Let $A=(a_{ij})\in M_n$ be an arbitrary matrix and let
$A_1=\begin{pmatrix}
a_{11}\\
a_{21}\\
\vdots\\
a_{n1}\\
\end{pmatrix}$
$A_2=\begin{pmatrix}
a_{12}\\
a_{22}\\
\vdots\\
a_{n2}\\
\end{pmatrix}\ldots$
$A_n=\begin{pmatrix}
a_{1n}\\
a_{2n}\\
\vdots\\
a_{nn}\\
\end{pmatrix}\in M_{n1}$ be columns of $A$. Prove that if the set$\{A_1,A_2,…,A_n\} $ is linearly dependent in vector space $M_{n1}$, then $\det A=0$.
I know this already has an answer here but I don't understand OP's solution.
$\lambda_1 A_1 + \ldots + \lambda_n A_n = 0$ where not all $\lambda_i$ are zero.
Suppose that $\lambda_1 \neq 0$. Then we get \begin{align*} A_1 = – \frac{\lambda_2}{\lambda_1} A_2 – \ldots – \frac{\lambda_n}{\lambda_1} A_n. \end{align*}
Now what happens after that, with the determinant?
Best Answer
Since exchanging two columns only switches sign to the determinant, it is not restrictive to assume that the last column is a linear combination of the previous $n-1$ columns: $$ A_n=\alpha_1A_1+\dots+\alpha_{n-1}A_{n-1} $$ By multilinearity of the determinant, you have $$ \det A= \det\begin{bmatrix} A_1 & \dots & A_{n-1} & \sum\limits_{i=1}^{n-1}\alpha_iA_i\end{bmatrix}= \sum_{i=1}^{n-1}\alpha_i \det\begin{bmatrix} A_1 & \dots & A_{n-1} & A_i\end{bmatrix}=0 $$ because a matrix with two equal columns has zero determinant again by the above mentioned property above that exchanging two columns changes the sign of the determinant.
With $\begin{bmatrix} v_1 & \dots & v_{n-1} & v_n\end{bmatrix}$ I denote the matrix whose columns are the column vectors $v_1,\dots,v_{n-1},v_n$.