# Finding eigenvalues and eigenvectors given sub-matrices

• ChiralSuperfields
In summary, the solution is to find the eigenvector and eigenvalue for the matrix, A, that corresponds to the desired eigenvalue.
ChiralSuperfields
Homework Statement
Relevant Equations
For this,

The solution is,

However, does someone please know what allows them to express the eigenvector for each of the sub-matrix in terms of t?

Many thanks!

From the definition a multiple of any eigenvector is also trivially an eigenvector. So who did this?

ChiralSuperfields
Any matrix operator is linear. So ##A (tv)= t A v ## for any vector, ##v##.
Now use that in your case with the eigenvector and eigenvalue.

Last edited:
ChiralSuperfields
You need to take into account that both conditions hold. If ##a\not = -1##, then there are two eigenvalues and the corresponding eigenvectors are multiples of the two given vectors. If ##a=-1##, then there is only one eigenvalue and all vectors are eigenvectors.

ChiralSuperfields
Thank you for your replies @hutchphd, @FactChecker and @martinbn !

So I am coming to think of it like this considering the most general case I could think of:

##A\vec v = \lambda\vec v##
##At\vec v = \lambda t\vec v## multiply both sides by a scalar ##t## which is a member of the reals
Therefore, the definition of eigenvector and eigenvalue, ##t\vec v## is an eigenvectors for ##\lambda## and ##\lambda## is the eigenvalue for A.

Many thanks!

ChiralSuperfields said:
##A\vec v = \lambda\vec v##
##At\vec v = \lambda t\vec v## multiply both sides by a scalar ##t## which is a member of the reals
Therefore, the definition of eigenvector and eigenvalue, ##t\vec v## is an eigenvectors for ##\lambda## and ##\lambda## is the eigenvalue for A.

Yes, but it would be better if you rephrased your statement to more clearly match the definition of an eigenvector and eigenvalue:
Suppose ##\lambda## and ##\vec v## are eigenvalue and eigenvector of ##A##, respectively. For any ##t \in \mathbb{R}##, where ##t \ne 0##, ##A( \vec {tv}) = t A\vec v = t \lambda \vec v = \lambda \vec {tv}##. So ##\vec {tv}## is also an eigenvector of ##A## with the eigenvalue ##\lambda##.

hutchphd and ChiralSuperfields
FactChecker said:
Yes, but it would be better if you rephrased your statement to more clearly match the definition of an eigenvector and eigenvalue:
Suppose ##\lambda## and ##\vec v## are eigenvalue and eigenvector of ##A##, respectively. For any ##t \in \mathbb{R}##, where ##t \ne 0##, ##A( \vec {tv}) = t A\vec v = t \lambda \vec v = \lambda \vec {tv}##. So ##\vec {tv}## is also an eigenvector of ##A## with the eigenvalue ##\lambda##.
Thank you for your help @FactChecker!

ChiralSuperfields said:
Thank you for your help @FactChecker!
Actually, I realize that your statement did match the definitions of eigenvector and eigenvalue, but I think that I have rephrased it more as a step-by-step proof.

ChiralSuperfields
FactChecker said:
Actually, I realize that your statement did match the definitions of eigenvector and eigenvalue, but I think that I have rephrased it more as a step-by-step proof.
Thank you for your reply @FactChecker! Yeah I only just realized the definition of eigenvector is for a value of ##\lambda## not the matrix A too!

ChiralSuperfields said:
Thank you for your reply @FactChecker! Yeah I only just realized the definition of eigenvector is for a value of ##\lambda## not the matrix A too!
It's for the combination of the matrix, ##A##, and the eigenvalue ##\lambda##

ChiralSuperfields
FactChecker said:
It's for the combination of the matrix, ##A##, and the eigenvalue ##\lambda##
Oh, thank you for you letting me know @FactChecker !

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