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I have seen in a few textbooks of proofs of $\det(A) = 0$ if the rows/columns of $A$ are linearly dependent from the properties of the determinant, but I have yet to see a proof that $\det(A) = 0$ from the definition of the determinant: $$\det(A) = \sum \text{sgn}(P)a_{1\alpha}a_{2\beta}...a_{n\omega}$$

where $(\alpha, \beta, ..., \omega)$ are permutations of the columns of $A$. And since all the properties of $\det$ follow from it's definition, I find that proving $\det(A) = 0$ if the rows/columns of $A$ are linearly dependent from the properties of the determinant, to be somewhat unsatisfactory.

Can someone provide a proof from the Leibniz Expansion definition of the determinant, that I provided above, of the result that $\det(A) = 0$ if the rows/columns of $A$ are linearly dependent?

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  • $\begingroup$ What is that "$\det(P)$" factor in your definition of $\;\det A\;$ ? $\endgroup$
    – DonAntonio
    Aug 28, 2016 at 11:11
  • $\begingroup$ @DonAntonio, $\det(P)$ is the determinant of the Permutation matrix given by the permutation of the columns $(\alpha, \beta, ..., \omega)$ $\endgroup$ Aug 28, 2016 at 11:13
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    $\begingroup$ Then the notation $\text{sgn}(P)$ may be better? $\endgroup$
    – Fei Li
    Aug 28, 2016 at 11:14
  • $\begingroup$ @Perturbative Thanks. Fei Li explains my point better as Libniz' definition of determinant uses the sygnum (sign) function, not other matrix's determinant within the definition of determinant, which imo could easily be seen as a circular definition. $\endgroup$
    – DonAntonio
    Aug 28, 2016 at 11:16
  • $\begingroup$ Then could you please edit your notation, cause otherwise I fear the question is prone to get downvote. $\endgroup$
    – Fei Li
    Aug 28, 2016 at 11:33

1 Answer 1

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Thanks to the comments below, this hint has been expanded to a complete solution.

Step 1.

Proposition. Suppose $A$ is a square matrix and $B$ is the matrix defined by interchanging two columns of $A$. Then $$\det A=-\det B.$$

Proof. Consider the definition. every term of $\det B$ in the sum is $(-1)$ times the corresponding term for $\det A$. Thus the whole $\det B$ differs from $\det A$ by $(-1)$.

Step 2.

Proposition.

  1. If $A$ is a square matrix that has two equal columns, then $\det A=0$.
  2. If we multiply a column of $A$ by $\lambda$, then $\det A=\lambda(\det A).$

Proof.

  1. Interchanging the two equal columns of $A$ gives the original matrix $A$. Then from step 1 we have $$\det A=-\det A,$$ which implies that $\det A=0.$
  2. This should be obvious from the definition.

Step 3.

Proposition. Let $A$ be a square matrix. Let $A'$ be obtained from $A$ by adding a multiple of one column to a different column. Then $\det A'=\det A$.

Proof. To convince oneself that the above is right, one first try some explicit calculations. Let's do that for $n=2$ and $n=3$.

For $n=2$, let $A=\begin{pmatrix}x_1&y_1\\x_2&y_2\end{pmatrix}$, and let $$A'=\begin{pmatrix}x_1+\lambda y_1&y_1\\x_2+\lambda y_2&y_2\end{pmatrix}.$$

Now $$\begin{split}\det A'&=(x_1+\lambda y_1)y_2-y_1(x_2+\lambda y_2)\\ &=x_1y_2+\lambda y_1y_2-x_2y_1-\lambda y_1y_2\\ &=x_1y_2-x_2y_1\\ &=\det A.\end{split}$$

For $n=3$, we let $A=\begin{pmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{pmatrix}$. We verify the claim for $$A'=\begin{pmatrix}x_1+\lambda z_1&y_1&z_1\\x_2+\lambda z_2&y_2&z_2\\x_3+\lambda z_3&y_3&z_3\end{pmatrix}.$$

Now $$\begin{split}\det A'= &(x_1+\lambda z_1)y_2z_3+(x_3+\lambda z_3)y_1z_2+(x_2+\lambda z_2)y_3z_1\\ &-(x_3+\lambda z_3)y_2z_1-(x_1+\lambda z_1)y_3z_2-(x_2+\lambda z_2)y_1z_3.\end{split}$$

Collecting terms that contain $\lambda$ and $-\lambda$:

$\begin{cases}\lambda z_1y_2z_3+\lambda z_3y_1z_2+\lambda z_2y_3z_1\\ -\lambda z_1y_2z_3-\lambda z_3y_1z_2-\lambda z_2y_3z_1\end{cases}$

so that those terms are canceled out, and what left is the $\det A$.

