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We have this, hopefully it is helpful: Case (I) $k\in \{3,-5\}$, in which case I am getting a free variable (you can check this for $k = 3$ and $k = -5$ individually via standard gaussian elimination), so infinitely many solutions. Case (II) $k = -\frac{1}{2}$, in which case I am getting a unique solution (you can check this via standard gaussian elimination)...

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I assume your matrices are over $\mathbb{R}$, although this approach can be modified using algebraic geometry to work over general fields. There is a nice two-step approach: first show that a matrix that commutes with all invertible matrices in fact commutes with all matrices, and then show that the only matrices that do that are scalar multiples of the ...

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The subdiagonal elements and the superdiagonal elements can be represented as the vector $(\beta_1, \beta_2, \dots \beta_{n-1})$. It has length $n-1$. The diagonal elements can be represented as the vector $(\alpha_1, \alpha_2, \dotsc, \alpha_n)$. It has length $n$. Tridiagonal solvers such dpttrs (based on the LDLT-factorization) from LAPACK capitalize on ...

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Assume the square matrix $A=(a_{ij})$ of size $n$ commutes with all invertible square matrices of the same size $n$. Take this particular matrix $B = {(a_{ij})}$ where $b_{ij} = i$ when $i=j$ and $b_{ij} = 0$ if $i \ne j$. $B$ is obviously invertible. Use the fact that it commutes with your matrix A. So we have $AB=BA$ When we write this down in details, ...

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Hint: Let $k\ge 0$, look at $f(t)=\text{Tr}((A+tB)^{k+1})$. What kind of function is this, what are its properties? If you need another hint, let me know!

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After extensive testing and some more investigation and research it became clear that this would not be the correct way to do the color correction (as stated) in a first place. Primarily due to the fact that an important dependency is missing in the proposed approach: namely the fact that $RGB$ trio depends on the specifics of the de-mosaicing. In short - ...

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Assume $n\ge 3$ so that eigenvalue $\lambda_3$ exists. $(A-\lambda_3 I)x = 0$ by definition of eigenvalue and eigenvector. $A$ and $I$ commute, so we can rearrange the terms. $$\left[\prod_{j\ne 3}(A-\lambda_j I)\right](A-\lambda_3 I) x = 0.$$ Okay, after posting this looked at the comments. @JMoravitz already gave the answer there!

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It may be useful to start with the case in which the matrix $A$ is also diagonal. In that case we have $$\left(\mathcal{D}\left\{ M_{2}^{-1}\iota\right\} \right)^{-1}M_{2}^{-1}\left(I-D_{0}\right)D_{0}\mathcal{D}\left\{ A^{T}M_{2}^{-1}\iota\right\} M_{2}^{-T}\mathcal{D}\left(M_{1}\iota\right) =M_{2}M_{2}^{-1}\left(I-D_{0}\right)D_{0}AM_{2}^{-1}M_{2}^{-1}M_{1}... 0 You can for an "infinitesimal" \delta build$$D_x = \begin{bmatrix}1& & \delta\\&1&\\&&1\end{bmatrix} , D_y = \begin{bmatrix}1& & \\&1&\delta\\&&1\end{bmatrix}$$Now$${D_x}^{n_x} {D_y}^{n_y} = \begin{bmatrix}1& & n_x\cdot \delta\\&1&n_y\cdot \delta\\&&1\end{bmatrix}$$With ... 1 Let X = \Phi\Sigma\Psi^\top be the SVD of X. Then UX = (U\Phi)\Sigma\Psi^\top. If this is the singular value decomposition for UX (i.e., if U\Phi is the matrix of left singular vectors for UX), then X and UX have the same singular values (i.e., they have the same \Sigma in their singular value decompositions; recall that the SVD is unique ... 0 It seemed to be a calculation error. I have now found that the previous row-echelon form that I got was incorrect 0 Translation is a technical term, and I don't see why you call the matrix, call it I+M, "translation". In any case, from the Mercator series for the logarithm, you see that$$ M^2=0, ~~~\leadsto ~~~ \log (I+M)= M, $$so that$$ M=a_1 \begin{bmatrix} 0 & 0 & 1\\ 0 & 0& 0\\ 0 & 0 & 0 \end{bmatrix} +a_2\begin{bmatrix} 0 & ...

