Given the Data of Eigenvalues, Determine if the Matrix is Invertible

Problem 686

In each of the following cases, can we conclude that $A$ is invertible? If so, find an expression for $A^{-1}$ as a linear combination of positive powers of $A$. If $A$ is not invertible, explain why not.

(a) The matrix $A$ is a $3 \times 3$ matrix with eigenvalues $\lambda=i , \lambda=-i$, and $\lambda=0$.

(b) The matrix $A$ is a $3 \times 3$ matrix with eigenvalues $\lambda=i , \lambda=-i$, and $\lambda=-1$.

Recall that the product of all the eigenvalues of $A$ is the determinant of $A$.

Solution.

(a) Eigenvalues $\lambda=i , \lambda=-i$, and $\lambda=0$.

The matrix $A$ is not invertible.

The determinant of $A$ is the product of its eigenvalues. In this case, that means that $\det(A) = i\cdot (-i)\cdot 0 = 0$.

Because the determinant is $0$, $A$ is singular and, hence, not invertible.

(b) Eigenvalues $\lambda=i , \lambda=-i$, and $\lambda=-1$

The determinant of $A$ is the product of its eigenvalues.
In this case, that means $\det(A) = i\cdot (-i)\cdot (-1)=-1$. Because the determinant is non-zero, the matrix $A$ is non-singular, and thus is invertible.

To find an expression for $A^{-1}$, we will use the Cayley-Hamilton theorem. First we find the characteristic polynomial of $A$, which is
\[ p(\lambda) = (\lambda-i)(\lambda+i)(\lambda+1) = \lambda^3 + \lambda^2 + \lambda + 1 . \]

The Cayley-Hamilton theorem says that $A$ must satisfy the equality
\[ A^3 + A^2 + A + I = \mathbf{0} , \]
where $\mathbf{0}$ is the zero matrix. Rewriting this, we have
\[ I = -A – A^2 – A^3 = A( -I – A – A^2 ) . \]

Multiplying on the left by $A^{-1}$ yields the desired equation,
\[ A^{-1} = -I – A – A^2 . \]

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