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Let $n \in \mathbb{N}$

True or false? If the column-vectors of $A \in \mathbb{R}^{n \times n}$ are linearly dependent, then we have that $\text{det(A)}=0$

The statement should be true and it shouldn't even matter if the column- or row-vectors are linearly dependent. Linearly dependent means we will always get $\text{det(A)}=0$.


I hope this is right? If it's right, can you tell me a better reasoning?

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It is right. If they are dependent, the rank isn't $n$, and a well-known theorem tells you that $det(A)=0$. That's the usual reasoning.

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If you add a multiple of one column to another column, the determinant stays the same. The determinant of a matrix with a zero column is zero, so you only need to prove that if the columns are linearly dependent it is possible to obtain a zero column.

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