# Given a vector can we find a set of Operators for which the vector is eigenvector with same eigenvalue.

Suppose we are given any arbitrary vector (say in $$\mathbb{C}^2$$ for simplicity), now I want to understand the set of (linear) operators for which the given vector is an eigenvector such that all eigenvalues are the same.

i.e. Given vector $$v$$ the set $$S=\{T \mid T \; \text{is an operator acting on v}\}$$ such that $$Tv=\lambda v$$ $$\forall \; T \in S .$$

It would be helpful if you can provide me relevant materials to understand this concept properly.

• One way to proceed, is to note that a linear operator acting on $\mathbb C^2$ is completely determined by a matrix of 4 complex numbers... May 28, 2021 at 5:22
• @CalvinKhor Can we show that the collection of such kinds of operators uniquely determines the vector? May 28, 2021 at 10:06
• No, you have no way to restrict eg the euclidean norm to $|v|$ instead of eg $2|v|$ May 28, 2021 at 11:31

It's an affine subspace of the space of matrices. Say $$v$$ is a non-zero vector of an $$n$$-dimensional space. The space of all matrices $$T$$ such that $$Tv = 0$$ is an $$n^2-n$$ dimensional linear subspace of the space of $$n\times n$$ matrices, because it's the kernel of a map from an $$n\times n$$-dimensional space onto an $$n$$-dimensional space. For given $$\lambda$$, the space you're looking at is an affine translation $$\lambda I + T$$ for $$T$$ such that $$Tv=0$$.

For $$\lambda=0$$, your space is the annihilator of $$v$$ (see here). You can recover the span of $$v$$ by taking annihilator again (and identifying the double dual with the original space with the canonical map).

That means that $$\{w \mid T(w) = 0\ \forall T\in S\}=span(v)$$.

This could probably be generalized to the affine case ($$\lambda\neq 0$$) using some tricks.

=== EDIT ===

In fact, if you know $$\lambda$$, since $$S_\lambda = S_0 + \lambda I$$. You can get $$S_0$$ from $$S_\lambda$$ (by subtracting $$\lambda I$$), and then recover the span of $$v$$ from $$S_0$$.

Now, to get $$\lambda$$ from $$S_\lambda$$ you can look at multiples of $$I$$ in $$S_\lambda$$...

Take any linear operator $$A$$ and modify it to $$A' = A P_\perp + \lambda P_v$$ where $$P_\perp = 1 - P_v$$ is the projection onto subspace orthogonal to $$v$$ and $$P_v$$ is the projection onto $$v$$. Then, obviously $$A' v = \lambda v$$ and $$A' w = A w$$ for any $$w \perp v$$.

All such $$A'$$ give the wanted class $$S$$. From the construction one directly can see that it is a co-dimension one subspace of all linear operators (in the given space).

• Nice answer, but why does this construction generate the entire class $S$? Jun 8, 2021 at 21:28
• @TomChen This is obvious from the construction -- we end up with $S$ having co-dimension one for one linear condition. Jun 8, 2021 at 21:49

This is essentially a more detailed version of the abstract answer given by @user932138. You may view the condition as an affine condition on the set of matrices. With $$T=(a_{ij})$$, you may rewrite $$\pmatrix{a_{11} & a_{12} \\ a_{21} & a_{22} } \pmatrix{v_1\\v_2} = \lambda \pmatrix{v_1\\v_2}$$ as $$M A := \pmatrix{v_1 & v_2 & 0 & 0 \\ 0&0& v_1 & v_2} \pmatrix{a_{11} \\ a_{12} \\ a_{21} \\ a_{22} } = \pmatrix{\lambda v_1\\\lambda v_2} =: B.$$ The $$2\times 4$$ matrix $$M$$ has rank 2 whenever $$(v_1,v_2)\neq (0,0)$$ so by standard linear algebra you know that $$MA=B$$ has solutions, and that the set of solutions form a 2 dimensional affine subspace of $${\Bbb C}^4$$.

More generally, if $$T\in M_n({\Bbb C})$$ and you are given $$d\leq n$$ linearly independent vectors $$v^1, ..., v^d$$ in $${\Bbb C}^n$$ and complex constants $$\lambda_1,...,\lambda_d$$ then the set of matrices $$T$$ that are subject to $$Tv^i=\lambda_1 v^i$$, $$1\leq i \leq d$$ will again give rise to a linear system $$MA=B$$ with $$M$$ a full rank $$(dn\times n^2)$$ matrix so the solutions will constitute an $$n^2-dn$$ dimensional affine subspace of $$M_n({\Bbb C}^n)$$.

Now, this gives you an affine algebraic description of the solution space.

A more geometric (abstract) description was given by other users. When given $$d$$ prescribed eigenvectors (as in my last example), simply choose a complement $$W$$ to the span of $$v^1,...,v^d$$, choose a basis $$w^1,...,w^{n-d}$$ of $$W$$. Then the action of a solution $$T$$ has prescribed values of the $$v$$'s and arbitrary values in $${\Bbb C}^n$$ on the $$w$$'s. Again this give you a solution space which is an $$n\times(n-d)$$ dimensional affine subspace of $$M_n({\Bbb C}^n)$$.