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September 27th, 2015, 09:23 PM   #1
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Question about invertible matrices

Need to show that if A is not invertible then exist B (nxn) such that AB=0 but B \= 0.

Thought about writing like A=E1...Ek*R where Ei are elementary matrices and R(different from I) the row reduced echelon matrice but doesnt help, any idea??

Last edited by bonildo; September 27th, 2015 at 09:37 PM.
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September 29th, 2015, 10:10 AM   #2
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anyone ??
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September 30th, 2015, 04:37 PM   #3
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Consider a matrix $A_{n\times n}$ which is not invertible. Then there exist $v=\left(a_1,\ldots,a_n\right)\in\mathbb R^n$ such that $v\neq 0$ and $Av=0$. Consider the matrix $B_{n\times n}=\left[\begin{array}{cccc}a_1 & a_1 & \ldots & a_1 \\a_2 & a_2 & \ldots & a_2\\\vdots & \vdots & \ldots & \vdots \\a_n & a_n & \ldots & a_n \end{array}\right]$. Since $v\neq 0$ we have that $B\neq 0$. We also have that $Be_i=v$ where $e_i=\left(0,0,\ldots,\underbrace{1}_{i^{th}}, \ldots,0\right)$ for $1\leq i\leq n$.\\

Consider a generic vector $w=\displaystyle\sum_{i=1}^n\alpha_ie_i$ for some $\alpha_i\in\mathbb R$. We have that $Bw=B\left(\displaystyle\sum_{i=1}^{n}\alpha_i e_i\right)=\displaystyle\sum_{i=1}^{n}\alpha_i Be_i=\displaystyle\sum_{i=1}^{n}\alpha_i v=\left(\displaystyle\sum_{i=1}^{n}\alpha_i\right) v$.\\

Hence $ABw=A\left(\left(\displaystyle\sum_{i=1}^n\alpha_ i\right)v\right)=\left(\displaystyle\sum_{i=1}^{n} \alpha_i\right)Av=0$ which give us that $AB=0$ for every vector in $\mathbb R^n$.
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