Why an Invariant Subspace Has an Eigenvector

[arthurhenry]

I am following a proof in the text "Algebras of Linear Transformations" and having problem justifying this line: ... M is an invariant subspace so it has an eigenvector. Why should an invariant subspace have an eigenvector? Thank you

I have a feeling this is a very simple result, if so I am sorry

 

[HallsofIvy]

A subspace, M, of vector space, V, is an "invariant subspace" for linear transformation T if and only if whenever u is in M, Tu is also in u. That means we can restrict T to M- think of T as a linear transformation on M alone. Now, if we are working over the complex numbers, every linear transformation has at least one eigenvector so T has at least one eigenvector in M.

 

[Monocles]

Considering the converse scenario may help as well, i.e., that eigenvectors span invariant subspaces. Consider that if u is an eigenvector of T, Tu = cu for some constant c. Thus, u and cu are collinear. Therefore, the subspace spanned by u is an invariant subspace of T.

 

[arthurhenry]

I thank you HallsofIvy,
Yes, it is clear now.

 

[blue_raver22]

, I am a beginner in this field

An invariant subspace is a subspace of a vector space that is preserved under the action of a linear transformation. This means that if we apply the transformation to any vector in the subspace, the resulting vector will still be in the subspace. This property is what makes invariant subspaces useful in linear algebra.

Now, let's consider an eigenvector of a linear transformation. An eigenvector is a vector that, when multiplied by the transformation, results in a scalar multiple of itself. In other words, the transformation only scales the vector, without changing its direction.

So, if we have an invariant subspace M, and we apply the linear transformation to any vector in M, the resulting vector will still be in M. This means that if we have an eigenvector in M, when we apply the transformation to it, we will still have an eigenvector in M. This is because the transformation only scales the eigenvector, without changing its direction.

Therefore, since an invariant subspace M contains vectors that are preserved under the action of a linear transformation, and an eigenvector is a vector that is only scaled by the transformation, it follows that an invariant subspace must contain at least one eigenvector. This is why an invariant subspace has an eigenvector.