# 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. – Hans Lundmark Jun 3 '14 at 12:46
• @HansLundmark What is the difference between "infinite" and "infinitely many" ? – dexterdev Jun 4 '14 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. – Hans Lundmark Jun 4 '14 at 14:32
• me too meant infinitely many eigen vectors, language problem :D . – dexterdev Jun 5 '14 at 11:40

## 2 Answers

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 $\mathbb{R}^N$.

This answers your first question, but let me point out an error in your thinking.

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. – dexterdev Jun 3 '14 at 9:41
• No worries. Let me know if there is anything that you want me to clarify. – Michael Albanese Jun 3 '14 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? – dexterdev Jun 3 '14 at 9:16
• Yes, thats right. – Ellya Jun 3 '14 at 9:20
• still the case of identity matrix is different, it have infinite family of eigen vectors, right? – dexterdev Jun 3 '14 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) – Ellya Jun 3 '14 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 $a,b\in\Bbb R$ – Ellya Jun 3 '14 at 9:32