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 $\lambda=-1$.

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,
\[ A^{-1} = -I – A – A^2 . \] Click here if solved 135The post Given the Data of Eigenvalues, Determine if the Matrix is Invertible first appeared on Problems in Mathematics.]]> https://yutsumura.com/given-the-data-of-eigenvalues-determine-if-the-matrix-is-invertible/feed/ 0 6813 https://yutsumura.com/a-recursive-relationship-for-a-power-of-a-matrix/ https://yutsumura.com/a-recursive-relationship-for-a-power-of-a-matrix/#respond Wed, 31 Jan 2018 04:53:07 +0000 https://yutsumura.com/?p=6808 Suppose that the $2 \times 2$ matrix $A$ has eigenvalues $4$ and $-2$. For each integer $n \geq 1$, there are real numbers $b_n , c_n$ which satisfy the relation \[ A^{n} = b_n...

Let $A$ and $B$ be $2\times 2$ matrices such that $(AB)^2=O$, where $O$ is the $2\times 2$ zero matrix.

Determine whether $(BA)^2$ must be $O$ as well. If so, prove it. If not, give a counter example.

Proof.

It is true that the matrix $(BA)^2$ must be the zero matrix as we shall prove now.

For notational convenience, let $C:=AB$.

As $C$ is a $2\times 2$ matrix, it satisfies the relation
\[C^2-\tr(C)C+\det(C)I=O \tag{*}\] by the Cayley-Hamilton theorem.
Here $I$ is the $2\times 2$ identity matrix.

We compute the determinant of $C$ as follows.
We have
\begin{align*}
\det(C)^2=\det(C^2)=\det((AB)^2)=\det(O)=0.
\end{align*}
Hence $\det(C)=0$.

Since $C^2=O$ and $\det(C)=0$, the Cayley-Hamilton relation (*) becomes
\[\tr(C)C=O.\] It follows that we have either $C=O$ or $\tr(C)=0$.
In either case, we have $\tr(C)=0$.

Note that
\begin{align*}
\det(BA)&=\det(AB)=\det(C)=0\\
\tr(BA)&=\tr(AB)=\tr(C)=0.
\end{align*}

Applying the Cayley-Hamilton theorem to the matrix $BA$, we obtain
\begin{align*}
O=(BA)^2-\tr(BA)\cdot BA+\det(BA)I=(BA)^2.
\end{align*}
Thus, we obtain $(BA)^2=O$ as claimed.

Let $A$ be a singular $2\times 2$ matrix such that $\tr(A)\neq -1$ and let $I$ be the $2\times 2$ identity matrix.
Then prove that the inverse matrix of the matrix $I+A$ is given by the following formula:
\[(I+A)^{-1}=I-\frac{1}{1+\tr(A)}A.\]

Using the formula, calculate the inverse matrix of $\begin{bmatrix}
2 & 1\\
1& 2
\end{bmatrix}$.

Proof.

We have
\begin{align*}
(I+A)\left(\, I-\frac{1}{1+\tr(A)}A \,\right)&=I-\frac{1}{1+\tr(A)}A+A-\frac{1}{1+\tr(A)}A^2\\[6pt] &=I-\frac{1}{1+\tr(A)}\left(\, A-(1+\tr(A))A +A^2\,\right)\\[6pt] &=I-\frac{1}{1+\tr(A)}\left(\, A^2-\tr(A)A \,\right) \tag{*}.
\end{align*}

The Cayley-Hamilton theorem for $2\times 2$ matrices yields that
\[A^2-\tr(A)A+\det(A)I=O.\] Since $A$ is singular, we have $\det(A)=0$.
Hence it follows that we have
\[A^2-\tr(A)A=O,\] and we obtain from (*) that
\[(I+A)\left(\, I-\frac{1}{1+\tr(A)}A \,\right)=I.\] Similarly,
\[\left(\, I-\frac{1}{1+\tr(A)}A \,\right)(I+A)=I.\]

Therefore, we conclude that the inverse matrix of $I+A$ is given by the formula
\[(I+A)^{-1}=I-\frac{1}{1+\tr(A)}A.\]

Find the inverse matrix of $\begin{bmatrix}
2 & 1\\
1& 2
\end{bmatrix}$ using the formula

Now let us find the inverse matrix of $\begin{bmatrix}
2 & 1\\
1& 2
\end{bmatrix}$ using the formula.

