# How many free variables can there be for an $n\times n$ matrix when solving for an eigenspace?

let's say you have an nxn matrix, call it $A$. If $A =$

$$\begin{matrix} a^1_1 & a^1_2 & \cdots & \cdots & \cdots & a^1_n \\ a^2_1 & a^2_2 & \cdots & \cdots & \cdots & a^2_n \\ \vdots & \vdots & \ddots & & & \vdots \\ \vdots & \vdots & & \ddots & & \vdots \\ \vdots & \vdots & & & \ddots & \vdots \\ a^n_1 & a^n_2 & \cdots & \cdots & \cdots & a^n_n \end{matrix}$$

then what is the dimension of the eigenspace? I know how to find the eigenvalues for any matrix, no matter whether those eigenvalues are real or complex, but, when you're solving for the eigenspace, you get at least one free variable. My question is, for an nxn matrix, how many free variables are there allowed to be when solving the equation

$A-\lambda I = 0$ where $\lambda$ is an eigenvalue? Is the number of free variables $n$? $n-1$? Thanks!

• I think the answer you're looking for is the following: The number of free variables in $\det(A - \lambda I) = 0$ is equal to the dimension of the null space of $A - \lambda I$. – Tom Apr 23 '17 at 21:07
• The dimension of an eigenspace is no greater than the algebraic multiplicity of the associated eigenvalue. E.g., if $\lambda$ has algebraic multiplicity $3$ then you might , two or three free variables. – amd Apr 24 '17 at 5:47

## 1 Answer

Expanding on my comment above: I think the answer you're looking for is the following: The number of free variables in $\det(A - \lambda I) = 0$ is equal to the dimension of the null space of $A - \lambda I$.

Here are some examples to play with: $$A_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ $$A_2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$$ $$A_1 = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$$ Setting $\lambda = 1$, you will find that the dimension of the null space of $A_3 - \lambda I$ is $3$, the dimension of the null space of $A_2 - \lambda I$ is $2$, and the dimension of the null space of $A_1 - \lambda I$ is $1$.

Edit: For any $m \times n$ matrix $B$, interpreted as a linear function (via right multiplication of a column vector), you have $B : \mathbb{R}^n \to \mathbb{R}^m$. The null space of $B$ is a subspace of the domain, and hence a subspace of $\mathbb{R}^n$, and therefore has dimension bounded above by $n$ and bounded below by $0$ (where the $0$ case means that the only vector ${\bf v}$ in $\mathbb{R}^n$ such that $B{\bf v} = {\bf 0}_m$ is ${\bf v} = {\bf 0}_n$).

• okay, so the dimension of A-$\lambda$I is at most $n$ then? – Matthew Graham Apr 23 '17 at 21:15
• @MatthewGraham See my edit! – Tom Apr 23 '17 at 21:20
• ah, I see it now, Thanks for all your help! I get it now. – Matthew Graham Apr 23 '17 at 21:22