Diagonalizable & Invertible from an 'eigenpicture' I was just wondering if it is possible to tell if a matrix is invertible or diagonalizable just by its eigenpicture? Say 
Example of an Eigenpicture
(Sorry I can't embed the picture in my post as I am quite new to the scene)
I have seen a post that is related to this particular problem which helps to identify both the eigenvalues and eigenvector. (Estimating Eigenvalues and Eigenvectors from an 'eignpicture') With the same example, just by its 'eigenpicture' is it possible for us to determine if it is diagonalizable, or even if it is invertible??
Thank you in advance
 A: I am not sure how much sense these eigenpictures make. $2\times 2$ matrices are the easiest to work with, so in the time you spend to draw it you can potentially solve invertability and diagonizability by hand...
Anyway. A square matrix is invertible iff the corresponding linear map is surjective iff the eigenpicture has two nonparallel blue lines (I assume the eigenpicture to have all outputs of unit vectors, in reality this is another limitation of this picture method).
Similarly a square matrix is diagonizable iff the the space admits a basis of eigenvectors. In our case this is equivalent to finding (at least) two nonparallel lines such that both the input vector and its blue output vector lie on the same line each. The matrix corresponding to the example is not diagonizable, since the only „line of eigenvectors“ is the diagonal in direction (1,1).
To be clear, I think eigenpictures are a great way to introduce and explain the concept of eigenvectors. But they don’t generalize well to higher dimensions (not even to 3d) and will never replace solid proofs, since they depend quite heavily on artistic skill and the matrix being „coarse“ enough such that its effects are visible in the picture...
A: The fundamental fact about diagonalizable maps and matrices is expressed by the following:
An {\displaystyle n\times n}n\times n matrix {\displaystyle A}A over a field {\displaystyle F}F is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to {\displaystyle n}n, which is the case if and only if there exists a basis of {\displaystyle F^{n}}F^{n} consisting of eigenvectors of {\displaystyle A}A. If such a basis has been found, one can form the matrix {\displaystyle P}P having these basis vectors as columns, and {\displaystyle P^{-1}!AP}{\displaystyle P^{-1}!AP} will be a diagonal matrix whose diagonal entries are the eigenvalues of {\displaystyle A}A. The matrix {\displaystyle P}P is known as a modal matrix for {\displaystyle A}A.
A linear map {\displaystyle T:V\mapsto V}{\displaystyle T:V\mapsto V} is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to {\displaystyle \operatorname {dim} (V)}{\displaystyle \operatorname {dim} (V)}, which is the case if and only if there exists a basis of {\displaystyle V}V consisting of eigenvectors of {\displaystyle T}T. With respect to such a basis, {\displaystyle T}T will be represented by a diagonal matrix. The diagonal entries of this matrix are the eigenvalues of {\displaystyle T}T.
Another characterization: A matrix or linear map is diagonalizable over the field {\displaystyle F}F if and only if its minimal polynomial is a product of distinct linear factors over {\displaystyle F}F. (Put another way, a matrix is diagonalizable if and only if all of its elementary divisors are linear.)
A: If all the eigenvalues are different then the matrix is diagonalizable.
If some eigenvalues are equal then we have to check the eigenvectors corresponding to that eigenvalues are linearly independant. 
