Let $C[-1, 1]$ be the vector space over $\R$ of all continuous functions defined on the interval $[-1, 1]$. Let
\[V:=\{f(x)\in C[-1,1] \mid f(x)=a e^x+b e^{2x}+c e^{3x}, a, b, c\in \R\}\] be a subset in $C[-1, 1]$.

(a) Prove that $V$ is a subspace of $C[-1, 1]$.

(b) Prove that the set $B=\{e^x, e^{2x}, e^{3x}\}$ is a basis of $V$.

(c) Prove that
\[B’=\{e^x-2e^{3x}, e^x+e^{2x}+2e^{3x}, 3e^{2x}+e^{3x}\}\] is a basis for $V$.

Proof.

(a) Prove that $V$ is a subspace of $C[-1, 1]$.

Note that each function in the subset $V$ is a linear combination of the functions $e^x, e^{2x}, e^{3x}$.
Namely, we have
\[V=\Span\{e^x, e^{2x}, e^{3x}\}\] and we know that the span is always a subspace. Hence $V$ is a subspace of $C[-1,1]$.

(b) Prove that the set $B=\{e^x, e^{2x}, e^{3x}\}$ is a basis of $V$.

We noted in part (a) that $V=\Span(B)$. So it suffices to show that $B$ is linearly independent.
Consider the linear combination
\[c_1e^x+c_2 e^{2x}+c_3 e^{3x}=\theta(x),\] where $\theta(x)$ is the zero function (the zero vector in $V$).
Taking the derivative, we get
\[c_1e^x+2c_2 e^{2x}+3c_3 e^{3x}=\theta(x).\] Taking the derivative again, we obtain
\[c_1e^x+4c_2 e^{2x}+9c_3 e^{3x}=\theta(x).\]

Evaluating at $x=0$, we obtain the system of linear equations
\begin{align*}
c_1+c_2+c_3&=0\\
c_1+2c_2+3c_3&=0\\
c_1+4c_2+9c_3&=0.
\end{align*}

We reduce the augmented matrix for this system as follows:
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 1 & 1 & 0 \\
1 &2 & 3 & 0 \\
1 & 4 & 9 & 0
\end{array} \right] \xrightarrow[R_3-R_1]{R_2-R_1}
\left[\begin{array}{rrr|r}
1 & 1 & 1 & 0 \\
0 &1 & 2 & 0 \\
0 & 3 & 8 & 0
\end{array} \right] \xrightarrow[R_3-3R_2]{R_1-R_2}\\[6pt] \left[\begin{array}{rrr|r}
1 & 0 & -1 & 0 \\
0 &1 & 2 & 0 \\
0 & 0 & 2 & 0
\end{array} \right] \xrightarrow{\frac{1}{2}R_3}
\left[\begin{array}{rrr|r}
1 & 0 & -1 & 0 \\
0 &1 & 2 & 0 \\
0 & 0 & 1 & 0
\end{array} \right] \xrightarrow[R_2-2R_2]{R_1+R_3}
\left[\begin{array}{rrr|r}
1 & 0 & 0 & 0 \\
0 &1 & 0 & 0 \\
0 & 0 & 1 & 0
\end{array} \right].
\end{align*}
It follows that the solution of the system is $c_1=c_2=c_3=0$.
Hence the set $B$ is linearly independent, and thus $B$ is a basis for $V$.

Anotehr approach.

Alternatively, we can show that the coefficient matrix is nonsingular by using the Vandermonde determinant formula as follows.
Observe that the coefficient matrix of the system is a Vandermonde matrix:
\[A:=\begin{bmatrix}
1 & 1 & 1 \\
1 &2 &3 \\
1^2 & 2^2 & 3^2
\end{bmatrix}.\] The Vandermonde determinant formula yields that
\[\det(A)=(3-1)(3-2)(2-1)=2\neq 0.\] Hence the coefficient matrix $A$ is nonsingular.
Thus we obtain the solution $c_1=c_2=c_3=0$.

(c) Prove that $B’=\{e^x-2e^{3x}, e^x+e^{2x}+2e^{3x}, 3e^{2x}+e^{3x}\}$ is a basis for $V$.

