Examples

Dependent columns implies non-invertible

Dependent columns implies non-invertible

If a square matrix $A$ has column vectors ${\mathbf{a}}_i$ which form a dependent set, then the square matrix is not invertible.

Proof: Let $x_i$ be numbers giving the dependency relation. These numbers can be combined into a vector $\mathbf{x}=(x_1,\dots ,x_n)$. Then:

$$\begin{array}{rl}A\mathbf{x}& =A(x_1{\mathbf{e}}_1+\cdots +x_n{\mathbf{e}}_n)\\ & =A(x_1{\mathbf{e}}_1)+\cdots +A(x_n{\mathbf{e}}_n)\\ & =x_1(A{\mathbf{e}}_1)+\cdots +x_n(A{\mathbf{e}}_n)\\ & =x_1{\mathbf{a}}_1+\cdots +x_n{\mathbf{a}}_n\\ & =0.\end{array}$$

However, since the $x_i$ are not all zero, we know $\mathbf{x}\ne 0$, and therefore $A$ sends something nonzero to zero, which means it cannot be invertible. (Otherwise, what would be the preimage of zero?)

Independence

Independence

Problem: Is the following set of vectors dependent or independent?

$$\left(\begin{array}{c}3\\ -1\\ 2\end{array}\right),\left(\begin{array}{c}0\\ 2\\ -1\end{array}\right),\left(\begin{array}{c}2\\ -4\\ 3\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right)$$

Solution: It must be a dependent set. These vectors live in ${\mathbb{R}}^3$, which is spanned by ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$. If these vectors were independent, then we could use Steinitz Exchange to write ${\mathbb{R}}^3=⟨{\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3⟩$ as the span of just three of them. Since the fourth is also in ${\mathbb{R}}^3$, it could then be written as a combination of the other three, contradicting independence.

3 vectors spanning 3D space must be independent

3 vectors spanning 3D space must be independent

In Problem 05-02, we saw that the span

$$⟨\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right),\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 2\\ 2\end{array}\right)⟩$$

is a 3-dimensional subspace of ${\mathbb{R}}^3$. (In fact it is ${\mathbb{R}}^3$, since the Steinitz Exchange will put all three ${\mathbf{e}}_i$ in the same span as this.) Therefore we know the vectors are independent, because there are three vectors spanning a 3D space.