Linear Independence Problems with solutions

The equation $c_1 u + c_2 v = \mathbf{0}$ yields the system:

Both equations reduce to $c_1 = -3c_2$. Since $c_2$ is a free variable, there are infinitely many solutions, for instance $c_1 = 3$, $c_2 = -1$.

Because the system has a nontrivial solution (not all coefficients zero), the set $\{u, v\}$ is linearly dependent. Explicitly, $3u - v = \mathbf{0}$, so $v = 3u$: one vector is a scalar multiple of the other.

Linearly dependent. There are 4 vectors in $\mathbb{R}^3$. Since the number of vectors exceeds the dimension of the space ($4 > 3$), the set must be linearly dependent.

Linearly dependent. Observe that $-6/4 = 3/(-2) = -9/6 = -3/2$, so the second vector equals $-\tfrac{3}{2}$ times the first. Two vectors are dependent if and only if one is a scalar multiple of the other.

Linearly dependent. Any set containing the zero vector is linearly dependent, because $0 \cdot v_1 + 1 \cdot \mathbf{0} + 0 \cdot v_3 = \mathbf{0}$ is a nontrivial combination equal to zero.

Column 3 is free ($c_3 = t$). From row 2: $-6c_2 = 6t \Rightarrow c_2 = -t$. From row 1: $c_1 = -3c_2 - 5c_3 = 3t - 5t = -2t$. Taking $t = 1$ gives the nontrivial solution $(c_1,c_2,c_3) = (-2,-1,1)$, so $-2v_1 - v_2 + v_3 = \mathbf{0}$.

There are 3 pivot positions, one in each column. The homogeneous system $Ax = \mathbf{0}$ has only the trivial solution, so $\{v_1, v_2, v_3\}$ is linearly independent.

There are 4 vectors in $\mathbb{R}^3$, and $4 > 3 = \dim(\mathbb{R}^3)$. Any set containing more vectors than the dimension of the ambient space is automatically linearly dependent.

Set $c_4 = 1$ (free after noting the system has a non-trivial kernel). From row 3: $6c_3 + 6 = 0 \Rightarrow c_3 = -1$. Row 2: $c_2 - 4(-1) - 3 = 0 \Rightarrow c_2 = -1$. Row 1: $c_1 + 2(-1) + 3 = 0 \Rightarrow c_1 = -1$. The dependence relation is:

Writing $A = [v_1|v_2|v_3|v_4]$:

Apply $R_4 \leftarrow R_4 - hR_1 - 2R_2 + R_3$:

The fourth pivot is $2 - 3h$. The set is linearly independent when $2 - 3h \neq 0$, i.e., $\boxed{h \neq \tfrac{2}{3}}$.

When $h = \tfrac{2}{3}$, row 4 vanishes and $c_4$ is free. Set $c_4 = 1$. Back-substitution gives $c_3 = -h = -\tfrac{2}{3}$, $c_2 = 1$, $c_1 = -2$. The dependence relation is:

Since $\det(A) = 23 \neq 0$, the matrix is invertible and the set $\{w_1, w_2, w_3\}$ is linearly independent.

Factoring: $\det(A(k)) = k(k^2 - 3)$.

The columns are linearly dependent exactly when $\det(A(k)) = 0$:

For any triangular matrix, $\det(M) = m_{11} m_{22} \cdots m_{nn}$. If all diagonal entries are nonzero, then $\det(M) \neq 0$. A square matrix with nonzero determinant is invertible; by the determinant test for independence, its columns are therefore linearly independent. $\square$

Let $M = \begin{pmatrix}0 & 1\\ 0 & 1\end{pmatrix}$. This is upper-triangular with $m_{11} = 0$. Its determinant is $0 \cdot 1 - 1 \cdot 0 = 0$, so its columns $\begin{pmatrix}0\\0\end{pmatrix}$ and $\begin{pmatrix}1\\1\end{pmatrix}$ are linearly dependent (the first column is the zero vector).

There are two pivot positions, so $\operatorname{rank}(A) = 2$.

$A$ has 3 columns but $\operatorname{rank}(A) = 2 < 3$. By the Rank-Nullity Theorem, $\operatorname{nullity}(A) = 3 - 2 = 1 > 0$, meaning $Ax = \mathbf{0}$ has a nontrivial solution. The columns are therefore linearly dependent.

Pivot columns: 1 and 4. So $\operatorname{rank}(B) = 2$ and $\operatorname{nullity}(B) = 4 - 2 = 2$.

Free variables: $x_2 = s$ and $x_3 = t$. From row 2: $x_4 = 0$. From row 1: $x_1 = x_2 - 2x_3 = s - 2t$. The general solution is

A basis for $\operatorname{null}(B)$ is $\left\{n_1 = (1,1,0,0)^T,\, n_2 = (-2,0,1,0)^T\right\}$. These are independent because the system $c_1 n_1 + c_2 n_2 = \mathbf{0}$ gives, from component 2: $c_1 = 0$, and from component 3: $c_2 = 0$.

