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16

Let $A = \left( \begin{array}{ccc} 1 & \frac{1}{2015} \\ 0 & 1\end{array} \right)$, then $A^{2015} = \left( \begin{array}{ccc} 1 & 1 \\ 0 & 1\end{array} \right)$.


7

$a+b=c+d$ implies that we should add up the first two columns of the matrix using a vector of the form $\begin{pmatrix} 1\\ 1\\ t \end{pmatrix}$. You can see that if $A$ is the matrix, then $A\begin{pmatrix} 1\\ 1\\ t \end{pmatrix}$=$\begin{pmatrix} a+b+t\\ c+d+t\\ 0 \end{pmatrix}$. Therefore, the suitable choice is of course $t=0$, and so the ...


5

$\require{begingroup} \begingroup$Note$\let\geq\geqslant\newcommand\norm[1]{\|#1\|}$ that $AA^T$ is symmetric (hence diagonalisable) and thus has real nonnegative eigenvalues. Indeed, if $v$ is a (non-zero) eigenvector corresponding to $\lambda$, then $$\lambda\norm v^2=\lambda v^Tv=v^TAA^Tv=\norm{A^Tv}^2\geq0\implies\lambda\geq0.$$ $AA^T+I$ is also ...


4

As @user153012 is asking for a proof in full detail, here is a brute-force approach using an explicit expression of a determinant of an $n$ by $n$ matrix, say $A = (a[i,j])$, $$\det A = \sum_{\sigma\in S_n}\operatorname{sgn}\sigma \prod_i a[{i,\sigma(i)}],$$ where $S_n$ is the symmetric group on $[n] = \{1,\dots, n\}$ and $\operatorname{sgn}\sigma$ denotes ...


4

Hint: If two rows (or columns) of a matrix are the same then its determinant is 0.


4

Thank you all for your replies. While I was just reflecting, I thought of the following solution, so, thought of sharing the following solution : If $A= \begin{pmatrix} a&b&1 \\ c&d&1 \\ 1&-1&0\\ \end{pmatrix}$, then : $A - \lambda I = \begin{pmatrix} a-\lambda&b&1 \\ c&d-\lambda&1 \\ 1&-1&-\lambda\\ ...


3

The characteristic polynomial is not difficult to solve. Setting $d=a+b-c$ it is $$ \chi (A)=t^3 + t^2( - 2a - b + c) + t(a^2 + ab - ac - bc)=t(a + b - t)(a - c - t), $$ so that $0$, $a+b$ and $d-b$ are the eigenvalues. This solution has the advantage that you obtain all eigenvalues.


3

If we let $x\in\mathbb R^3$ be a vector and name $$E = \pmatrix{0&1&1\\1&0&1\\1&1&0}, e = \pmatrix{1\\1\\1}$$ Then your system is $$(E + kI)x = e$$ This system is uniquely solvable iff $\det(E+kI) \ne 0$. $$\det(E+kI) = k^3 + 1 + 1 - k - k - k = k^3 - 3k + 2 = (k-1)^2(k+2)$$ Thus we reach the same conclusion: $k\ne 1,-2$ A side note: ...


3

$$\det(I+ x A) = x^n \det (A + \frac 1 x I) = x^n\left(\lambda_1 + \frac 1 x\right)\left(\lambda_2 + \frac 1 x\right)\cdots \left(\lambda_n + \frac 1 x\right) $$


3

This is a fundamental result about determinants, and like most of such results it holds for matrices with entries in any commutative (unitary) ring. It is therefore good to have a proof that does not rely on the coefficients being in a field; I will use the Leibniz formula as definition of the determinant rather than a characterisation as alternating ...


3

Your idea that $AA^T$ is positive definite is a good idea. More precisely, you can show that $AA^T$ is positive semidefinite. This is easiest to show using definitions, because $M$ is positive semidefinite if for all $x$, you have $$x^TMx \geq 0,$$ and in your case, this inequality is fairly simple to show.


3

Replace the first column by the sum of all the columns: the wanted determinant is equal to $$\Delta_n:=\det\begin{pmatrix} 1 & -1 & \cdots & -1 \\ 1 & n-1 & \cdots & \vdots \\ \vdots & \vdots & \ddots & \vdots \\ 1 & \cdots & \cdots & n-1 \end{pmatrix}.$$ Now, for each $j\in\{2,\dots,n\}$, take the column ...


3

I'll use $v$ here to be $v^T$ in what you have. The Matrix inversion lemma tells you for $A$ invertible and $u,v$ vectors. Then (provided $u v^T$ doesn't make $A$ singular when added to it), $(A+ u v^T)^{-1} = A^{-1} - \frac{A^{-1} u v^T A^{-1}}{1+ v^T A^{-1} u}$. The Matrix determinant lemma tells you under the same conditions $det(A+u v^T) =(1+ v^T ...


3

You have two variables and two equations, $d=ab^2$ and $t=a+2b$ where $d,t$ are some constants. Solve this system of equations: $a=t-2b \rightarrow d=(t-2b)b^2 \rightarrow 2b^3-tb^2+d=0$ So you have a cubic in $b$ which has a generic solution with 3 roots $b_1,b_2,b_3$. Each of these determines a corresponding value for $a$, so in general your ...


