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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?

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    $\begingroup$ Actually no. If $A$ and $B$ have your given dimensions, then $$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} $$ $\endgroup$ – Patrick Da Silva Nov 23 '12 at 11:52
  • $\begingroup$ But i should prove that it is not. Let's take them as A 365x54 and B is 54x365 $\endgroup$ – Yigit Can Nov 23 '12 at 11:55
  • $\begingroup$ @AD.: Yes, that's precisely what I mean. $\endgroup$ – Patrick Da Silva Nov 23 '12 at 11:56
  • $\begingroup$ @YigitCan So, can you specify exactly what the problem is? $\endgroup$ – AD. Nov 23 '12 at 11:57
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    $\begingroup$ @Yigit : In your new example you switched the conditions ; if $A$ has size $m \times n$ and $B$ has size $n \times m$, the statement is true for all matrices only when $n > m$, i.e. when the transformation $T_B$ in my answer cannot be surjective. $\endgroup$ – Patrick Da Silva Nov 23 '12 at 11:57
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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,

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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

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[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.

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