Do $N \times N$ identity matrices have N identical eigen values and infinite eigen vectors?

Suppose I have a $N \times N$ square matrix $A$. We know that for any matrix $A$, we have $N$ eigen values and corresponding eigen vectors. So a $2 \times 2$ matrix have two eigen vectors and corresponding eigen values. Now I am in a confusion. Its regarding identity matrices. For those matrices any vector can be eigen vector right? So there is a possibility of infinite eigen vectors. But what about eigen values? They are still $\lambda_1 =1 , \lambda_2 =1$ ,right?

EDIT : I am considering vectors $(x,y)$ and $(a.x,a.y)$ as same family. $(a.x,a.y)$ = $a \times (x,y)$, where a is a scalar.

• There are no "infinite eigenvectors", but infinitely many eigenvectors. Jun 3, 2014 at 12:46
• @HansLundmark What is the difference between "infinite" and "infinitely many" ? Jun 4, 2014 at 8:14
• "Infinite" means "infinitely big". For example, a set can be finite or infinite. "Infinite sets" and "infinitely many sets" are completely different things. To say that an eigenvector is infinite doesn't really make sense. Jun 4, 2014 at 14:32
• me too meant infinitely many eigen vectors, language problem :D . Jun 5, 2014 at 11:40

If an $N\times N$ matrix $A$ has an eigenvalue $\lambda$, there are infinitely many vectors $v$ satisfying $Av = \lambda v$; that is, $A$ has infinitely many eigenvectors for the eigenvector $\lambda$. In fact,

$$E_{\lambda} := \{v \in \mathbb{R}^N \mid Av = \lambda v\}$$

is a subspace of $\mathbb{R}^N$.

When we say $A$ has $N$ eigenvalues, we mean that the characteristic equation for $A$, $|\lambda I - A| = 0$, has $n$ zeroes; this is always true by the fundamental theorem of algebra, but some of the eigenvalues may be complex (e.g. a $2\times 2$ rotation matrix). Note, we count eigenvalues with multiplicity, so $\lambda$ could be repeated multiple times. We call the order of the zero $\lambda$ the algebraic multiplicity of $\lambda$.

For any eigenvalue $\lambda$, some say that $\lambda$ has $k$ corresponding eigenvectors if $\dim E_{\lambda} = k$ (this terminology is not often defined but is instead used in verbal communication). If $A$ has distinct eigenvalues $\lambda_1, \dots, \lambda_M$, one might say that $A$ has $\dim E_{\lambda_1} + \dots + \dim E_{\lambda_M}$ corresponding eigenvectors. The dimension of $E_{\lambda}$ is called the geometric multiplicity of $\lambda$.

We have the following result relating the two notions of multiplicity:

The geometric multiplicity is less than or equal to the algebraic multiplicity.

There are cases where the geometric multipicity of $\lambda$ is strictly less than the algebraic multiplicity, and therefore $A$ has less than $N$ corresponding eigenvectors. For example,

$$A = \left[\begin{matrix}1 & 1\\ 0 & 1\end{matrix}\right]$$

has a repeated eigenvalue of $1$ but only has a one-dimensional eigenspace.

To reconcile the difference between geometric multiplicity and algebraic multiplicity, one can consider generalised eigenvectors.

• This will take some time for me to completely understand, but I appreciate your effort. Jun 3, 2014 at 9:41
• No worries. Let me know if there is anything that you want me to clarify. Jun 3, 2014 at 10:21

there are always $n$ eigenvalues for an $n\times n$ matrix, but there are infinite eigenvectors.

• you mean scalar multiples of $n$ eigen vectors.If I am considering $(x , y)$ and $(a.x , a.y)$ as same family, there are only two families of eigen vectors for a $2 \times 2$ matrix, right? Jun 3, 2014 at 9:16
• Yes, thats right. Jun 3, 2014 at 9:20
• still the case of identity matrix is different, it have infinite family of eigen vectors, right? Jun 3, 2014 at 9:23
• in that case, the families would be $(1,1)^T$, $(0,1)^T$, $(1,0)^T$ (of course taking all scalar multiples) Jun 3, 2014 at 9:24
• I realised you are correct, there are an infinite set of families, because $(a,b)^T$ is an eigenvector, for all combinations of $a,b\in\Bbb R$ Jun 3, 2014 at 9:32