How can I calculate the determinant without knowing every element of the matrix?

Question

I was given the following matrix and asked only to find the determinant. As far as I know we need to know every element of the matrix in order to do that. But maybe I'm missing something. Any help is appreciated. Thanks in advance.

Answer 1

I'm assuming the pattern continues - 1's on the lower-left-upper-right diagonal and 0's everywhere else. Since the only nonzero product is the lower-left-upper-right diagonal, and that gets subtracted, the determinant is -1. (Try it with square matrices of various sizes on a calculator.)

Answer 2

First, as noted above, the problem statement actually gives you all the elements: it is implied that the matrix has $1$ at the main skew diagonal and 0 everywhere else.

Second, you can formally prove that the answer is given by the recurrence: denote as $d_{n}$ the value of determinant of $n$-th matrix of the form indicated in the problem statement. Then $$d_{1}=1$$ $$d_{n}=(-1)^{n-1}d_{n-1}$$ Indeed, perform the Laplace expansion of the determinant by the last column. The only nonzero term will be exactly $(-1)^{k-1}d_{n-1}$. This is a linear recurrence of the first order. There are many ways to solve such recurrences, my favorite is the method of generating function. Consider the function (or, in general, the formal power series) defined by the power series expansion $$D(z)=\sum_{k=0}^{\infty}d_{n}z^{n}=d_0 + d_1 \cdot z + d_2 \cdot z^2 + \cdots $$

Choose, for convenience, $d_{0}=1$ (this won't break the recurrence: $d_{1}=(-1)^{1-1}d_{0}=(-1)^{0}1=1$). Then multiply both parts of the recurrence by $z^{n}$: $$d_{n}z^{n}=(-1)^{n-1}d_{n-1}z^{n}$$ Sum over all $n$'s for which the expression on both sides makes sense: $$\sum_{n=1}^{\infty}d_{n}z^{n}=\sum_{n=1}^{\infty}(-1)^{(n-1)}d_{n-1}z^{n}$$

On the left hand side, we have $D(z)-d_{0}=D(z)-1$. On the right hand side, we have $zD(-z)$. Therefore, we obtain an equation on the generating function $D(z)$: $$D(z)-1=zD(-z)$$ with the condition that $D(0)=1$ (since $D(0)$ is the value of the zeroth coefficient of the power series). Hence $$\frac{D(z)-1}{z}=D(-z)$$ Then, changing $-z\to z$ in both sides, we obtain $$\frac{D(-z)-1}{-z}=\frac{1-D(-z)}{z}=D(z)$$ At the other hand, $$D(z)=zD(-z)+1$$ Hence, $$zD(-z)+1=\frac{1-D(-z)}{z}$$ $$z^2D(-z)+z=1-D(-z)$$ $$z^2D(z)-z=1-D(z)$$ and we have a linear equation on $D(z)$: $$z^2D(z)+D(z)=1+z$$ $$D(z)=\frac{1+z}{z^2+1}$$ We've obtained the generating function. Remember that we defined it as $\sum_{n=0}^{\infty}d_{n}z^{n}$ Hence, to obtain the explicit formula for $d_n$, all we need is to obtain the coefficient before the $n$-th term of the Taylor series of this function. To do that, it is essential to remember the following simple formula: $$\frac{1}{1-z}=1+z+z^2+...$$ This identity is true only when $|z|<1$, if we consider it as a functional identity, but it is convenient to consider it as the definition of the generating function $\frac{1}{1-z}$ as a formal power series: it has the same algebraic properties in the ring of formal power series, as $\frac{1}{1-z}$ has in the ring of analytic functions on the domain $|z|<1$. Then $$\frac{1}{1+z^2}=\sum_{n=0}^{\infty}(-1)^{n}z^{2n}$$ and $$\frac{1+z}{1+z^2}= \frac{1}{1+z^2}+\frac{z}{1+z^2} = \sum_{n=0}^{\infty}(-1)^{n}z^{2n}+\sum_{n=0}^{\infty}(-1)^{n}z^{2n+1}$$. Hence, the even coefficients of the Taylor series of $D(z)$ are $+1, -1, +1, ...$, and the odd coefficients are $+1, -1, +1, -1, \cdots$, as expected.

Answer 3

Your matrix $A\in\mathbb F^{n\times n}$ is an involution, i.e. $A^2=I$ (check: it is a symmetric permutation matrix), so all eigenvalues are $\in\big\{-1,1\big\}$.

If $n$ is even then $n=2r$ so
$A=\begin{bmatrix}\mathbf 0 & I_r \\ I_r & \mathbf 0\end{bmatrix}\implies \det\big(A\big)=(-1)^r$ since $\text{trace}\big(A\big)=0$
i.e. this implies $A$ has $r$ eigenvalues $=1$ and $r$ eigenvalues $=-1$ (why?).

If $n$ is odd then $n=2r+1$, $\implies \det\big(A\big)=(-1)^r$ since $\text{trace}\big(A\big)=1$
i.e. by the matrix setup of 0's and 1's we know $\text{trace}\big(A\big)\in \big\{0,1\big\}$ but trace of $0$ would imply $\frac{2r+1}{2}$ eigenvalues of 1 and $\frac{2r+1}{2}$ eigenvalues of $-1$ which is impossible. So $\text{trace}\big(A\big)=1 \implies A$ has $r+1$ eigenvalues $=1$ and $r$ eigenvalues $=-1$.