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$a_{ij}=u_iv_j$ is equivalent to saying $A=uv^T$. Hence $$A^2=uv^Tuv^T=(v^T u)A,\quad\ldots\quad A^n=(v^T u)^{n-1} A$$ Given that $A^5=16A$ then shows $v^T u=\pm2$ or $\pm2i$, or $A=0$. Note that the trace of $A$ is $\sum_iv_iu_i=v^Tu$. [Note that $A$ is not invertible, so one cannot multiply by $A^{-1}$.]

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The fact that $a_{ij} = u_iv_j$ means that $A$ has rank $\leq 1$ (as $A = U^TV$). Note that this implies that $A$ cannot be inverted unlike what you wrote. Then, we can proceed by looking at the eigenvalues of $A$ (as the trace of $A$ is the sum of these eigenvalues). The fact that $A$ has rank $0$ or $1$ means that $0$ is an eigenvalue. $A^5 = 16 A$ could ...

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Consider, say,$$A=\begin{bmatrix}1&1\\1&1\end{bmatrix}.$$Its column space is $\left\{\left[\begin{smallmatrix}x\\x\end{smallmatrix}\right]\,\middle|\,x\in\Bbb R\right\}$, right? But the reduced row echelon form of $A$ is $\left[\begin{smallmatrix}1&1\\0&0\end{smallmatrix}\right]$, and the column space of this matrix is $\left\{\left[\begin{... 2 For the first part, multiply$\alpha x^T+\beta y^T+\gamma z^T=0$by$x$,$y$, and$zto obtain \begin{align} 3\alpha+2\beta+\gamma &=0\\ 2\alpha+3\beta+2\gamma &=0\\ \alpha+2\beta+3\gamma &=0\\ \end{align}. Show that the only solution is(0,0,0)$, hence$\{x,y,z\}$is a basis. (Notice that the coefficient matrix is the Gram-matrix of$x$,$...