We first write
\[\begin{bmatrix}
2 & 1\\
1& 2
\end{bmatrix}=I+A,\] where
\[A=\begin{bmatrix}
1 & 1\\
1& 1
\end{bmatrix}.\] Then $A$ is a singular matrix with $\tr(A)=2$.

The formula yields that
\begin{align*}
\begin{bmatrix}
2 & 1\\
1& 2
\end{bmatrix}^{-1}&=(I+A)^{-1}\\[6pt] &=I-\frac{1}{3}A\\[6pt] &=\frac{1}{3}\begin{bmatrix}
2 & -1\\
-1& 2
\end{bmatrix}.
\end{align*}

Related Question.

There is a similar formula for inverse matrices of certain $n\times n$ matrices, called Sherman-Woodberry formula.

See the post ↴
Sherman-Woodbery Formula for the Inverse Matrix
for the statement of the Sherman-Woodberry formula and its proof.

Find the inverse matrix of the $3\times 3$ matrix
\[A=\begin{bmatrix}
7 & 2 & -2 \\
-6 &-1 &2 \\
6 & 2 & -1
\end{bmatrix}\] using the Cayley-Hamilton theorem.

Solution.

To apply the Cayley-Hamilton theorem, we first determine the characteristic polynomial $p(t)$ of the matrix $A$.
Let $I$ be the $3\times 3$ identity matrix.
We have
\begin{align*}
p(t)&=\det(A-tI)\\
&=\begin{vmatrix}
7-t & 2 & -2 \\
-6 &-1-t &2 \\
6 & 2 & -1-t
\end{vmatrix}\\[6pt] &=(7-t)\begin{vmatrix}
-1-t & 2\\
2& -1-t
\end{vmatrix}
-2\begin{vmatrix}
-6 & 2\\
6& -1-t
\end{vmatrix}+(-2)\begin{vmatrix}
-6 & -1-t\\
6& 2
\end{vmatrix}\\[6pt] &\text{(by the first row cofactor expansion)}\\[6pt] &=-t^3+5t^2-7t+3.
\end{align*}
(You may also use the rule of Sarrus to compute the $3\times 3$ determinant.)

Thus, we have obtained the characteristic polynomial
\[p(t)=-t^3+5t^2-7t+3\] of the matrix $A$.

The Cayley-Hamilton theorem yields that
\[O=p(A)=-A^3+5A^2-7A+3I,\] where $O$ is the $3\times 3$ zero matrix.
(Here, don’t forget to put the identity matrix $I$.)

Rearranging terms, we have
\begin{align*}
&A^3-5A^2+7A=3I\\[6pt] &\Leftrightarrow A(A^2-5A+7I)=3I\\[6pt] &\Leftrightarrow A\left(\frac{1}{3}(A^2-5A+7I)\right)=I.
\end{align*}
Similarly, we have
\[\left(\frac{1}{3}(A^2-5A+7I)\right)A=I.\] It follows from these two equalities that the matrix
\[\frac{1}{3}(A^2-5A+7I)\] is the inverse matrix of $A$.

Therefore, we have
\begin{align*}
A^{-1}&=\frac{1}{3}(A^2-5A+7I)\\[6pt] &=\frac{1}{3}\left(\, \begin{bmatrix}
25 & 8 & -8 \\
-24 &-7 &8 \\
24 & 8 & -7
\end{bmatrix}-5\begin{bmatrix}
7 & 2 & -2 \\
-6 &-1 &2 \\
6 & 2 & -1
\end{bmatrix}+7\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 1
\end{bmatrix} \,\right)\\[6pt] &=\frac{1}{3}\begin{bmatrix}
-3 & -2 & 2 \\
6 &5 &-2 \\
-6 & -2 & 5
\end{bmatrix}.
\end{align*}

In summary, the inverse matrix of $A$ is
\[A^{-1}=\frac{1}{3}\begin{bmatrix}
-3 & -2 & 2 \\
6 &5 &-2 \\
-6 & -2 & 5
\end{bmatrix}.\]

More Exercise

Test whether you understand how to find the inverse matrix using the Cayley-Hamilton theorem by the next problem.

The solution is given in the post “Find the Inverse Matrix Using the Cayley-Hamilton Theorem“.

More Problems about the Cayley-Hamilton Theorem

Problems about the Cayley-Hamilton theorem and their solutions are collected on the page:

The Cayley-Hamilton Theorem

Let $A$ be a $3\times 3$ real orthogonal matrix with $\det(A)=1$.

