Invertibility and Eigenvalues

There is a direct connection between eigenvalues and invertibility: a matrix $A$ is invertible if and only if none of its eigenvalues are zero. Here is why: if $\lambda = 0$ is an eigenvalue, then $A\mathbf{u} = 0$ for some nonzero $\mathbf{u}$, meaning $A$ has a nontrivial null space, which means $A$ is not invertible.

Conversely, if all eigenvalues $\lambda_1, \ldots, \lambda_n$ are nonzero, we can use the spectral decomposition $A = U \Lambda U^\top$ to build the inverse: $A^{-1} = U \Lambda^{-1} U^\top$, where $\Lambda^{-1} = \text{diag}(1/\lambda_1, \ldots, 1/\lambda_n)$. This works precisely because each $\lambda_i \neq 0$.

Also, $\det(A) = \prod_{i=1}^n \lambda_i$ — the determinant equals the product of all eigenvalues. So $A$ is invertible $\iff \det(A) \neq 0 \iff$ no eigenvalue is zero.