2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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... $\endgroup$ May 28, 2021 at 5:22
  • $\begingroup$ @CalvinKhor Can we show that the collection of such kinds of operators uniquely determines the vector? $\endgroup$
    – IamKnull
    May 28, 2021 at 10:06
  • $\begingroup$ No, you have no way to restrict eg the euclidean norm to $|v|$ instead of eg $2|v|$ $\endgroup$ May 28, 2021 at 11:31

4 Answers 4

1
$\begingroup$

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 $Tv=0$.

$\endgroup$
1
$\begingroup$

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 $S_\lambda$...

$\endgroup$
1
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ Nice answer, but why does this construction generate the entire class $S$? $\endgroup$
    – Tom Chen
    Jun 8, 2021 at 21:28
  • $\begingroup$ @TomChen This is obvious from the construction -- we end up with $S$ having co-dimension one for one linear condition. $\endgroup$
    – Nikodem
    Jun 8, 2021 at 21:49
0
$\begingroup$

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 $M_n({\Bbb C}^n)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .