Does a zero eigenvalue mean that the matrix is not invertible regardless of its diagonalizability?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

1. 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.

2. 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.

Answer 4

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.

Answer 5

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.