We consider the coordinate vectors of vectors in $B’$ with respect to the basis $B$.
The coordinate vectors with respect to basis $B$ are
\[[e^x-2e^{3x}]_B=\begin{bmatrix}
1 \\
0 \\
-2
\end{bmatrix}, [e^x+e^{2x}+2e^{3x}]_B=\begin{bmatrix}
1 \\
1 \\
2
\end{bmatrix}, [3e^{2x}+e^{3x}]_B=\begin{bmatrix}
0 \\
3 \\
1
\end{bmatrix}.\] Let $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ be these vectors and let $T=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$.
Then we know that $B’$ is a basis for $V$ if and only if $T$ is a basis for $\R^3$.

We claim that $T$ is linearly independent.
Consider the matrix whose column vectors are $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$:
\begin{align*}
\begin{bmatrix}
1 & 1 & 0 \\
0 &1 &3 \\
-2 & 2 & 1
\end{bmatrix}
\xrightarrow{R_3+2R_1}
\begin{bmatrix}
1 & 1 & 0 \\
0 &1 &3 \\
0 & 4 & 1
\end{bmatrix}
\xrightarrow[R_3-4R_1]{R_1-R_2}\\[6pt] \begin{bmatrix}
1 & 0 & -3 \\
0 &1 &3 \\
0 & 0 & -11
\end{bmatrix}
\xrightarrow{-\frac{1}{11}R_3}
\begin{bmatrix}
1 & 0 & -3 \\
0 &1 &3 \\
0 & 0 & 1
\end{bmatrix}
\xrightarrow[R_2-3R_3]{R_1+3R_3}
\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 1
\end{bmatrix}.
\end{align*}

Thus, the matrix is nonsingular and hence the column vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly independent.
As $T$ consists of three linearly independent vectors in the three-dimensional vector space $\R^3$, we conclude that $T$ is a basis for $\R^3$.
Therefore, by the correspondence of the coordinates, we see that $B’$ is a basis for $V$.

Related Question.

If you know the Wronskian, then you may use the Wronskian to prove that the exponential functions $e^x, e^{2x}, e^{3x}$ are linearly independent.

See the post
Using the Wronskian for Exponential Functions, Determine Whether the Set is Linearly Independent for the details.

Try the next more general question.

The solution is given in the post ↴
Exponential Functions are Linearly Independent

The post Exponential Functions Form a Basis of a Vector Space first appeared on Problems in Mathematics.

Let \[A=\begin{bmatrix}
a_0 & a_1 & \dots & a_{n-2} &a_{n-1} \\
a_{n-1} & a_0 & \dots & a_{n-3} & a_{n-2} \\
a_{n-2} & a_{n-1} & \dots & a_{n-4} & a_{n-3} \\
\vdots & \vdots & \dots & \vdots & \vdots \\
a_{2} & a_3 & \dots & a_{0} & a_{1}\\
a_{1} & a_2 & \dots & a_{n-1} & a_{0}
\end{bmatrix}\] be a complex $n \times n$ matrix.
Such a matrix is called circulant matrix.
Then prove that the determinant of the circulant matrix $A$ is given by
\[\det(A)=\prod_{k=0}^{n-1}(a_0+a_1\zeta^k+a_2 \zeta^{2k}+\cdots+a_{n-1}\zeta^{k(n-1)}),\] where $\zeta=e^{2 \pi i/n}$ is a primitive $n$-th root of unity.

Proof.

Let $\omega$ be any $n$-th root of unity.
Consider the vector
\[\mathbf{v}=\begin{bmatrix}
1 \\
\omega \\
\omega^2 \\
\vdots \\
\omega^{n-1}
\end{bmatrix}.\] We show that the vector $\mathbf{v}$ is an eigenvector of $A$.
We compute
\begin{align*}
A\mathbf{v}=\
\begin{bmatrix}
a_0 & a_1 & \dots & a_{n-2} &a_{n-1} \\
a_{n-1} & a_0 & \dots & a_{n-3} & a_{n-2} \\
a_{n-2} & a_{n-1} & \dots & a_{n-4} & a_{n-3} \\
\vdots & \vdots & \dots & \vdots & \vdots \\
a_{2} & a_3 & \dots & a_{0} & a_{1}\\
a_{1} & a_2 & \dots & a_{n-1} & a_{0}
\end{bmatrix}
\begin{bmatrix}
1 \\
\omega \\
\omega^2 \\
\vdots \\
\omega^{n-1}\\
\end{bmatrix}.
\end{align*}

The first entry of the vector $A\mathbf{v}$ is
\[a_0+a_1\omega+a_2\omega^2+\cdots a_{n-2}\omega^{n-2}+a_{n-1}\omega^{n-1}=:\lambda.\] We define $\lambda$ to be this number.

