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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(b) Let
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Let $A, B$ be complex $2\times 2$ matrices satisfying the relation
\[A=AB-BA.\]
Prove that $A^2=O$, where $O$ is the $2\times 2$ zero matrix.
Hint.
Find the trace of $A$.
Use the Cayley-Hamilton theorem
Proof.
We first calculate the […]

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Then for each positive integer $n$ find $a_n$ and $b_n$ such that
\[A^{n+1}=a_nA+b_nI,\]
where $I$ is the $2\times 2$ identity matrix.
Solution.
Since $-1, 3$ are eigenvalues of the […]

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Let
\[A=\begin{bmatrix}
1 & -1\\
2& 3
\end{bmatrix}.\]
Find the eigenvalues and the eigenvectors of the matrix
\[B=A^4-3A^3+3A^2-2A+8E.\]
(Nagoya University Linear Algebra Exam Problem)
Hint.
Apply the Cayley-Hamilton theorem.
That is if $p_A(t)$ is the […]

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\[
A=
\begin{bmatrix}
-2 & 0 & 1 \\
-5 & 3 & a \\
4 & -2 & -1
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\]
for some variable $a$. Find all values of $a$ which will guarantee that $A$ has eigenvalues $0$, $3$, and $-3$.
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Let $A=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}$ be an $2\times 2$ matrix.
Express the eigenvalues of $A$ in terms of the trace and the determinant of $A$.
Solution.
Recall the definitions of the trace and determinant of $A$:
\[\tr(A)=a+d \text{ and } […]

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Then prove that the inverse matrix of the matrix $I+A$ is given by the following formula:
\[(I+A)^{-1}=I-\frac{1}{1+\tr(A)}A.\]
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Proof.
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