Proving that the matrix is not invertible.

Question

A is a 2x3 matrix and B is a 3x2. How can i prove that the matrix D = AB is not invertible. I could not go further in this problem. The only thing that i have found is the multiply of these two matrix will be 2x2 matrix but how can i find it is not invetible?

Answer 1

The statement would be true if you considered $D = BA$.

You can see that the matrix $A$ gives rise to a transformation $T_A : \mathbb R^3 \to \mathbb R^2$. Similarly, the matrix $B$ gives rise to $T_B : \mathbb R^2 \to \mathbb R^3$ and $T_D = T_B \circ T_A : \mathbb R^3 \to \mathbb R^3$. The problem with $T_D$ is that $$ \mathrm{Im}(T_D) \subseteq \mathrm{Im}(T_B) $$ and $T_B$ cannot be surjective because the image of a basis in $\mathbb R^2$ can span at most a subspace of $\mathbb R^3$ of dimension $2$, not $3$.

Hope that helps,

Answer 2

in general if $A$ is a $m\times n$ matrix and $B$ is a $n\times m$ matrix with $n < m$ then $AB$ cannot be invertible.

results used:

a matrix $A$ is invertible iff $Ax = 0$ has only trivial solution.

$A$ is a $m\times n$ matrix with $m < n$ then $Ax=0$ has non trivial solution.

there is nontrivial $x_0$ such that such that $Bx_0=0$ hence $AB(x_0) = 0$ as a result $AB$ cannot be invertible

Answer 3

[assuming $n<m$] Recall that $\text{rank}( A ),\text{rank}(B)\leqslant\min\{n,m\} $, and $\text{rank}(AB)\leqslant\min\{\text{rank}A,\text{rank}B\}$. The product is a $m\times m$ matrice with rank $\leqslant n < m$, and therefore cannot be invertible.