# Tag Info

6

Yes indeed, a direct consequence of the shoelace formula is that a triangle (convex polygon) having all its vertices with rational coordinates has a rational area. It is a nice technique for showing that there is no equilateral triangle in $\mathbb{Z}\times\mathbb{Z}$, too.

5

If we do row reduction, i.e. $R_n-R_{n-1}$, $R_{n-1}-R_{n-2}$, up to $R_2-R_1$, (twice) we get \begin{align} &\begin{vmatrix} 0 & 1 & 2 & 3 & 4 & \cdots & n-1 \\ 1 & 0 & 1 & 2 & 3 & \cdots & n-2 \\ 2 & 1 & 0 & 1 & 2 & \cdots & n-3 \\ \vdots & \vdots & \vdots & \vdots ...

5

The determinant of such tridiagonal matrices of order $n$ are computed with the linear recurrence of order $2$: $$D_n=-\lambda D_{n-1}-\frac\lambda4 D_{n-2}$$ and the initial conditions $\; D_0=1,\enspace D_1=\lambda$. How to find a closed formula for $D_n$ : We look for basic solutions that are geometric progressions $r^n\ (r\ne 0)$; this leads to ...

5

Let $x$ be an eigenvector of the orthogonal matrix $C$ corresponding to the eigenvalue $\lambda\neq 0$. Then $$Cx = \lambda x.$$ Applying $C^T$ to both sides from the left, and recalling that $C^TC=I$, we get $$x = \lambda C^T x,$$ or $$\frac{1}{\lambda} x = C^Tx.$$

4

If $A$ is invertible then $\exists \, M \in \mathcal{R}^{n \times n}$ such that $AM=MA=I_n$. Then determinant being a homomorphism gives $$\det(AM)=\det(A)\det(M)=\det(I)=1_{\mathcal{R}}.$$ Thus $\det(A)$ must be a unit in $\mathcal{R}$. For the other way: Assume $\det(A)$ is invertible. Since $$\det(A)I_n = A\operatorname{adj}(A) = ... 4 If n=1, \det A=a_{1,1}, which is trivially a convex function of A. If n>1, take two diagonal matrices, with a_{i,i}=1 if i\leq n/2 and 0 otherwise, and b_{i,i}=0 if i\leq n/2 and 1 otherwise. Then \det A=\det B=0. However, for t\in]0,1[, \det [tA+(1-t)B] > 0, hence the function is not convex. 4 The characteristic polynomial of a 2 \times 2 matrix can be written as: $$p(\lambda) = \lambda^2 - \textrm{tr}(A)\lambda + \textrm{det}(A)$$ (Check here). If a matrix A is symmetric then it is diagonalizable such that: A = Q \Lambda Q^T \quad \textrm{ where } \quad \Lambda= \textrm{diag}(\lambda_1,\lambda_2) ... 4 A simple parity argument will do. By assumption, every element of S=\{1,2,\ldots,n\} must occur an odd number of times (or more precisely, n times). Yet, if some element of S does not appear on the diagonal, it must appear in M an even number of times, because the matrix is symmetric and off-diagonal elements occur in pairs. 4 Suppose that A is m\times m. Then$$\det\left(\frac{1}{n}A^n\right)=\frac{1}{n^m}\det(A)^n$$On the other hand,$$ \frac{1}{n-k}\det(kA^{n-k})=\frac{k^m}{n-k}\det(A)^{n-k} $$Assuming that \det(A) is positive, the inequality means that$$ \det(A)^k\leq\frac{(kn)^m}{n-k} $$Since \det(A) can be arbitrarily large and the RHS does not depend on ... 3 Hint: \det(AB)=\det(A)\det(B). Apply this to PP^{-1}=I and to P^{-1}AP 3 One has:$$A(A^2+4I)=2I.$$Therefore:$$A\left(\frac{1}{2}A^2+2I\right)=I.$$Hence A is right invertible and thus invertible. N.B. You have to assume that 2 is invertible in the ring where A takes its entries. 3 Observe that your relation is equivalent to$$A\left(\frac{1}{2}A^2+2A\right)=I.$$2 You can see that$$\det(A^3+4A)=\det(A)\times\det(A^2+4I)=\det(2I)\neq 0,which implies \det(A)\neq0 and so A is invertible. 2 Hint: Determinants of tridiagonal matrices can be calculated with a linear recurrence of order 2. In the present case, denoting D_n this determinant of order n, we have: \begin{align*} D_n&=\lambda D_{n-1}- D_{n-2}\\ \text{with initials conditions:}\quad D_0&=1,\enspace D_1=\lambda \end{align*} The solutions are linear combinations of ... 2 It is easier to solve the linear system. Let (x_1,..,x_n) be a non zero vector in the kernel, and conisder an index k such that \vert x_k\vert is maximal. As \sum _{j=1}^n a_{jk}x_j=0, we have a_{kk} x_k=-\sum _{j\not =j}a_{jk}x_j. Therefore \vert a_{kk}\vert \vert x_k\vert \leq \sum _{j\not =k}\vert a_{jk}\vert \vert x_j\vert \leq (\sum _{j\not ... 2 From recursion, we have\det(A_n) = (-1)^{n-1} \det(A_{n-1})$$Hence,$$\det(A_n) = (-1)^{(n-1)+(n-2) + \cdots + 1}\det(A_1) = (-1)^{n(n-1)/2} = \begin{cases} 1 & \text{ if }n \equiv0,1\pmod4\\ -1 & \text{ if }n \equiv 2,3\pmod4\end{cases}$$2 For 2 by 2 matrices and for 3 by 3, there is a fairly satisfactory cookbook method that you can do the same way every time, in order to find the characteristic polynomial (the eigenvalues are the roots of that). For a 2 by 2, call the trace \sigma_1 and the determinant \sigma_2. Then the characteristic polynomial is$$ \color{blue}{\lambda^2 - ...

