# Eigenvalue and eigenvectors explanation for transformation.

What do eigenvalues and eigenvectors of a Matrix A tell us about a transformation T (T: R^3 -> R^3)?

I want to know if I'm on the right page, close to, or not correct about any of these. Let's say I have a 3x3 matrix and its eigenvalues are 0,2, and 5.

1) Eigenvalues can tell us if A is an invertible matrix if and only if zero is not an eigenvalue of A, but because it is in this example we know that A is not an invertible matrix. That means that the transformation of A is not invertible as well.

2) If the vectors are eigenvectors that correspond with the eigenvalues of an nxn matrix A then the set of vectors are linearly independent. Since the eigenvalues of A and eigenvectors correspond to each other then the set of vectors is linearly independent. Meaning the vectors of the transformation of A are linearly independent.

3) If A has 'n' linearly independent eigenvectors then the nxn matrix A is diagonalizable. Because A has all 3 linearly independent eigenvectors we can conclude that A is diagonalizable and the Transformation of A is diagonalizable.

• The second is only true when the eigenvalues are distinct.
– amd
May 4 '20 at 17:21

2. As stated, it is not true. If $$v$$ is an eigenvector, then so is $$2v$$. But $$\{v,2v\}$$ is linearly dependent. However, if $$\lambda_1,\ldots,\lambda_n$$ are distinct eigenvalues and if, for each $$k\in\{1,\ldots,n\}$$, $$v_k$$ is an eigenvector corresponding to the eigenvaulue $$\lambda_k$$, then the set $$\{v_1,\ldots,v_n\}$$ is linearly independent.
• Yes, it will.${}$ May 4 '20 at 17:35
• The null space consists of the null vector together with the eigenvectors corresponding to the eigenvalue $0$. And I see nothing that I could add concerning the column space. May 4 '20 at 18:16