The second entry is
\begin{align*}
&a_{n-1}+a_0\omega+\cdots+a_{n-3}\omega^{n-2}+a_{n-2}\omega^{n-1}\\
&=(a_{n-1}\omega^{n-1}+a_0+\cdots+a_{n-3}\omega^{n-3}+a_{n-2}\omega^{n-2})\omega \\
&=(a_0+\cdots+a_{n-3}\omega^{n-3}+a_{n-2}\omega^{n-2}+a_{n-1}\omega^{n-1})\omega \\
&=\lambda \omega
\end{align*}

Similarly the $i$-th entry of the vector $A\mathbf{v}$ is
\begin{align*}
&a_{n-i+1}+a_{n-i+2}\omega +\cdots+ a_{n-i}\omega^{n-1}\\
&= (a_{n-i+1}\omega^{n-i+1}+a_{n-i+2}\omega^{n-i+2} +\cdots+ a_{n-i}\omega^{n-i})\omega^{i-1}\\
&=\lambda \omega^{i-1} .
\end{align*}
Therefore we obtain
\[A\mathbf{v}=\begin{bmatrix}
\lambda \\
\lambda\omega \\
\lambda \omega^2 \\
\vdots \\
\lambda\omega^{n-1}
\end{bmatrix}=\lambda \mathbf{v}.\]

Since $\mathbf{v}$ is a nonzero vector, it follows that $\lambda$ is an eigenvalue of $A$ and $\mathbf{v}$ is an eigenvector corresponding to $\lambda$.

The above argument holds for any $n$-th root of unity $\omega$.
We take $\omega=\zeta^k$, where $k$ runs from $0$ to $n-1$.
It follows that the vector
\[\mathbf{v}_k:=\begin{bmatrix}
1 \\
\zeta^k \\
\zeta^{2k} \\
\vdots \\
\zeta^{k(n-1)}
\end{bmatrix}\] is eigenvector corresponding to the eigenvalue
\[\lambda_k:=a_0+a_1\zeta^k+a_2\zeta^{2k}+\cdots a_{n-2}\zeta^{k(n-2)}+a_{n-1}\zeta^{k(n-1)}\] for each $k=0,1, \dots, n-1$.

We claim that the vectors $\mathbf{v}_k$ are linearly independent.

To see this, form a matrix whose column vectors are these vectors. That is, we consider
\[B=\begin{bmatrix}
1& 1 & 1 & \dots & 1 &1 \\
1&\zeta & \zeta^{2} & \dots & \zeta^{n-2} & \zeta^{n-1} \\
1&\zeta^{2} & \zeta^{4} & \dots & \zeta^{2(n-2)} & \zeta^{2(n-1)} \\
1& \zeta^{3} & \zeta^{6} & \dots & \zeta^{3(n-2)} & \zeta^{3(n-1)} \\
\vdots& \vdots & \vdots & \dots & \vdots & \vdots \\
1& \zeta^{n-1} & \zeta^{2(n-1)} & \dots & \zeta^{(n-1)(n-2)} &\zeta^{(n-1)(n-1)}
\end{bmatrix}.\]

This is the Vandermonde matrix and its determinant is
\[\det(B)=\prod_{i < j }(\zeta^j-\zeta^i)\neq 0.\] Thus, the matrix $B$ is nonsingular, hence its column vectors are linearly independent. It follows that $\lambda_k$, $k=0, 1, \dots, n$ are all the eigenvalues of $A$. Since the determinant is the product of all the eigenvalues of $A$, we have \begin{align*} \det(A)&=\prod_{k=0}^{n-1}\lambda_k\\ &=\prod_{k=0}^{n-1}(a_0+a_1\zeta^k+a_2 \zeta^{2k}+\cdots+a_{n-1}\zeta^{k(n-1)}), \end{align*} as required. This completes the proof.

The post Determinant of a General Circulant Matrix first appeared on Problems in Mathematics.

Let $c_1, c_2,\dots, c_n$ be mutually distinct real numbers.