2

$AB = 0$ does not exclude $A = B = 0$. You can simply rearrange the equations: $$\begin{vmatrix} 1 & 1 & 1 \\ \cos\theta_2 & \cos\theta_3 & \cos\theta_1 \\ \sin\theta_2 & \sin\theta_3 & \sin\theta_1 \\ \end{vmatrix}=0$$ And use an analogous proof to show that ${\beta = \alpha} \vee {\beta = \gamma}$. And rearrange a final time to ...

2

Assume that $\alpha = \beta \not= \gamma$. We have $$(\gamma - \alpha)z =(\alpha)(x+y+z) + (\gamma - \alpha)z = \alpha x+ \beta y + \gamma z = 0$$ Since we assumed $\alpha \not= \gamma$, we have $z=0$, a contradiction. Therefore, we have $\alpha = \beta = \gamma$, as desired. $\blacksquare$

2

Since $AA'$ is positive semidefinite, in the eigendecomposition of $AA' = S^{-1}JS$, the diagonal matrix $J$ only has positive values on the diagonal, hence $$\det(AA') = \det(S^{-1}JS) = \det(S^{-1})\det(J)\det(S)=\det(J) \geq 0$$

2

I find induction and Laplace expansion on the first column or row to be the easiest way.

2

If $M$ is an $n\times n$ matrix, then you can consider the determinant to be a homogeneous polynomial function in the entries of the matrix. So if you take the limit of the determinant, you're basically taking the limit of a function of $n$ that is determined by its coordinates. So provided you have a sequence of square matrices $M(n)$ in which each entry is ...

2

One has: $$J:=\begin{pmatrix}0 & A\\B & 0\end{pmatrix}=\begin{pmatrix}A& 0\\0 & B\end{pmatrix}\times\begin{pmatrix}0 & I_n\\I_n & 0\end{pmatrix}.$$ You only have to compute: $$\varepsilon:=\det\left(\begin{pmatrix}0 & I_n\\I_n & 0\end{pmatrix}\right).$$ Indeed, using the first equality, one has: ...

2

Well I think we could proceed like this: $$(I-\lambda P)(I+\lambda P+\lambda^{2}P^{2}+\lambda^{3}P^{3}+...)=(I-\lambda^{n}P^{n})$$ Which if $\lambda<1$ gives you $$(I-\lambda P)(I+\lambda P+\lambda^{2}P^{2}+\lambda^{3}P^{3}+...)=I$$ from there you see that since $P^n=...=P^2=P$ then: ...

2

Counterexample Suppose that $$k=1, \;\;\;\; n=2, \;\;\;\; \text{and} \;\;\;\; \mathbf{A}=\left[ \begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}\right]$$ then $$\det\left(\frac{1}{n}\mathbf{A}^n\right)=\frac{9^2}{2^2} \gt 9 = \frac{1}{n-k}\det\left(k\mathbf{A}^{n-k}\right)$$ which violates the proposed inequality.

2

The rank two condition on $A$ means that the Gauss eliminations gives $$A\sim \left(\matrix{* & * & *\\ 0 & * & *\\ 0 & 0 & 0}\right).$$ The Gauss eliminations does not change the determinant, therefore $$\det A=\det\left(\matrix{* & * & *\\ 0 & * & *\\ 0 & 0 & 0}\right)=0.$$

1

Instead of proceeding in terms of adjoints, it would be a great deal easier to proceed as follows. First of all, note that each of $a,c,f$ is nonzero. Next, consider the augmented matrix $[A\mid I_3],$ where $I_3$ is the $3\times3$ identity matrix. Perform row operations on the augmented matrix until it is in the form $[I_3\mid B].$ The matrix $B$ is the ...

1

Your work is fine but you forgot to transpose the adjoint! The inverse is essentially given by the adjugate matrix, which is the transpose of the cofactor matrix.

1

As $\det(A)+\det(B)=0$, we can suppose wlog that $\det(A)=1$ and $\det(B)=-1$. Now $$\det(A+B)=\det({}^tA+{}^tB)=\det(A^{-1}+B^{-1})$$ Hence $$\det(A+B)=\det(A)\det(A^{-1}+B^{-1})=\det(I+A B^{-1})$$ and $$\det(A+B)=\det(I+AB^{-1})(-\det(B))=-\det(B+A)$$ Hence $\det(A+B)=0$.

1

I work in the field $C$, let $P$ be a matrix such that $PAP^{-1}$ is sup triangular, $c_1,...c_n$ the eigenvalues of $A$, $det(E+xA)=det(P(E+xA)P^{-1})=det(E+xPAP^{-1})=(1+xc_1)...(1+xc_n)$, this implies that $det(E+xA)=1+x(c_1+...+c_n)+x^2U(x)$ and henceforth ${d\over{dx}}_{x=0}=c_1+...+c_n=tr(A)$.

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