It is indeed possible for a matrix to have an eigenvalue that is equal to zero. If a square matrix has eigenvalue zero, then it means that the matrix is *singular* (not invertible).

Recall that we say that a matrix A has eigenvalue \lambda, if there exists a nonzero vector v such that,

Av = \lambda vIn particular, the vector v\neq 0 is called an eigenvector for the matrix. Note that it is by definition impossible for an *eigenvector* to be zero. On the other hand, it is possible that the *eigenvalue* is zero.

If an eigenvalue of A is zero, it means that the kernel (nullspace) of the matrix is nonzero. This means that the matrix has determinant equal to zero. Such a matrix will not be invertible.

The converse of the above statement is also true. If a matrix A has determinant equal to 0, it means that 0 is an eigenvalue for the matrix. We now give an example of a matrix for which we have zero eigenvalue.

**Example:**

Consider the matrix,

A = \begin{bmatrix} 1 & 2 \\ 3 & 6 \end{bmatrix}The characteristic polynomial of the matrix is given as,

P(\lambda) = det(A-\lambda I) = det\left(\begin{bmatrix} 1-\lambda & 2 \\ 3 & 6-\lambda \end{bmatrix}\right) = \lambda(\lambda - 7)Solving P(\lambda) = 0 we see that, the matrix A has eigenvalues 7 and 0. Notice that the matrix A is not invertible (it has determinant equal to 0).