Column picture of $AB$: $A$ times each column of $B$

Column $j$ of $AB$ equals $A$ times column $j$ of $B$. Thus the column space of $AB$ lies inside the column space of $A$ .

This is the fast way to build products mentally: track where $B$'s basis lands, then ask $A$ to move those destinations. If columns of $B$ are $\mathbf{b}_1,\mathbf{b}_2$, columns of $AB$ are $A\mathbf{b}_1, A\mathbf{b}_2$ in order.

If $B$ has rank $r$ and $A$ is invertible, rank of $AB$ equals $r$. Invertible $A$ preserves rank when multiplying on the left. Identity $I$ acts as right and left multiplicative neutral element for compatible shapes.

Outputs of $AB$ are always $A$ applied to vectors already in the span of columns of $A$. The column picture keeps geometry visible through composition .

Check your understanding. The tasks below rest on these ideas: Correct: every column of $AB$ is $A$ applied to something, so all outputs lie in $\mathrm{Col}(A)$. Not quite: it is bounded by $A$'s column space, not $B$'s, and certainly not by a null space. Correct: multiplying by an invertible matrix neither creates nor destroys independent directions. Not quite: the rank does not collapse to $0$, jump to full, or grow by the matrix size. Correct: multiplying by $I$ leaves a matrix unchanged on either side. Not quite: $I$ is not a zero divisor, not a projection, and is its own inverse only, not everyone's. Correct: column $j$ of $AB$ is $A$ applied to column $j$ of $B$, order preserved. Not quite: you multiply on the left by $A$, not on the right, the order is not reversed, and the result is columns, not a dot product.