Non-invertible maps collapse at least one direction

Dependent columns mean several inputs share outputs; the kernel captures those differences. Geometry squishes space onto a smaller-dimensional image .

If $A\mathbf{x}=\mathbf{0}$ for nonzero $\mathbf{x}$, the map is not injective. Full column rank implies kernel $\{\mathbf{0}\}$ in the domain. A $2\times 2$ matrix with determinant $0$ sends the plane to at most a line, or a point if the matrix is zero.

Teachers sometimes say "information loss"; linear algebra makes that precise via rank and nullity. Injective maps give at most one $\mathbf{x}$ for any $\mathbf{b}$, so solutions are unique when they exist.

Watch collapse visually: two inputs differing by a kernel vector land on the same output. That overlap is why non-invertible maps fail to be reversible .

Check your understanding. The tasks below rest on these ideas: Correct: a nonzero vector in the null space means $\mathbf{x}$ and $\mathbf{0}$ both map to $\mathbf{0}$, so the map is not one-to-one. Not quite: this rules out invertibility, says nothing about orthogonality, and does not force $A$ to be zero. Correct: independent columns admit no nontrivial relation, so only $\mathbf{0}$ maps to $\mathbf{0}$. Not quite: a nontrivial null space would mean dependence, and the null space is generally not the whole domain or the column space. Correct: zero determinant collapses area, dropping the image to a line or, for the zero matrix, a point. Not quite: it cannot preserve the full plane, never produces a circle, and only collapses to a point in the degenerate zero case. Correct: one-to-one maps give uniqueness, though existence (whether $\mathbf{b}$ is in the image) is a separate question. Not quite: injectivity is about uniqueness, not guaranteed existence, multiplicity, or forcing $\mathbf{b}=\mathbf{0}$.