Show that exponential functions
\[e^{c_1x}, e^{c_2x}, \dots, e^{c_nx}\] are linearly independent over $\R$.

Hint.

1. Consider a linear combination \[a_1 e^{c_1 x}+a_2 e^{c_2x}+\cdots + a_ne^{c_nx}=0.\]
2. Differentiate this equality $n-1$ times and you will get $n$ equations.
3. Write a matrix equation for the system. You will see the Vandermonde matrix.

Proof.

Suppose that we have a linear combination of these functions that is zero.
Namely, suppose we have
\[a_1 e^{c_1 x}+a_2 e^{c_2x}+\cdots + a_ne^{c_nx}=0\] for some real numbers $a_1, a_2, \dots, a_n$.

We want to show that the coefficients $a_1, a_2, \dots, a_n$ are all zero.

By differentiating the equation, we obtain
\[a_1c_1e^{c_1x}+a_2c_2e^{c_2x}+\cdots +a_n c_n e^{c_n x}=0.\] Differentiating repeatedly we further obtain the equalities
\begin{align*}
& a_1c_1^2e^{c_1x}+a_2c_2^2e^{c_2x}+\cdots +a_n c_n^2 e^{c_n x}=0\\
& a_1c_1^3e^{c_1x}+a_2c_2^3e^{c_2x}+\cdots +a_n c_n^3 e^{c_n x}=0\\
& \dots \\
& a_1c_1^{n-1}e^{c_1x}+a_2c_2^{n-1}e^{c_2x}+\cdots +a_n c_n^{n-1}
e^{c_n x}=0\\
\end{align*}

We rewrite these $n$ equations into the following matrix equation.
\[\begin{bmatrix}
1 & 1 & \dots &1 \\
c_1 & c_2 & \dots & c_n \\[3pt] c_1^2 & c_2^2 & \dots & c_n^2 \\[3pt] \vdots & \vdots & \vdots & \vdots \\[3pt] c_1^{n-1} & c_2^{n-1} & \dots & c_n^{n-1}
\end{bmatrix}
\begin{bmatrix}
a_1 e^{c_1x} \\
a_2 e^{c_2x} \\
\vdots \\
a_n e^{c^nx}
\end{bmatrix}=\begin{bmatrix}
0 \\
\vdots \\
0
\end{bmatrix} \tag{*}
\]

The determinant of the left matrix is
\[\det \begin{bmatrix}
1 & 1 & \dots &1 \\
c_1 & c_2 & \dots & c_n \\[3pt] c_1^2 & c_2^2 & \dots & c_n^2 \\[3pt] \vdots & \vdots & \vdots & \vdots \\[3pt] c_1^{n-1} & c_2^{n-1} & \dots & c_n^{n-1}
\end{bmatrix}=\prod_{i<j}(c_j-c_i)\] by the Vandermonde determinant.

Since by assumption $c_1,\dots, c_n$ are distinct, the determinant is not zero.
Therefore by multiplying equality (*) by the inverse on the left, we obtain
\[\begin{bmatrix}
a_1 e^{c_1x} \\
a_2 e^{c_2x} \\
\vdots \\
a_n e^{c^nx}
\end{bmatrix}=\begin{bmatrix}
0 \\
\vdots \\
0
\end{bmatrix}. \] Since the functions $e^{c_i x}$ are always positive, we must have $a_1=a_2=\cdots=a_n=0$ as required.
Therefore the functions $e^{c_1 x}, \dots, e^{c_n x}$ are linearly independent.

Comment.

The determinant that we considered above is called the Wronskian for the set of functions $\{e^{c_1x}, e^{c_2x}, \dots, e^{c_nx}\}$.

Related Question.

The following problems are more concrete versions of the current problem.

See the post ↴
Exponential Functions Form a Basis of a Vector Space
for the solution.

The solutions is given in the post ↴
Using the Wronskian for Exponential Functions, Determine Whether the Set is Linearly Independent

The post Exponential Functions are Linearly Independent first appeared on Problems in Mathematics.

Let $A$ be an $n \times n$ matrix such that $\tr(A^n)=0$ for all $n \in \N$.
Then prove that $A$ is a nilpotent matrix. Namely there exist a positive integer $m$ such that $A^m$ is the zero matrix.

Steps.