(a) If $\frac{-1+\sqrt{3}i}{2}$ is one of the eigenvalues of $A$, then find the all the eigenvalues of $A$.

(b) Let
\[A^{100}=aA^2+bA+cI,\] where $I$ is the $3\times 3$ identity matrix.
Using the Cayley-Hamilton theorem, determine $a, b, c$.

(Kyushu University, Linear Algebra Exam Problem)
 

Solution.

(a) Find the all the eigenvalues of $A$.

Since $A$ is a real matrix and $\frac{-1+\sqrt{3}i}{2}$ is a complex eigenvalue, its conjugate $\frac{-1-\sqrt{3}i}{2}$ is also an eigenvalue of $A$.
As $A$ is a $3\times 3$ matrix, it has one more eigenvalue $\lambda$.

Note that the product of all eigenvalues of $A$ is the determinant of $A$.
Thus, we have
\[\frac{-1+\sqrt{3}i}{2} \cdot \frac{-1-\sqrt{3}i}{2}\cdot \lambda =\det(A)=1.\] Solving this, we obtain $\lambda=1$.
Therefore, the eigenvalues of $A$ are
\[\frac{-1+\sqrt{3}i}{2}, \frac{-1-\sqrt{3}i}{2}, 1.\]

(a) Using the Cayley-Hamilton theorem, determine $a, b, c$.

To use the Cayley-Hamilton theorem, we first need to determine the characteristic polynomial $p(t)=\det(A-tI)$ of $A$.
Since we found all the eigenvalues of $A$ in part (a) and the roots of characteristic polynomials are the eigenvalues, we know that
\begin{align*}
p(t)&=-\left(\, t-\frac{-1+\sqrt{3}i}{2} \,\right)\left(\, t-\frac{-1-\sqrt{3}i}{2} \,\right)(t-1) \tag{*}\\
&=-(t^2+t+1)(t-1)\\
&=-t^3+1.
\end{align*}
(Remark that if your definition of the characteristic polynomial is $\det(tI-A)$, then the first negative sign in (*) should be omitted.)

Then the Cayley-Hamilton theorem yields that
\[P(A)=-A^3+I=O,\] where $O$ is the $3\times 3$ zero matrix.

Hence we have $A^3=I$.
We compute
\begin{align*}
A^{100}=(A^3)^{33}A=I^{33}A=IA=A.
\end{align*}

Thus, we conclude that $a=0, b=1, c=0$.

Comment.

Observe that we did not use the assumption that $A$ is orthogonal.

Find the inverse matrix of the matrix
\[A=\begin{bmatrix}
1 & 1 & 2 \\
9 &2 &0 \\
5 & 0 & 3
\end{bmatrix}\] using the Cayley–Hamilton theorem.

Solution.

To use the Cayley-Hamilton theorem, we first compute the characteristic polynomial $p(t)$ of the matrix $A$. We have
\begin{align*}
&p(t)=\det(A-tI)\\
&\begin{vmatrix}
1-t & 1 & 2 \\
9 &2-t &0 \\
5 & 0 & 3-t
\end{vmatrix}\\[6pt] &=(-1)^{3+1}5\begin{vmatrix}
1 & 2\\
2-t& 0
\end{vmatrix}+(-1)^{3+2}\cdot 0 \begin{vmatrix}
1-t & 2\\
9& 0
\end{vmatrix}+(-1)^{3+3}(3-t)\begin{vmatrix}
1-t & 1\\
9& 2-t
\end{vmatrix} \\[6pt] & \text{(by the 3rd row cofactor expansion)}\\
&=5(2t-4)+0+(3-t)\left(\, (1-t)(2-t)-9 \,\right)\\
&=-t^3+6t^2+8t-41.
\end{align*}

Then the Cayley-Hamilton theorem yields that $p(A)=O$, the zero matrix. That is, we have
\begin{align*}
O=p(A)=-A^3+6A^2+8A-41I.
\end{align*}
Thus, we have
\[41I=-A^3+6A^2+8A=A(-A^2+6A+8I),\] or equivalently
\[I=A\left(\, \frac{1}{41}(-A^2+6A+8I) \,\right).\] It follows that the inverse matrix is given by
\[A^{-1}=\frac{1}{41}(-A^2+6A+8I).\]

