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?
Proving that the matrix is not invertible.
-
3Actually no. If $A$ and $B$ have your given dimensions, then $$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} $$ – 2012-11-23
-
0But i should prove that it is not. Let's take them as A 365x54 and B is 54x365 – 2012-11-23
-
0@AD.: Yes, that's precisely what I mean. – 2012-11-23
-
0@YigitCan So, can you specify exactly what the problem is? – 2012-11-23
-
3@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. – 2012-11-23
3 Answers
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,
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
[assuming $n