3

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}$Since volume of a parallelipiped spanned by a set of vectors is invariant under the operation of adding a scalar multiple of one vector to another, it suffices to compute the volume of the parallelipiped spanned by $$ \left[\begin{array}{@{}c@{}} 1 \\ 0 \\ 0 \\ 0 \\ ...


3

The determinant you want to calculate is equal to zero because the third line is 2 times the first line. It doesn't matter the determinant of the first matrix you wrote.


2

The are already some nice proofs in the other answer and the comments. Here is an alternative proof by reductio ad absurdum. A square matrix $M$ is nonsingular if and only if the equation $Mx=0$ has only the trivial solution. So, suppose $(A-I)x=0$, i.e. $x=Ax$. It suffices to prove that $x=0$. Since $A^3=0$, from $x=Ax$ we get $A^2x=A^3x=0$. Since ...


2

In Step 2 to Step 3, $t-5$ is factored out of the first row and the third row. The multiplies the determinant by the same factor, each time the factoring is done. In Step 4 to Step 5, expansion by minors is done on the first column. This is easy since there is a $1$ in only one position and $0$'s in the other positions in that column. What's more, adding ...


2

This is not true in general. Knowing the roots of two polynomials are the same is not enough to conclude they are identical. Indeed, consider $$p(X) = (1+X)(1-X)$$ and $$q(X) = \frac{1}{2}(1+X)(1-X)$$ over $\mathbb{Q}$, say. These polynomials have the same roots, but $p \neq q$. update: (thank you @quid for pointing this out) Since you are working with ...


2

$A$ is symmetric and therefore has $n$ orthogonal real eigenvectors with eigenvalue $\lambda_1,\ldots, \lambda_n$ and we have $\det A=\lambda_1\cdot\ldots\cdot \lambda_n$. Guessing eigenvectors might help. Unfortunately this guessing (apart from the "all ones" vector) is only really simple in the complexification: If $\zeta\in\mathbb C$ is an $n$th root ...


2

Another way of looking at the $\det(A)$: by swapping pairs of rows $(1,n-1),(2,n-2)\dots$ this matrix is transformed into circulant matrix $C$ with the same $0$-th row $(1,\dots,n)$, which has a well known explicit formula for determinant, \begin{align} \det(C)&=\prod_{j=0}^{n-1}\sum_{k=0}^{n-1} (k+1)\exp\left(2\pi i\frac{jk}{n}\right). \end{align} ...


2

Add the first $k$ rows to the last row. This makes your matrix upper triangular and shows that $\det J = y_{k+1}^k$.


2

If you form the parallelepiped by $\{\sum_i c_i v_i \, | \, 0 \leq c_i \leq 1\}$ then the geometric volume of this object in $n$ dimensions is precisely $\det V$ where $V$ is the matrix with column vectors given by the $v_i$. If you'd like an intuitive understanding, note that it works for a cube and it also works for a "sheared" cube where you shift one ...


2

Let $B$ be the adjoint matrix of $A$. Then $ABu=\det(A)u$ for all $u\in R^n$. Thus $Bu$ is an element of $R^n$ such that $A(Bu)=\det(A)u$.


2

If $A$ is invertible, then the factorization $$M:=\left(\begin{array}{cc} A & B \\ B^T & C\end{array}\right)= \left(\begin{array}{cc} A & 0 \\ B^T & 1\end{array}\right) \left(\begin{array}{cc} 1 & A^{-1}B \\ 0 & C-B^TA^{-1}B\end{array}\right)$$ implies that $$\operatorname{det}M=\operatorname{det}A\cdot ...


2

we will derive a three term recurrence relations for the determinant $a_n$ of the matrix of size $n.$ expanding by the first row, we get $$a_n = 3a_{n-1} - 2a_{n-2}, a_0 = 1, a_1 = 3, a_2 = 7, a_3= 15, \cdots$$ the characteristic equation is $$\lambda^2 - 3\lambda + 2= 0\to \lambda = 1, 2$$ therefore $$ a_n = C \, 2^n + D, C + D = 1, 2C + D = 3$$ gives $C ...


2

See the step before the final step. Expand ( calculate ) the determinant along the first column. You'll see that only one 3*3 determinant survives. Again expand along first column . You'll see that only one 2*2 determinant survives.


2

This is because at the last but one step, the matrix is block-triangular so its determinant is the product of the determinants of the diagonal blocks: $$\begin{vmatrix}1&2&1&1\\ 0&1&1&2\\ 0&0&4&5\\ 0&0&3&1\end{vmatrix} =\begin{vmatrix}1&2\\ 0&1\end{vmatrix} \cdot\begin{vmatrix} 4&5\\ ...


2

Let $W = XX^T + C$ then your 2 functions are $$\eqalign { f &= {\rm log}({\rm det}(W)) = {\rm tr}({\rm log}(W) \cr g &= aa^T:W^{-1} = A:W^{-1} \cr }$$ The differentials with respect to $W$ can be found in the cookbook as $$\eqalign { df &= d\,{\rm tr}({\rm log}(W) \cr &= W^{-T}:dW\cr dg &= A:dW^{-1} \cr &= ...


2

What is $U_y$? The jacobian matrix is $J = \begin{pmatrix} 1 & 0 \\ 3v & 3u\end{pmatrix}$ It's determinant is $3u$



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