Eigenvalues and null space I would like to understand the relationships between the null space and the eigenvalues of a matrix better.
First of all, we know that an $n \times n$ matrix will have $n$ eigenvalues, though the eigenvalues can be complex and repeated.
Next, we know that if $A$ has the eigenvalue 0, then the corresponding eigenvector is in the null space $N(A)$, since $A\textbf{x}=0\textbf{x}=\textbf{0}$. This implies that all eigenvectors that correspond to the eigenvalue 0 exactly span $N(A)$.
Using the above-mentioned two conclusions, and assume we have an $n \times n$ matrix with rank $r$, now we know the dimension of the null space is $n-r$. From this, can we conclude that there will be at least $n-r$ eigenvalues that equal to 0? and exact $n-r$ independent eigenvectors to span the null space?
 A: If $A$ has full rank, then the dimension of the null space is exactly $0$.
Now, if $A_{n×n}$ has rank $r\lt n $, then the dimension of the null space $=(n-r)$. This $(n-r)$ will be the geometric multiplicity of the eigenvalue $0$.
But we know that, algebraic multiplicity $\ge$ geometric multiplicity.
So, algebraic multiplicity of eigenvalue $0$ should be at least $(n-r)$. This means that there will be at least $(n-r)$ numbers of $0$'s, as the eigenvalues of $A$.
And, since geometric multiplicity of an eigenvalue $=$ the number of linearly independent eigenvectors corresponding to that eigenvalue, we can conclude that there are exactly $(n-r)$ numbers of linearly independent eigenvectors corresponding to the eigenvalue $0$.
A: Given a matrix $A\in\mathbb{R}^{n\times n}$:

*

*A vector $x$ is an eigenvector of $A$ if $Ax = \lambda x$ where $\lambda$ is the eigenvalue.


*The kernel (null space) of $A$ is the set $\{v | Av=0\}$, i.e., all $v$ that have an eigenvalue $0$.


*The eigenspace, $E_{\lambda}$, is the null space of $A-\lambda I$, i.e., $\{v | (A-\lambda I)v = 0\}$. Note that the null space is just $E_{0}$.


*The geometric multiplicity of an eigenvalue $\lambda$ is the dimension of $E_{\lambda}$, (also the number of independent eigenvectors with eigenvalue $\lambda$ that span $E_{\lambda}$)


*The algebraic multiplicity of an eigenvalue $\lambda$ is the number of times $\lambda$ appears as a root to $det(A-x I)$.


*algebraic multiplicity $\geq $ geometric multiplicity.
Consider the following Example, $A = \begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}$.
Then $n = 2$ and the rank of $rank(A) = 1$. The $det(A-x I) = x^{2}$ and the roots are $x = \{0,0\}$. We see that the eigenvalue $0$ has algebraic multiplicity $2$. But, the geometric multiplicity is the dimension of $E_{0} = span\left(\begin{bmatrix}1 \\ 0\end{bmatrix}\right)$ which is $1$. So from this example we see that $n-r = 1$, which is equal to the geometric multiplicity of $\lambda = 0$.
Therefore, we conclude that $\lambda = 0$ will have an algebraic multiplicity of at least $n−r$ and a geometric multiplicity of $n−r$. This is obvious from the definition of rank and geometric multiplicity.
