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Let $v1,v2,...,v_n$ column vectors of invertible matrix $A \in \mathcal{M}_{n\times n}$ with coefficients in $\mathbb{R}$ and $M \in \mathcal{M}_{n\times n}$ such that the i-th column vector of $M$ is $v_1+v_2+...+v_i$. Prove the matrix $M$ is invertible.

I try to use the fact if A is invertible then $v_1,v_2,..,v_n$ are linearly independent but I don't know if linearly independence implies $A$ is invertible for square matrix ¿there is another way to solve it?

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2 Answers 2

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I think I can help you get started.

Let's look at a 3 by 3 case ok?

You have the matrix $A=(v_1,v_2,v_3)$ where $v_i$ are vectors in $\mathbb R^3$

Do you agree with me that $A*\begin{pmatrix} 1 & 1 & 1\\ 0 & 1& 1\\0 & 0 & 1\end{pmatrix}$ will yield the matrix $M=(v_1,v_1+v_2,v_1+v_2+v_3)$?

Using that determinant of product is product of determinant, we can infer $\det(M)=\det(A)$

and so $M$ is invertible.

I leave it to you to prove that this argument still stands for all $n$.

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Using column operations, you can prove that the determinant of the new matrix is the same as the old one's:

First reduce C2-C1 to get C2 to be v2+v1-v1=v2, next do C3-C2-C1 to get C3=v1+v2+v3-v2-v1=v3, and so on until you reach detA.

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