1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ What are you allowed to assume about Det? Can you assume that it is invariant under the addition/subtraction of columns? $\endgroup$ – Paul Jan 24 '17 at 14:17
  • 1
    $\begingroup$ These (columns of $A$ linearly dependent, $\det(A) = 0$) are two conditions on square matrix $A$ of several that are commonly shown to be equivalent in introductory linear algebra courses; the path to proving equivalence might depend on previous "machinery". In particular it is often shown beforehand that elementary row operations do not affect whether $\det(A)=0$, and this previous Question shows that such operations conserve linear dependence relations among columns. $\endgroup$ – hardmath Jan 24 '17 at 14:21
4
$\begingroup$

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 $v_1,\dots,v_{n-1},v_n$.

$\endgroup$
  • $\begingroup$ I don't understand. What is $\begin{bmatrix} A_1 & \dots & A_{n-1} & A_i\end{bmatrix}$? $\endgroup$ – lmc Jan 25 '17 at 13:41
  • $\begingroup$ @Lewis The matrix with the stated columns, I added some explanation. $\endgroup$ – egreg Jan 25 '17 at 15:42
2
$\begingroup$

I'll give a variant of the proof, hoping you'll understand better.

Suppose there's a non-trivial linear relation between the columns: $$\lambda_1A_1+\lambda_2A_2+\dots+\lambda_nA_n=0.$$ Say $\lambda_1\ne 0$. By linearity w.r.t. the 1st column, $\;\det( \lambda_1A_1,A_2,\dots,A_n)=\lambda_1\det(A_1,A_2,\dots,A_n)$. Also, the determinant is alternating and linear in each column, so $$ \lambda_1\det A=\det(\lambda_1A_+\lambda_2A_2+\dots+\lambda_nA_n,A_2,\dots,A_n)= \det(0,A_2,\dots,A_n)=0,$$ whence $\det A=0$.

$\endgroup$
  • $\begingroup$ Can you please explain this part a bit: $ \lambda_1\det A=\det(\lambda_1A_+\lambda_2A_2+\dots+\lambda_nA_n,A_2,\dots,A_n)= \det(0,A_2,\dots,A_n)=0$ $\endgroup$ – lmc Jan 25 '17 at 13:37
1
$\begingroup$

The determinant of $A$ is the product of the elements of the main diagonal when $A$ is converted to row echelon form. For a linearly dependent set of columns, when $A$ is converted to row echelon form, there will be a $0$ in the main diagonal of the matrix corresponding to the column of the free variable and hence, the determinant is $0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.