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## Hot answers tagged determinant

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


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

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