By a direct computation, we have
\[A^2=\begin{bmatrix}
20 & 3 & 8 \\
27 &13 &18 \\
20 & 5 & 19
\end{bmatrix}\] and
\begin{align*}
-A^2+6A+8I&=-\begin{bmatrix}
20 & 3 & 8 \\
27 &13 &18 \\
20 & 5 & 19
\end{bmatrix}+6\begin{bmatrix}
1 & 1 & 2 \\
9 &2 &0 \\
5 & 0 & 3
\end{bmatrix}+8\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 1
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
-6 & 3 & 4 \\
27 &7 &-18 \\
10 & -5 & 7
\end{bmatrix}.
\end{align*}

Therefore the inverse matrix is
\[A^{-1}=\frac{1}{41}\begin{bmatrix}
-6 & 3 & 4 \\
27 &7 &-18 \\
10 & -5 & 7
\end{bmatrix}.\]

More Exercise

Test whether you understand how to find the inverse matrix using the Cayley-Hamilton theorem by the next problem.

The solution is given in the post “How to use the Cayley-Hamilton Theorem to Find the Inverse Matrix“.

More Problems about the Cayley-Hamilton Theorem

Problems about the Cayley-Hamilton theorem and their solutions are collected on the page:

The Cayley-Hamilton Theorem

Let $n>1$ be a positive integer. Let $V=M_{n\times n}(\C)$ be the vector space over the complex numbers $\C$ consisting of all complex $n\times n$ matrices. The dimension of $V$ is $n^2$.
Let $A \in V$ and consider the set
\[S_A=\{I=A^0, A, A^2, \dots, A^{n^2-1}\}\] of $n^2$ elements.
Prove that the set $S_A$ cannot be a basis of the vector space $V$ for any $A\in V$.

Proof.

We prove that the set $S_A$ is linearly dependent, hence it cannot be a basis of $V$.
Since $A$ is an $n\times n$ matrix, its characteristic polynomial $p(t)=\det(tI-A)$ is a degree $n$ polynomial.

(Your preferred definition of the characteristic polynomial might be $\det(A-tI)$. It is straight forward to modify the following proof with this definition.)

Let us write it as
\[p(t)=t^n+a_{n-1}t^{n-1}+\cdots+a_1x+a_0.\] Then the Cayley-Hamilton theorem states that
\[p(A)=A^n+a_{n-1}A^{n-1}+\cdots+a_1A+a_0I=O\] is the zero matrix.

Since the coefficient of $A^n$ is $1$, this gives a non-trivial linear combination of $I, A, \dots, A^n$. Therefore the set
\[T:=\{I, A, \dots, A^n\}\] is linearly dependent.

As $T$ is a subset of $S_A$, the set $S_A$ is also linearly dependent.
Therefore, $S_A$ is not a basis of $V$. This completes the proof.

Let $A$ be an $n\times n$ complex matrix.
Let $p(x)=\det(xI-A)$ be the characteristic polynomial of $A$ and write it as
\[p(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0,\] where $a_i$ are real numbers.

Let $C$ be the companion matrix of the polynomial $p(x)$ given by
\[C=\begin{bmatrix}
0 & 0 & \dots & 0 &-a_0 \\
1 & 0 & \dots & 0 & -a_1 \\
0 & 1 & \dots & 0 & -a_2 \\
\vdots & & \ddots & & \vdots \\
0 & 0 & \dots & 1 & -a_{n-1}
\end{bmatrix}=
[\mathbf{e}_2, \mathbf{e}_3, \dots, \mathbf{e}_n, -\mathbf{a}],\] where $\mathbf{e}_i$ is the unit vector in $\C^n$ whose $i$-th entry is $1$ and zero elsewhere, and the vector $\mathbf{a}$ is defined by
\[\mathbf{a}=\begin{bmatrix}
a_0 \\
a_1 \\
\vdots \\
a_{n-1}
\end{bmatrix}.\]

Then prove that the following two statements are equivalent.

1. There exists a vector $\mathbf{v}\in \C^n$ such that
   \[\mathbf{v}, A\mathbf{v}, A^2\mathbf{v}, \dots, A^{n-1}\mathbf{v}\] form a basis of $\C^n$.
2. There exists an invertible matrix $S$ such that $S^{-1}AS=C$.
   (Namely, $A$ is similar to the companion matrix of its characteristic polynomial.)

Proof.

Before proving the equivalence of the statements, let us first note a simple but useful fact.
Let $B$ be $n\times n$ matrix and let $\mathbf{e}_i$ be the unit vector of $\C^n$ as defined in the problem. Then the matrix product $B\mathbf{e}_i$ is the $i$-th column vector of the matrix $B$.

