Properties of Determinants: Adding Columns Question

Question

Hello Mathematics community,

I came across two properties of determinants that are causing me great confusion. For the first one, say $x_1$, $x_2$, $x_3$ are column 3-vectors. Then, $ \text{det} [ x_1+x_2\ x_2+x_3\ x_3+x_1] = 2\, \text{det} [ x_1\ x_2\ x_3]$.

For the second one, say $x_1$, $x_2$, $x_3$, $x_4$ are column 4-vectors. Then, $\text{det} [ x_1+x_2\ x_2+x_3\ x_3+x_4\ x_4+x_1] = 0$.

Understanding the technicalities of these two equations is causing me a bit of a headache. I understand the first one makes sense. Since one is essentially adding each column twice, the determinant would be twice as large. However, I cannot understand why the second equation holds true.

Answer 1

By the multilinearity of the determinant, applied to the first column, we have $$\det[x_1 + x_2,x_2 + x_3,\dots,x_n + x_1] = \det[x_1,x_2 + x_3,\dots,x_n + x_1] + \det[x_2,x_2 + x_3,\dots,x_n + x_1].$$

Now, each of these terms can be simplified a lot by subtracting one column from another. We have \begin{align} \det[x_1,x_2 + x_3,\dots,x_{n-1} + x_n, x_n + x_1] &= \det[x_1,x_2 + x_3,\dots,x_{n-1} + x_n, x_n] \\ &= \det[x_1,x_2 + x_3,\dots,x_{n-1}, x_n] \\ &= \dots \\ &= \det[x_1,x_2,\dots,x_{n-1}, x_n] \end{align} by subtracting the first column from the last, the last from the next to last, the next to last from the one before it, and so on. Similarly, we have \begin{align} \det[x_2,x_2 + x_3,\dots,x_{n-1} + x_n, x_n + x_1] &= \det[x_2,x_3,\dots,x_{n-1} + x_n, x_n + x_1] \\ &= \dots \\ &= \det[x_2,x_3,\dots,x_n, x_n + x_1]\\ &= \det[x_2,x_3,\dots,x_{n}, x_1] \end{align} by subtracting the first column from the second, then the second from the third, and so on.

As a result, we get $$\det[x_1 + x_2,x_2 + x_3,\dots,x_n + x_1] = \det[x_1,x_2,\dots,x_{n-1}, x_n] + \det[x_2,x_3,\dots,x_{n}, x_1].$$

Now, the parity of $n$ becomes important. If $n$ is odd, the cyclic shift that takes us from $\det[x_1,x_2,\dots,x_{n-1}, x_n]$ to $\det[x_2,x_3,\dots,x_{n}, x_1]$ requires an even number of column swaps, so these two determinants have the same sign, and add. If $n$ is even, the cyclic shift requires an odd number of column swaps, so these two determinants have opposite signs, and cancel.

Answer 2

The first one is because\begin{multline}\det[x_1+x_2\ \ x_2+x_3\ \ x_3+x_1]=\det[x_1\ \ x_2\ \ x_3]+\det[x_1\ \ x_2\ \ x_1]+\det[x_1\ \ x_3\ \ x_3]+\\+\det[x_2\ \ x_2\ \ x_3]+\det[x_1\ \ x_3\ \ x_1]+\det[x_2\ \ x_2\ \ x_1]+\det[x_2\ \ x_3\ \ x_3]+\\+\det[x_2\ \ x_3\ \ x_1].\end{multline}Of these determinants, only the first and the last one aren't necessarily $0$. And the last one is equal to $\det[x_1\ \ x_2\ \ x_3]$.

The other determinant is $0$ because the columns are not linearly independent: $$(x_1+x_2)+(x_3+x_4)=(x_2+x_3)+(x_4+x_1).$$

Answer 3

In your notation, $$[x_1+x_2\ x_2+x_3\ x_3+x_4\ x_4+x_1 ] =[x_1\ x_2\ x_3\ x_4]\pmatrix{1&0&0&1\\1&1&0&0\\0&1&1&0\\0&0&1&1}.$$ The matrix $\pmatrix{1&0&0&1\\1&1&0&0\\0&1&1&0\\0&0&1&1}$ has determinant zero.