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Let: $ b_1=\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix} b_2=\begin{bmatrix} 1 \\ 2 \\ 0 \\ 0 \\ \end{bmatrix} b_3=\begin{bmatrix} 1 \\ 2 \\ 3 \\ 0 \\ \end{bmatrix} b_4=\begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ \end{bmatrix}$

Show that a matrix $ A \in \mathcal{M}_{4 \times 4} $ is invertible if only if exist $x_1,x_2,x_3,x_4 \in \mathbb{R}^4$ such that $Ax_i=bi $ for $i=1,2,3,4$

The first implication es trivial since you can take $A^{-1}b_i$ as a solution but i can't prove the second implication. I've tried to use the fact if A is non invertible. Then its column vectors are Linearly dependent ¿Can you help me?

1 Answers1

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Let $X$ be a matrix whose columns are $x_i,i=1,2,3,4$ and let $B$ a matrix whose columns are $b_i, i=1,2,3,4$ then you get the equation $AX=B$, but $B$ is invertible , and this implies that $X$ is also invertible(why ?), and how can this imply that $A$ is invertible.

  • So $(B^{-1}A)X=I$ then $X(B^{-1}A)=I$, therefore $X^{-1}=(B^{-1}A)$ next $XB^{-1}=A^{-1}$ – Pablo Herrera Nov 30 '14 at 14:22
  • $X$ is invertible, because if not, then the columns are linearly dependent, i.e. there exist $\alpha_1,\alpha_2,\alpha_3,\alpha_4$ with $\alpha_1x_1+\alpha_2x_2+\alpha_3x_3+\alpha_4x_4=0$ which implies $A(\alpha_1x_1+\alpha_2x_2+\alpha_3x_3+\alpha_4x_4)=0$ i.e. $A\alpha_1x_1+A\alpha_2x_2+A\alpha_3x_3+A\alpha_4x_4=0$. So $\alpha_1b_1+\alpha_2b_2+\alpha_3b_3+\alpha_4b_4=0$. This means $b_1,b_2,b_3,b_4$ are linearly dependent, but this contradicts the fact that $B$ is invertible. So $X$ is now invertible and we can write $A= BX^{-1}$, and multiplication of two invertible matrices is invertible. –  Nov 30 '14 at 17:17
  • w.r.t your comment. please note, that $D.C=I$ doesn't always imply that $C.D=I$ –  Nov 30 '14 at 17:34
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    The reason why I say DC=I then CD=I is D and C are square matrices so I can use a theorem see this topic: http://math.stackexchange.com/questions/3852/if-ab-i-then-ba-i – Pablo Herrera Nov 30 '14 at 18:02
  • nice new information –  Nov 30 '14 at 18:25