By the Rank-Nullity Theorem applied to $A$ (with 6 columns): $\operatorname{nullity}(A) = 6 - 4 = 2$. $A^T$ is $6 \times 7$ with $\operatorname{rank}(A^T) = 4$, so $\operatorname{nullity}(A^T) = 7 - 4 = 3$. The four fundamental subspaces have dimensions:

Columns: $A$ has 6 columns but $\operatorname{rank}(A) = 4 < 6$. The columns are not linearly independent ($\operatorname{nullity}(A) = 2 > 0$).

Rows: The row space has dimension 4, but $A$ has 7 rows. Since $4 < 7$, the rows are also not linearly independent.

Since $W[f,g](x) = -3e^x \neq 0$ for every $x \in \mathbb{R}$, the functions $f$ and $g$ are linearly independent on $\mathbb{R}$.

Compute derivatives: $f_1' = 0$, $f_2' = -\sin(2x)$, $f_3' = \sin(2x)$, $f_2^{\prime\prime} = -2\cos(2x)$, $f_3{\prime\prime} = 2\cos(2x)$. Expanding along the first column:

The Wronskian is identically zero. For a general set of functions this is inconclusive, but dependence is confirmed directly below.

The Pythagorean identity states $\cos^2 x + \sin^2 x = 1$ for all $x \in \mathbb{R}$, so:

The coefficients $(c_1,c_2,c_3)=(1,-1,-1)$ are not all zero, so $\{f_1,f_2,f_3\}$ is linearly dependent.

Since $f_i^{(k)}(x) = \lambda_i^k e^{\lambda_i x}$, at $x=0$ we have $f_i^{(k)}(0) = \lambda_i^k$. The Wronskian matrix at $x=0$ is therefore:

which is the Vandermonde determinant $V(\lambda_1,\lambda_2,\lambda_3)$.

By the Vandermonde formula:

Since the $\lambda_i$ are distinct, each factor is nonzero, so $\det V \neq 0$. Therefore $W[f_1,f_2,f_3](0) \neq 0$, and $\{e^{\lambda_1 x}, e^{\lambda_2 x}, e^{\lambda_3 x}\}$ is linearly independent on $\mathbb{R}$. $\square$

Equating coefficients: $c_1+c_2=0$, $c_1-c_2=0$, $c_3=0$. The first two equations give $c_1=c_2$ and $c_1=-c_2$, so $c_1=c_2=0$. Combined with $c_3=0$, the only solution is trivial.

$\{p_1,p_2,p_3\}$ is linearly independent. Since $\dim(\mathcal{P}_2)=3$ and the set contains 3 independent vectors, it also spans $\mathcal{P}_2$ and is therefore a basis.

Suppose $c_1 A_1+c_2 A_2+c_3 A_3 = \mathbf{0}_{2\times2}$. Equating entries:

Only the trivial solution exists, so $\{A_1,A_2,A_3\}$ is linearly independent.

$\{A_1,A_2,A_3,A_4\}$ is linearly dependent since $\dim(M_{2\times2})=4$ and we can verify directly that $A_4 = A_1 + A_3$:

The dependence relation is $A_1 + A_3 - A_4 = \mathbf{0}$.

Suppose $c_1 T(v_1)+\cdots+c_k T(v_k)=\mathbf{0}_W$. By linearity of $T$:

So $\sum c_i v_i \in \ker(T) = \{\mathbf{0}_V\}$, giving $c_1 v_1+\cdots+c_k v_k=\mathbf{0}_V$. Since the $v_i$ are linearly independent in $V$, all $c_i=0$. Therefore $\{T(v_1),\ldots,T(v_k)\}$ is linearly independent. $\square$

Let $T:\mathbb{R}^2\to\mathbb{R}$ be the projection $T(x,y)=x$. Then $\ker(T)=\{(0,y)\}\neq\{\mathbf{0}\}$. The set $\{v_1,v_2\}=\{(1,0),(1,1)\}$ is linearly independent in $\mathbb{R}^2$, but $T(v_1)=1=T(v_2)$, so $\{T(v_1),T(v_2)\}=\{1,1\}$ is a repeated singleton, linearly dependent in $\mathbb{R}$.

Suppose $c_1 v_1+\cdots+c_k v_k = \mathbf{0}$. Fix any index $j$ and take the inner product of both sides with $v_j$:

By linearity and orthogonality ($\langle v_i,v_j\rangle=0$ for $i\neq j$):

Since $v_j \neq \mathbf{0}$, we have $\|v_j\|^2 > 0$, so $c_j = 0$. Since $j$ was arbitrary, all coefficients are zero and the set is linearly independent. $\square$