1. Use the Jordan canonical form of the matrix $A$.
2. We want to show that all eigenvalues are zero. (Review Nilpotent matrix and eigenvalues of the matrix)
3. Seeking a contradiction, assume some eigenvalues are not zero.
4. Using  $\tr(A^n)=0$, create a system of linear equations.
5. Calculate the determinant of the coefficient matrix of the system using the Vandermonde matrix formula.
6. Find a contradiction.

Proof.

We first want to prove that all the eigenvalues of $A$ must be zero.
Seeking a contradiction, assume that some of the eigenvalues of $A$ are not zero.
So assume that $\lambda_i$, $i=1, \dots, r$ are distinct nonzero eigenvalues of $A$ and each $m_i \geq 1$ is a multiplicity of $\lambda_i$.

We use the Jordan canonical form of the matrix $A$.
There exists an invertible matrix $S$ such that $S^{-1}AS=T$, where $T$ is an upper triangular matrix.The diagonal entries are eigenvalues of $A$.

Then we have for any positive integer $n$,
\begin{align*}
0 &= \tr(A^n)=\tr((STS^{-1})^n)=\tr(ST^n S^{-1})=\tr(T^{n}) \\
&=m_1 \lambda_1^n +m_2 \lambda_2^n+ \cdots + m_r \lambda_r^n.
\end{align*}

Note that since $T$ is an upper triangular matrix, the nonzero diagonal entries of $T^n$ are $\lambda_i^n$ appearing $m_i$ times.
Changing $n$ from $1$ to $r$, we obtain the system of linear equation. (Think $m_i$ as variables.)
\begin{align*}
m_1 \lambda_1^1 +m_2 \lambda_2^1+ \cdots + m_r \lambda_r^1 &=0 \\
m_1 \lambda_1^2 +m_2 \lambda_2^2+ \cdots + m_r \lambda_r^2 &=0 \\
& \vdots \\
m_1 \lambda_1^r +m_2 \lambda_2^r+ \cdots + m_r \lambda_r^r &=0 \\
\end{align*}
Equivalently, we have the matrix equation
\begin{align*}
\begin{bmatrix}
\lambda_1^1 & \lambda_2^1 & \cdots & \lambda_r^1 \\
\lambda_1^2 & \lambda_2^2& \cdots & \lambda_r^2 \\
\vdots & \vdots & \vdots & \vdots \\
\lambda_1^r & \lambda_2^r&\cdots & \lambda_r^r \\
\end{bmatrix}
\begin{bmatrix}
m_1 \\
m_2 \\
\vdots \\
m_r
\end{bmatrix}
=
\begin{bmatrix}\tag{*}
0 \\
0 \\
\vdots \\
0
\end{bmatrix}.
\end{align*}

Let $B$ denote the matrix above whose entries are powers of eigenvalues $\lambda_i$.
We calculate the determinant of $B$.
\begin{align*}
\det(B)&=\lambda_1 \lambda_2 \cdots \lambda_r
\begin{bmatrix}
1 & 1 & \cdots & 1 \\
\lambda_1 & \lambda_2& \cdots & \lambda_r \\
\vdots & \vdots & \vdots & \vdots \\
\lambda_1^{r-1} & \lambda_2^{r-1}&\cdots & \lambda_r^{r-1} \\
\end{bmatrix} \\[6pt] &=\lambda_1 \lambda_2 \cdots \lambda_r \prod_{1 \leq i<j \leq n}(\lambda_j-\lambda_i) \neq 0.
\end{align*}
(Note that the matrix above is a Vandermonde matrix.)

Thus the matrix $B$ is invertible. This means that the matrix equation (*) has unique solution
\begin{align*}
\begin{bmatrix}
m_1 \\
m_2 \\
\vdots \\
m_r
\end{bmatrix}
=
\begin{bmatrix}\tag{*}
0 \\
0 \\
\vdots \\
0
\end{bmatrix}.
\end{align*}
But this is a contradiction because each multiplicity $m_i$ is grater than zero.

Thus we proved that all eigenvalues of $A$ are zero.

Recall that a matrix is nilpotent if and only if its eigenvalues are zero.
See the post ↴
Nilpotent Matrix and Eigenvalues of the Matrix
for a proof of this fact.

Hence it follows from this fact that $A$ is a nilpotent matrix.

The post If Every Trace of a Power of a Matrix is Zero, then the Matrix is Nilpotent first appeared on Problems in Mathematics.