$( 1 \implies 2)$

Suppose that the statement 1 is true. That is, there exists a vector $\mathbf{v} \in \C^n$ such that
\[\mathbf{v}, A\mathbf{v}, \dots, A^{n-1}\mathbf{v}\] form a basis of $\C^n$. In particular, these vectors are linearly independent.

Form a matrix
\[S:=[\mathbf{v}, A\mathbf{v}, \dots, A^{n-1}\mathbf{v}],\] then the matrix $S$ is invertible since the columns vectors are linearly independent.

We prove that $AS=SC$.
We have
\begin{align*}
AS&=A[\mathbf{v}, A\mathbf{v}, \dots, A^{n-1}\mathbf{v}],\\
&=[A\mathbf{v}, A^2\mathbf{v}, \dots, A^{n}\mathbf{v}].
\end{align*}

On the other hand, we have
\begin{align*}
SC&=S [\mathbf{e}_2, \mathbf{e}_3, \dots, \mathbf{e}_n, -\mathbf{a}]\\
&=[S\mathbf{e}_2, S\mathbf{e}_3, \dots, S\mathbf{e}_n, -S\mathbf{a}].
\end{align*}

As noted at the beginning, the vector $S\mathbf{e}_2$ is the second column vector of $S$, which is $A\mathbf{v}$. Similarly, $S\mathbf{e}_i$ is the $i$-th column vector of $S$, and thus we have
\[S\mathbf{e}_i=A^{i-1}\mathbf{v} \tag{*}.\] Hence we have
\begin{align*}
SC&=[A\mathbf{v}, A^2\mathbf{v}, \dots, A^{n-1}\mathbf{v}, -S\mathbf{a}].
\end{align*}

Thus, it remains to prove that $-S\mathbf{a}=A^n\mathbf{v}$.
Write
\[\mathbf{a}=\begin{bmatrix}
a_0 \\
a_1 \\
\vdots \\
a_{n-1}
\end{bmatrix}=a_0\mathbf{e}_1+a_1\mathbf{e}_2+\cdots+a_{n-1}\mathbf{e}_n.\] Then we have
\begin{align*}
S\mathbf{a}&=S(a_0\mathbf{e}_1+a_1\mathbf{e}_2+\cdots+a_{n-1}\mathbf{e}_n)\\
&=a_0S\mathbf{e}_1+a_1S\mathbf{e}_2+\cdots+a_{n-1}S\mathbf{e}_n\\
&\stackrel{(*)}{=} a_0\mathbf{v}+a_1A\mathbf{v}+\cdots+ a_{n-1}A^{n-1}\mathbf{v}\\
&=(a_0I+a_1 A+\cdots +a_{n-1}A^{n-1})\mathbf{v}\\
&=-A^n\mathbf{v},
\end{align*}
where the last step follows from the Cayley-Hamilton theorem
\[O=p(A)=A^n+a_{n-1}A^{n-1}+\cdots a_1A+a_0I.\]

This proves that $-S\mathbf{a}=A^n\mathbf{v}$ and hence $AS=SC$.
Since $S$ is invertible, we obtain
\[S^{-1}AS=C.\]

$( 2 \implies 1)$

Next suppose that the statement 2 is true. That is, there exists an invertible matrix $S$ such that $S^{-1}AS=C$. Then we have $AS=CS$.
Define $\mathbf{v}:=S\mathbf{e}_1$ to be the first column vector of $S$.

We prove that with this choice of $\mathbf{v}$, the vectors
\[\mathbf{v}, A\mathbf{v}, A^2\mathbf{v}, \dots, A^{n-1}\mathbf{v}\] form a basis of $\C^n$.