For a general proof, let's look again at the $n=3$ case. After one understand this, other $n$'s are just a play of notations.

Let $$A=\begin{pmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{pmatrix}.$$ To maximize our simplification of the messy notations in the definition of determinant, we have chosen different alphabets $x,y,z$ to name each column, so that we don't have to use numbers to name those columns. Furthermore, let's use $\sigma$ to denote the permutation, so that the definition of determinant of $A$ becomes: $$\det A=\sum_{\sigma}(\text{sgn}\sigma)x_{\sigma(1)}y_{\sigma(2)}z_{\sigma(3)},$$ where the sum is over all permutations of $\{1,2,3\}$ (Think of the $1,2,3$ in the $\sigma(1),\sigma(2),\sigma(3)$ as the domain of the function $\sigma$). Using this notation, it is pretty clear what the effect of adding a multiple of one column to another is:

Let $$A'=\begin{pmatrix}x_1+\lambda z_1&y_1&z_1\\x_2+\lambda z_2&y_2&z_2\\x_3+\lambda z_3&y_3&z_3\end{pmatrix}.$$ Then $$\begin{split}\det A' &=\sum_{\sigma}(\text{sgn}\sigma)\left(x_{\sigma(1)}+\lambda z_{\sigma(1)}\right)y_{\sigma(2)}z_{\sigma(3)}\\ &=\sum_{\sigma}(\text{sgn}\sigma)x_{\sigma(1)}y_{\sigma(2)}z_{\sigma(3)}+\sum_{\sigma}(\text{sgn}\sigma)\lambda z_{\sigma(1)}y_{\sigma(2)}z_{\sigma(3)}\\ &=\sum_{\sigma}(\text{sgn}\sigma)x_{\sigma(1)}y_{\sigma(2)}z_{\sigma(3)}+\lambda\sum_{\sigma}(\text{sgn}\sigma)z_{\sigma(1)}y_{\sigma(2)}z_{\sigma(3)}.\end{split}$$

The second summand is, by definition, the determinant of the following matrix

$$\begin{pmatrix}z_1&y_1&z_1\\z_2&y_2&z_2\\z_3&y_3&z_3\end{pmatrix}$$

which is $0$ by our step 2. What left is, again by definition, the determinant of $A$. Thus $\det A'=\det A$, and the proof is complete.

There is really no need for me to write out the general proof for arbitrary $n$, which add nothing to the above proof other than heavy notations. You can write that out by yourself if you wish.

Step 4.

Suppose the columns (or rows, but it's more natural for me to think about columns) of a square matrix $A$, denoted by $(C_1,C_2,\ldots,C_n)$, are linearly dependent. This means that one of $C_i$ can be expressed as a nontrivial linear combination of the other columns. For concreteness, let's say $C_1=\lambda_2C_2+\cdots+\lambda_nC_n$, where not all of $\lambda_2,\ldots,\lambda_n$ are zero. Then since by step 3, adding a multiple of one column to another does not change the determinant of $A$, we can add $-\lambda_nC_n,\ldots,-\lambda_3C_3$ to $C_1$ without changing the determinant. This leaves us $C_1=\lambda_2C_2$, from which $\det A=0$ follows by step 2.

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  • $\begingroup$ The question is not about one row being a multiple of another row, but about some linear dependency existing among all the rows. $\endgroup$ Aug 28, 2016 at 12:37
  • $\begingroup$ Christian Blatter is right, but since adding a multiple of one column/row to another different one does not change the determinant, one can eventually reduce the case to two linearly dependent columns/rows. To see this, Carsten S's suggestion is useful: just write out the calculations for $2\times 2$ or $3\times 3$ matrices explicitly. $\endgroup$
    – Fei Li
    Aug 28, 2016 at 13:05
  • $\begingroup$ It's not at all obvious from the OP's definition of the determinant that row operations leave it unchanged. That's why I made a comment to your answer. $\endgroup$ Aug 28, 2016 at 13:09

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