Linear Algebra Proofs

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Q7. Linear Dependence and Independence Definitions: Linear Dependence: A set of vectors $\{\vec{v}_1, \vec{v}_2, \ldots, \vec{v}_k\}$ in a vector space $V$ is said to be linearly dependent if there exist scalars $c_1, c_2, \ldots, c_k$, not all zero, such that their linear combination equals the zero vector: $c_1\vec{v}_1 + c_2\vec{v}_2 + \ldots + c_k\vec{v}_k = \vec{0}$ This implies that at least one vector in the set can be expressed as a linear combination of the others. Linear Independence: A set of vectors $\{\vec{v}_1, \vec{v}_2, \ldots, \vec{v}_k\}$ in a vector space $V$ is said to be linearly independent if the only way to form the zero vector as a linear combination of these vectors is by setting all scalars to zero: $c_1\vec{v}_1 + c_2\vec{v}_2 + \ldots + c_k\vec{v}_k = \vec{0} \quad \text{implies that} \quad c_1 = c_2 = \ldots = c_k = 0$ This means that no vector in the set can be expressed as a linear combination of the others. Test for Linear Dependence/Independence: Given the set of vectors: $\vec{v}_1 = (1,2,3)$, $\vec{v}_2 = (2,4,6)$, $\vec{v}_3 = (3,6,9)$. We need to determine if there exist scalars $c_1, c_2, c_3$, not all zero, such that: $c_1\vec{v}_1 + c_2\vec{v}_2 + c_3\vec{v}_3 = \vec{0}$ Substitute the given vectors: $c_1(1,2,3) + c_2(2,4,6) + c_3(3,6,9) = (0,0,0)$ This expands into a system of linear equations: $1c_1 + 2c_2 + 3c_3 = 0$ $2c_1 + 4c_2 + 6c_3 = 0$ $3c_1 + 6c_2 + 9c_3 = 0$ We can observe a relationship between the vectors directly. Notice that: $\vec{v}_2 = (2,4,6) = 2(1,2,3) = 2\vec{v}_1$ $\vec{v}_3 = (3,6,9) = 3(1,2,3) = 3\vec{v}_1$ From $\vec{v}_2 = 2\vec{v}_1$, we can rearrange this to $2\vec{v}_1 - \vec{v}_2 + 0\vec{v}_3 = \vec{0}$. Here, we have scalars $c_1=2$, $c_2=-1$, $c_3=0$. Since not all these scalars are zero (e.g., $c_1=2 \neq 0$ and $c_2=-1 \neq 0$), the set of vectors is linearly dependent. Alternatively, using the system of equations: We can write the augmented matrix for this system: $\begin{pmatrix} 1 & 2 & 3 & | & 0 \\ 2 & 4 & 6 & | & 0 \\ 3 & 6 & 9 & | & 0 \end{pmatrix}$ Perform Row Operations: $R_2 \to R_2 - 2R_1$: $R_3 \to R_3 - 3R_1$: $\begin{pmatrix} 1 & 2 & 3 & | & 0 \\ 0 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & | & 0 \end{pmatrix}$ The system reduces to a single equation: $c_1 + 2c_2 + 3c_3 = 0$. This system has infinitely many solutions (non-trivial solutions). We can choose $c_2$ and $c_3$ as free variables. Let $c_2 = s$ and $c_3 = t$, where $s, t$ are any real numbers (scalars). Then $c_1 = -2s - 3t$. For example, if we choose $s=1$ and $t=0$, then $c_1 = -2$, $c_2 = 1$, $c_3 = 0$. This gives the linear combination: $-2(1,2,3) + 1(2,4,6) + 0(3,6,9) = (-2,-4,-6) + (2,4,6) + (0,0,0) = (0,0,0)$. Since we found scalars $c_1=-2, c_2=1, c_3=0$ (not all zero) that satisfy the equation, the vectors are linearly dependent. Conclusion: The given set of vectors $\{(1,2,3), (2,4,6), (3,6,9)\}$ is linearly dependent because there exist non-zero scalars (e.g., $c_1=2, c_2=-1, c_3=0$) such that their linear combination equals the zero vector.