# Can an eigenvector contain free variables?

I have the matrix $$B=\begin{pmatrix}-6&4&-17\\0&-3&-4\\0&2&3\\\end{pmatrix}$$ which I found to have eigenvalues $$\lambda_1=-6, \lambda_2=-1, \lambda_3=1$$.

When attempting to find the eigenvectors for $$\lambda_2$$ and $$\lambda_3$$ I found the solution to contain free variables and I was unsure whether an eigenvector can contain free variables or not.

Thanks.

• What do you mean by "free variables"? Note that you should get a one-dimensional eigenspace (after adding $0$) corresponding to each eigenvalue. So your answer would be of the form $tv$ for some fixed $v \neq 0$ and $t \in \mathbb{R}\setminus\{0\}.$ Jan 5, 2020 at 13:40

Notice that if $$v$$ is an eigenvector, then for any non-zero number $$t$$, $$t\cdot v$$ is also an eigenvector.

If this is the free variable that you refer to, then yes.

Edit:

In general, if $$v_i\neq 0$$ satisfies $$Av_i = \lambda v_i$$, then $$A\left( \sum_{i=1}^k \alpha_i v_i\right) = \lambda \left( \sum_{i=1}^k \alpha_i v_i\right)$$

That is if $$\sum_{i=1}^k \alpha_i v_i \ne 0$$, then it is an eigenvector with respect to the same eigenvalue.

• And, of course, in generalm the eigenspace might be of dimension greater than one, in which case there are more than one "free variable". Jan 5, 2020 at 22:13

There seems to be some confusion regarding the terminology.

Recall that, given a matrix $$A$$, an eigenvector is a non-zero vector $$v$$ such that $$Av = \lambda v$$ for some number $$\lambda$$, the corresponding eigenvalue.

Hence, an eigenvector is, fundamentally, nothing but an ordinary vector -- it might be $$(1, 4, -2)$$ or $$(1, 0, 0)$$ or $$(0, 1, -1)$$. Hence, a given eigenvector doesn't "contain" anything but its coordinates, which are numbers.

However, given an eigenvalue, there are always infinitely many eigenvectors corresponding to that eigenvalue. Indeed, if $$v$$ is an eigenvector associated with the eigenvalue $$\lambda$$, then so is $$t v$$ for every non-zero value of $$t$$. This is easy to prove (and a proof is given in Wizact's answer).

In general, the set of all eigenvectors corresponding to a given eigenvalue, together with the zero vector, is called the eigenspace of that eigenvalue. This is a vector space itself. It might be one-dimensional, or it might be of any higher dimension.

For instance, consider the matrix

$$A=\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\\\end{pmatrix}.$$

This corresponds to a linear transformation $$\mathbf{R}^3 \to \mathbf{R}^3$$ which is particularly easy to visualise: it is orthogonal projection onto the $$xy$$ plane. So each vector in $$\mathbf{R}^3$$ is mapped to its shadow on the $$xy$$ plane when illuminated from above or below (put informally):

Geometrically, it is clear that each vector already in the $$xy$$ plane, that is, each vector of the form $$(x, y, 0)$$, is mapped to itself. Hence, every such vector besides the zero vector is an eigenvector corresponding to the eigenvalue $$1$$. Also, no other vector is mapped to itself.

It is also clear that each vector on the $$z$$ axis, that is, each vector of the form $$(0, 0, z)$$, is mapped to the zero vector. Thus, each non-zero such vector is an eigenvector with eigenvalue $$0$$. Also, no other vector is mapped to the zero vector.

We therefore say that the $$xy$$ plane is the eigenspace corresponding to the eigenvalue $$1$$. This is clearly a two-dimensional vector space. A basis in this vector space is $$\left\{(1, 0, 0), (0, 1, 0)\right\}$$, so every vector in this eigenspace is a linear combination of these basis vectors. (More precisely, every vector in this space can be written as a linear combination of these basis vectors in a unique way.) Hence, every element in this space is of the form

$$s (1, 0, 0) + t (0, 1, 0)$$

for some coordinates $$s$$ and $$t$$. Every non-zero vector in this eigenspace is an eigenvector (for this eigenvalue).

Examples of eigenvectors: $$(1, 0, 0)$$, $$(0, 1, 0)$$, $$(1, 1, 0)$$, $$(7, -4, 0)$$.

Similarly, the $$z$$ axis is the eigenspace corresponding to the eigenvalue $$0$$. This is a one-dimensional vector space. A basis in this vector space is $$\left\{ (0, 0, 1) \right\}$$, so every vector in this eigenspace is a linear combination of this vector. Hence, every vector in this space is of the form

$$t (0, 0, 1)$$

for some coordinate $$t$$. Every non-zero vector in this eigenspace is an eigenvector (for this eigenvalue).

Examples of eigenvectors: $$(0, 0, 1)$$, $$(0, 0, -7)$$, $$(0, 0, \pi)$$, $$(0, 0, \mathrm{arcsinh}(0.3)^\pi)$$.

Hence, the expression for general vector in a given eigenspace -- typically, as a linear combination of the vectors in a basis for this eigenspace -- contains one or more variables, but each eigenvector is a pure numeric entity.

Yes. In fact, you always get (at least) one free variable when you are finding eigenvectors. This is because if $$\lambda$$ is an eigenvalue of $$B$$, then by definition $$|B - \lambda I| = 0,$$ so when you do row reduction on this matrix, you will always get at least one row of zeros. As Siong pointed out, that corresponds to the fact that if $$v$$ is an eigenvector corresponding to $$\lambda$$, then $$t \cdot v$$ is also an eigenvector corresponding to the same eigenvalue, for any $$t \neq 0$$. Indeed: $$B (t \cdot v) = t \cdot Bv = t \cdot \lambda v = \lambda (t \cdot v).$$