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

Problems and Solutions of Eigenvalue, Eigenvector in Linear Algebra

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 $\lambda=-1$.

 
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Hint.

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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