We have
\begin{align*}
AS&=A[S\mathbf{e}_1, S\mathbf{e}_2, \dots, S\mathbf{e}_n]\\
&=[AS\mathbf{e}_1, AS\mathbf{e}_2, \dots, AS\mathbf{e}_n] \end{align*}
and
\begin{align*}
SC&=S [\mathbf{e}_2, \mathbf{e}_3, \dots, \mathbf{e}_n, -\mathbf{a}]\\
&=[S \mathbf{e}_2, S\mathbf{e}_3, \dots, S\mathbf{e}_n, -S\mathbf{a}].
\end{align*}
Since $AS=SC$, comparing column vectors, we obtain
\begin{align*}
AS\mathbf{e}_1&=S\mathbf{e}_2\\
AS\mathbf{e}_2&=S\mathbf{e}_3\\
\vdots\\
AS\mathbf{e}_{n-1}&=S\mathbf{e}_n\\
AS\mathbf{e}_n&=-S\mathbf{a}.\\
\end{align*}

It follows that we have inductively
\begin{align*}
A\mathbf{v}&=S\mathbf{e}_2\\
A^2\mathbf{v}&=A(A\mathbf{v})=AS\mathbf{e}_2=S\mathbf{e}_3\\
\vdots\\
A^{n-1}\mathbf{v}&=A(A^{n-2}\mathbf{v})=AS\mathbf{e}_{n-1}=S\mathbf{e}_n.
\end{align*}
Namely we obtain
\[A^i\mathbf{v}=S\mathbf{e}_{i+1}\] for $i=0, 1, \dots, n-1$.
Now, suppose that we have the linear combinaiton
\[c_1\mathbf{v}+c_2A\mathbf{v}+ \dots c_n A^{n-1}\mathbf{v}=\mathbf{0},\] for some coefficient scalars $c_1, \dots, c_n \in \C$.

Then we have using the above relations
\begin{align*}
c_1S\mathbf{e}_1+c_2S\mathbf{e}_2+\cdots +c_nS\mathbf{e}_n=\mathbf{0}.
\end{align*}
Since $S$ is invertible, the column vectors $S\mathbf{e}_i$ are linearly independent.
Thus, we must have $c_1=c_2=\cdots=c_n=0$.

Therefore the vectors
\[\mathbf{v}, A\mathbf{v}, A^2\mathbf{v}, \dots, A^{n-1}\mathbf{v}\] are linearly independent, and they form a basis of $\C^n$ since we have $n$ linearly independent vectors in the $n$-dimensional vector space $\C^n$.

Related Question.

For a basic question about the companion matrix of a polynomial, check out the post “Companion matrix for a polynomial“.

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.

1. Find the trace of $A$.
2. Use the Cayley-Hamilton theorem

Proof.

We first calculate the trace of the matrix $A$ as follows. We have
\begin{align*}
\tr(A)&=\tr(AB-BA)\\
&=\tr(AB)-\tr(BA)\\
&=\tr(AB)-\tr(AB)=0.
\end{align*}

Thus $\tr(A)=0$ and it follows from the Cayley-Hamilton theorem (see below) for the $2\times 2$ matrix $A$ that
\begin{align*}
O&=A^2-\tr(A)A+\det(A)I\\
&=A^2+\det(A)I,
\end{align*}
where $I$ is the $2\times 2$ identity matrix.

Thus, we obtain
\[A^2=-\det(A)I. \tag{*}\]

Next, we compute $A^2$ in two ways.
We have
\begin{align*}
A^2=A(AB-BA)=A^2B-ABA
\end{align*}
and
\begin{align*}
A^2=(AB-BA)A=ABA-BA^2.
\end{align*}
Adding these two, we have
\begin{align*}
2A^2&=A^2B-BA^2\\
& \stackrel{(*)}{=} (-\det(A)I)B-B(-\det(A)I)\\
&=-\det(A)B+\det(A)B=O.
\end{align*}

As a result, we obtain $A^2=O$. This completes the proof.

The Cayley-Hamilton theorem for a $2\times 2$ matrix

Let us add the proof of the fact we used in the proof about the Cayley-Hamilton theorem.
Let $A=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}$ be a $2\times 2$ matrix.

Then its characteristic polynomial is
\begin{align*}
p(x)&=\det(A-xI)\\
&=\begin{vmatrix}
a-x & b\\
c& d-x
\end{vmatrix}\\
&=(a-x)(d-x)-bc\\
&=x^2-(a+d)x+ad-bc\\
&=x^2-\tr(A)x+\det(A),
\end{align*}
since $\tr(A)=a+d$ and $\det(A)=ad-bc$.

The Cayley-Hamilton theorem says that the matrix $A$ satisfies its characteristic equation $p(x)=0$.
Namely we have
\[A^2-\tr(A)A+\det(A)I=O.\] This is the equality we used in the proof.

Variation

As a variation of this problem, consider the following problem.

Let $A, B$ be $2\times 2$ matrices satisfying $A=AB-BA$.
Then prove that $\det(A)=0$.