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[SOLVED] Eigenvalues and Eigenvectors over a Polynomial Ring

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hi everyone, :)

Here's another question that I solved. Let me know if you see any mistakes or if you have any other comments. Thanks very much. :)

Problem:

Prove that the eigenvector \(v\) of \(f:V\rightarrow V\) over a field \(F\), with eigenvalue \(\lambda\), is an eigenvector of \(P(f)\) where \(P(t)\in F[t]\). What is the eigenvalue of \(v\) with respect to \(P(f)\)?

My Solution:

Let \(A\) be the matrix representation of the linear transformation \(f\). Then we can write, \(Av=\lambda v\). Now let, \(P(t)=a_0+a_1 t+a_2 t^2+\cdots+a_n t^n\) where \(a_i\in F\,\forall\,i\). Then,

\[P(A)=a_0+a_1 A+a_2 A^2+\cdots+a_n A^n\]

\[\Rightarrow P(A)(v)=a_0 v+a_1 (Av)+a_2 (A^2 v)+\cdots+a_n (A^n v)\]

\[\Rightarrow P(A)(v)=a_0 v+a_1 \lambda v+a_2 \lambda^2 v+\cdots+a_n \lambda^n v\]

\[\therefore P(A)(v)=(a_0 +a_1 \lambda +a_2 \lambda^2 +\cdots+a_n \lambda^n )(v)\]

Hence \(v\) is an eigenvector for \(P(f)\) and the corresponding eigenvalue is, \(a_0 +a_1 \lambda +a_2 \lambda^2 +\cdots+a_n \lambda^n\).
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Hi everyone, :)

Here's another question that I solved. Let me know if you see any mistakes or if you have any other comments. Thanks very much. :)

Problem:

Prove that the eigenvector \(v\) of \(f:V\rightarrow V\) over a field \(F\), with eigenvalue \(\lambda\), is an eigenvector of \(P(f)\) where \(P(t)\in F[t]\). What is the eigenvalue of \(v\) with respect to \(P(f)\)?

My Solution:

Let \(A\) be the matrix representation of the linear transformation \(f\). Then we can write, \(Av=\lambda v\). Now let, \(P(t)=a_0+a_1 t+a_2 t^2+\cdots+a_n t^n\) where \(a_i\in F\,\forall\,i\). Then,
Here's a conceptual error. You have written $Av=\lambda v$. Here $A$ is a matrix and $v$ is a vector. Here you should have used the column vector representation of $v$ with respect to the same basis you used to represent $f$ as the matrix $A$.

\[P(A)=a_0+a_1 A+a_2 A^2+\cdots+a_n A^n\]

\[\Rightarrow P(A)(v)=a_0 v+a_1 (Av)+a_2 (A^2 v)+\cdots+a_n (A^n v)\]

\[\Rightarrow P(A)(v)=a_0 v+a_1 \lambda v+a_2 \lambda^2 v+\cdots+a_n \lambda^n v\]

\[\therefore P(A)(v)=(a_0 +a_1 \lambda +a_2 \lambda^2 +\cdots+a_n \lambda^n )(v)\]

Hence \(v\) is an eigenvector for \(P(f)\) and the corresponding eigenvalue is, \(a_0 +a_1 \lambda +a_2 \lambda^2 +\cdots+a_n \lambda^n\).
This is fine (keeping in mind the above comment) although you didn't need to use the matrix representation of $f$. You could have done what you have done with $f$ itself. The solution would be shorter too.

:)
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Here's a conceptual error. You have written $Av=\lambda v$. Here $A$ is a matrix and $v$ is a vector. Here you should have used the column vector representation of $v$ with respect to the same basis you used to represent $f$ as the matrix $A$.

This is fine (keeping in mind the above comment) although you didn't need to use the matrix representation of $f$. You could have done what you have done with $f$ itself. The solution would be shorter too.

:)
Thank you very much for the ideas. I was too lazy to write down "let \(v\) be the column vector representation of ......." and thought it was rather implied when I used the matrix representation of \(f\).

Your other idea sounds good, but for some weird reason, I always like to deal with matrices rather than the functional form of linear transformations. :)

Thanks again for your generous help. :)
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
... for some weird reason, I always like to deal with matrices rather than the functional form of linear transformations. :)
There is no problem if $V$ has dimension $n$ (finite), because we can use the well known isomophism of algebras $\phi:\operatorname{End}(V)\to \mathbb{K}^{n\times n},$ $\phi (f)=A=[f]_B,$ where $B$ is a fixed basis of $V.$
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
I could be wrong, but I think what caffeinemachine was getting at is this:

Suppose $P(t) = a_0 + a_1t +\cdots + a_nt^n$.

By definition:

$P(f)$ is defined as the linear transformation:

$P(f)(v) = (a_0I + a_1f + \cdots a_nf^n)(v) = a_0v + a_1f(v) + \cdots + a_nf^n(v)$

(in other words, we use the "point-wise" sum for linear transformations, and composition for multiplication, as is usual for a ring of endomorphisms of an abelian group. The scalar multiplication is also "pointwise": $(af)(v) = a(f(v))$).

Presumably, you have already proved that if $\lambda$ is an eigenvalue for $f$ with eigenvector $v$, then (for natural numbers $k$):

$f^k(v) = \lambda^kv$

(note the exponent on the left refers to k-fold composition, and the exponent on the right refers to exponentiation in the field). If you have not done so, it's easy to prove using induction (you may wish to use the common convention that $f^0 = I = \text{id}_V$, the identity function on $V$).

Thus, for an eigenvector $v$ with eigenvalue $\lambda$, we have that $v$ is likewise an egienvector for $P(f)$ with eigenvalue $P(\lambda)$.

The advantage to this is that no mention is made of the dimensionality of the vector space $V$, and no assumptions are made about any basis.