy Jan 5 '13 at 12:50

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.

  • $\begingroup$ Great! could you explain a contradiction with what? $\endgroup$ – Zizo Jan 6 '13 at 15:46
  • $\begingroup$ @Zizo A contradiction to the assumption that $(1+\lambda)^m$ is a simple eigenvalue of $(I+A)^m$. $\endgroup$ – user1551 Jan 6 '13 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 $(I+A)^m.$ $\endgroup$ – Zizo Jan 7 '13 at 2:23
  • $\begingroup$ @Zizo Exactly. So we have a proof for your statement, haven't we? $\endgroup$ – user1551 Jan 7 '13 at 8:22
  • $\begingroup$ YES. \=D/ .. THANKS. $\endgroup$ – Zizo Jan 7 '13 at 8:57

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