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5

Note: This answer is wrong, as indicated by the comments below. Let $M \oplus N$ denote the block-diagonal matrix $$ M \oplus N = \pmatrix{M&0\\0&N} $$ Then one solution with $m = 4n$ is $$ E = \pmatrix{A \oplus I & I \oplus B\\I \oplus C & D \oplus I} $$ where $I$ denotes the $n \times n$ identity matrix.


5

In general, $A \geq 0$ is area and $A^2= (ad-bc)^2$, then $A=\vert ad-bc \vert$, so in fact area of a parallelogram is absolute value of the determinant. In your case, where $(c,d)$ is between 0 and 180 degrees ccw of $(a,b)$, call this angle $\theta \in $[0, 180]. Note $(d,-c)$ is a rotation of $(c,d)$ 90 degrees cw, so $(a,b)$ and $(d,-c)$ are $90-\theta$...


4

The problem is Sarru's rule only works when the dimension is $3$. Indeed, one formal definition of the determinant is $$\sum_{\sigma \in S_n} \varepsilon(\sigma) \prod_{i=1}^n a_{i, \sigma (i)}$$ which looks scary (and might be to some extent), but all it's doing is taking all possible permutations of the row and column indices and multiplying those ...


3

$A^TA$ is positive semidefinite, hence $\det(A^TA)\geq 0.$ Proof of $A^TA$ is positive semidefinite: $x^TA^TAx=\left\|Ax\right\|^2 \geq 0.$


2

Suppose the statement holds for $n\geq2$, and consider integers $a_1,\ldots,a_{n+1}$ whose GCD is $1$. Let $d=\gcd(a_1,a_2,\ldots,a_n)$, and let $a_i'=a_i/d$ for $i\leq n$. Note that $$ 1=\gcd(a_1,\ldots,a_{n+1})=\gcd(d, a_{n+1}) $$ and $$ 1=\gcd(a_1',\ldots,a_n'). $$ By induction, there are matrices $X\in SL_2(\mathbb Z)$ and $Y\in SL_n(\mathbb Z)$ ...


2

We have $A_1=\frac{3}{2}I$, so $A_2=\frac{13}{12}I$, so $A_3=\frac{313}{312}I$, and $A_3^{-1}=\frac{312}{313}I$. Hence its determinant is between 0 and 1, so the limit is 0.


2

The formula you are talking about is an area for any polygon in terms of determinants: $$\left\lvert \frac 1 2 \left(\left[\begin{matrix}x_1 & y_1 \\ x_2 & y_2\end{matrix}\right]+\left[\begin{matrix}x_2 & y_2 \\ x_3 & y_3\end{matrix}\right]+...+\left[\begin{matrix}x_{n-1} & y_{n-1} \\ x_n & y_n\end{matrix}\right]+\left[\begin{matrix}...


1

Definition: A set of vectors $\{\mathbf u_1, \dots, \mathbf u_k\}\subset \Bbb R^n$ is linearly independent if $$a_1\mathbf u_1 + \cdots + a_k\mathbf u_k = \mathbf 0 \implies a_1 = \cdots = a_k=0$$ So let's check if $\{(-6,2)\}$ is a linearly independent set. For we write down $\mathbf 0$ as a linear combination of the set: $$(0,0) = a(-6,2)$$ Are there ...


1

Note that it can be easily shown that $$ A^{-1} \;\; =\;\; \left [ \begin{array}{cc} 2 & -1\\ -1 & 1 \\ \end{array} \right ]. $$ This implies that $$ A_1 \;\; =\;\; \left [ \begin{array}{cc} 3/2 & 0 \\ 0 & 3/2 \\ \end{array} \right ] $$ Implying that $$ A_2 \;\; =\;\; \left [ \begin{array}{cc} 3/4 & 0 \\ 0 & 3/4 \\ \end{array} ...


1

Let the SVD of $A \in \mathbb R^{m \times n}$ be $$A = U \Sigma V^T = \begin{bmatrix} U_1 & U_2\end{bmatrix} \begin{bmatrix} \hat\Sigma & O\\ O & O\end{bmatrix} \begin{bmatrix} V_1^T\\ V_2^T\end{bmatrix}$$ where the zero matrices may be empty. The eigendecomposition of $A^T A$ is, thus, $$A^T A = V \Sigma^T U^T U \Sigma V^T = V \Sigma^T \Sigma ...


1

It is, of course, possible to note that $A^TA$ is positive semidefinite as in the other answer. Another method is to apply the Cauchy-Binet formula, which allows us to see that (when $A$ is $m \times n$ with $m \leq n$) $$ \det(A^TA) = \left(\sum_{S \in \binom{[m]}{n}} \det A_{S,[n]}\right)^2 \geq 0 $$


1

Three facts: Symmetric semi positive definite matrix has all eigenvalues real and greater or equal zero. Symmetric positive definite matrix has all eigenvalues real and greater than zero. Determinant of any matrix is equal to product of all eigenvalues. Hence, if determinant of symmetric semi positive definite matrix $A$ is nonzero, then $A$ is positive ...


1

Let $s$ be the product of the diagonal entries of $A$ and the diagonal entries of $D$, let $t$ be the product of the diagonal entries of $B$ and the diagonal entries of $C$ and consider the $m\times m$ matrix $E=\begin{pmatrix}\det{(A)}\det{(D)}-\det{(B)}\det{(C)}-s+t&s-t\\-1&1\\&&1\\&&&\ddots\\&&&&1\end{pmatrix}$....


1

I can get close; consider the block matrix $$\left(\begin{matrix} a_1x_1 & r\\ c&I\end{matrix}\right)$$ where $r$ is the row vector $(\begin{matrix} a_2 & a_3 & \cdots & a_n\end{matrix})$, $c$ is the column vector $(\begin{matrix} -x_2 & -x_3 & \cdots & -x_n\end{matrix})^T$, and $I$ is the $(n-1)\times (n-1)$ identity matrix.



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