Does a zero eigenvalue mean that the matrix is not invertible regardless of its diagonalizability? If the matrix $A$ is diagonalizable, then we know that its similar diagonal matrix $D$ has determinant $0$, so the matrix $A$ itself is invertible? However, if $A$ not diagonalizable, how are we sure that the matrix $A$ which has $0$ as eigenvalue is not invertible?
Here I have another confusion, does the degree of the characteristic polynomial  determine the size of matrix. i.e. $\lambda (\lambda+2)^3 (\lambda-1)^2$ has $6\times 6$ matrix?    
 A: *

*I am unsure of the exact wording in your first paragraph so let me use the interpretation, "If a matrix has zero as an eigenvalue, what does this say about the invertibility?"


Note that by definition, $\lambda$ is an eigenvalue to $A$ if $Ax = \lambda x$ for some $x \ne 0$. But this means $(A - \lambda I)x = 0$. Now if we took $\lambda = 0$, then the above becomes $Ax = 0$ for some nonzero $x$. There are several ways to interpret this (e.g. the nullspace is nontrivial) which will all lead to the same conclusion, namely that the matrix is noninvertible. 


*Again by definition, the characterstic polynomial is defined to be $\chi_A (x) = \textrm{det} (xI - A)$ which obviously has degree $n$ if $A$ is a square $n \times n$ matrix. 

A: If $0$ is an eigenvalue, then there is a nonzero vector $v$ with $Mv = 0$. Then the kernel of $M$ is not trivial (it is at least one-dimensional), and so it is not one-to-one viewed as a linear transformation. Then it is not invertible.
The geometric picture: $M$ 'collapses' the subspace spanned by $v$, and so maps its domain into a hyperplane in its codomain. This picture actually says: $M$ is not onto, but this is another way to assert non-invertibility.
Note that $M$ can still be diagonalizable, as it can still have a basis of eigenvectors, independent of whether or not it 'collapses' some.
A: The determinant of a matrix is the product of its eigenvalues. So, if one of the eigenvalues is $0$, then the determinant of the matrix is also $0$. Hence it is not invertible.
A: Suppose $M$ is an invertible matrix, with nonzero eigenvector $v$ corresponding to the eigenvalue $0$.  Then we would have
$$v = M^{-1}Mv = M^{-1}(0v) = 0$$
But $v \ne 0$.  This shows that if $0$ is an eigenvalue of $M$, $M$ cannot be invertible.
A: The characteristic polynomial of matrix A is the determinant of $A- \lambda I$.  Since that has one copy of $\lambda$ in each row and column, yes, the degree of the characteristic polynomial is equal to the order of the matrix.