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You don't have to solve it backward. See this solution, please. Given $\mathbf {detB} = −202$, and $\mathbf{adj} B = \left[ \begin{array}{ccc} -45& 33& -28 \\ 32& -10& -16\\ -4& -24& 2\\ \end{array} \right]$, then, $\mathbf{B^{-1}} = \frac{1}{detB} . adjB$. This implies that, $\mathbf {B^{-1}} = -\frac{1}... 1 The adjugate satisfies the relation $$B\operatorname{adj}B=(\det B)I\qquad(*)$$ and therefore, if$B$is invertible, $$B=(\det B)(\operatorname{adj}B)^{-1}$$ Your final linear system is part of the relation$(*)$, if you notice, but you forgot some equations. With some patience, you get $$(\operatorname{adj}B)^{-1}= \begin{bmatrix} -1/101 & 3/202 &... 5 B^{-1} = \frac{adj(B)} {\det B} \implies B = \frac{(adj(B))^{-1}}{\det B} Can you do now? 0 As you will see, \beta=\{e_1,e_2\} is a basis for R^2. So, what you define bilinear form f as [f]_\beta and correspondingly you find the respective coordinates of the matrix that is, for example, You say f(e_1,e_1) is first element of first row and you calculated that f(e_1,e_1)=2. Likewise, you calculate other entries!! Hope it helps!! 1 A more intuitive argument (why for a square (nxn) matrix ,if it's rows are linearly dependent it implies that so are also it's columns) Given that rows are linearly dependent,performing Gauss elimination must produce row of all zeroes (possibly more than one) . Therefore the columns of the row reduced echelon form matrix are linearly dependent. That's so ... 1 Notice that this linear map/matrix is basically two R^2\to R^2 maps joined together: One map consists of a linear map from span\{v_1,v_2\} to itself, the other a linear map from span\{v_3,v_4\}. Respectively, these have matrix representations$$ \begin{bmatrix}{-2 \: -1 \\\quad 0 \quad 0}\end{bmatrix} $$and$$ $$\begin{bmatrix}{\quad 0 \: +3 \\ -3 \... 2 Let (e_1,...,e_n) a basis of F^n and (f_1,...,f_m) a basis of F^m. Write T(e_i)=\sum a_{ij}f_j. Let x\in F^n, x=\sum_i x_ie_i implies that T(x)=T(\sum_i x_ie_i)=\sum x_iT(e_i)=\sum_i x_i\sum_j a_{ij}f_j =\sum_j\sum_i a_{ij}x_if_j. This implies that the coordinates of T(x) in the basis (f_1,...,f_m) are the line of AX where A where A ... 0 You seem to already understand what you are asking. The column space of a matrix is just the set of all linear combinations of the columns of A. It's the subspace spanned by those columns. So, if d^T is contained in this subspace, then the system is consistent for that specific d^T, otherwise it is not. All of this is just another way of asking if row ... 4 No, it is not necessarily the case that the eigenvalues of A (or D) are eigenvalues of M. Note that \lambda is an eigenvalue of A iff A - \lambda I fails to be invertible. If \lambda is not an eigenvalue of A, then the characteristic polynomial of M can be written in the form$$ \det(M - \lambda I) = \det(A - \lambda I) \cdot \det([D - C(A -... 2 The function $$B\in GL_{n+1}(\mathbb R)\mapsto b_{11}\in\mathbb R$$ is continuous and constant on equivalent classes, so it factors thru the quotient giving a continuous unbounded function. Therefore the quotient is not compact. 0 The determinant of a square matrix is unchanged by adding one row to another row, since the determinant of a square matrix containing two identical rows is zero. the determinant is multilinear (i.e. linear with respect to each row). The solution consists of the row operation$R_3 \leftarrow R_3 + R_1$, so the determinant of the given matrix is unchanged by ... 0 Another option is to think about the matrix$X_r$as a$2n \times 2n$complex matrix. Motivated by the$2 \times 2$case, over the complex numbers, we have the identity $$\begin{pmatrix} a - ib & 0 \\ 0 & a + ib \end{pmatrix} = P^{-1} \begin{pmatrix} a & -b \\ b & a \end{pmatrix} P$$ where $$P = \frac{1}{\sqrt{2}} \begin{pmatrix} -iI_n &... 0 A basis for the space is \{t^{3}+3t^{2}+4t, -5t^{2}+t+1\} and the correct dimension is 2. 1 The strategy here is to manipulate the expression for the projection of a onto b into the form (\text{some matrix})a. Note that a \cdot b can be written as b^Ta. Also, any scalar "commutes" with a vector, so (b^Ta)b = bb^Ta. Hence,$$a-\dfrac{a \cdot b}{\|b\|^2}b = a - b\dfrac{b^Ta}{\|b\|^2} = Ia - \dfrac{bb^T}{\|b\|^2}a = \left(I - \... 0 First write your null space basis as a matrix: $$N = \begin{bmatrix} 4 & 1 \\ (1) & 0 \\ 0 & 2 \\ 0 & (1) \end{bmatrix}$$ Then modify the basis (without changing the nullspace) so that each vector's "last" element is a "1" and in a different row from all other basis vectors. That is already done in your problem. Then pad ... 0 I believe that the accepted solution does not have full generality and will note work if, say, the first two columns of the matrix are linearly dependent. If the columns of matrix$N$represent the null space, then the RREF can be reconstructed by the following matlab command rref(transpose(null(transpose(N)))) See this lesson on Lemma for a discussion of ... 0 Hint. The first condition says that $$\pmatrix{A&-B\\ -C&D}\pmatrix{D^T&B^T\\ C^T&A^T}=I.$$ 2 One may view this as a determinant of a block matrix. In general (cf. John Silvester, Determinants of Block Matrices), if$R$is a commutative ring and$C$is a commutative subring of$M_m(R)$, each matrix$Z\in M_n(C)$can be viewed as a matrix$Y\in M_{mn}(R)$and $$\det{}_RY=\det{}_R\left(\det{}_CZ\right).\tag{1}$$ In your case, let$m=2,\, R=\mathbb R$... 2 While I don't quite understand why you only want to use integers, you can prove this by induction. Let$d_n$be the determinant of your matrix of size$n \times n$. Then$d_1 = 2$,$d_2 = 3$. So assume that$d_k = k+1$for all$k < n$. By Laplace expansion and the induction hypothesis, we get $$d_n = 2d_{n-1} - d_{n-2} = 2n - (n-1) = n+1$$ and we are done.... 0 One way would be to define the perturbation$\epsilon V = (\hat Q - Q)^T Q$, so that you are interested in finding eigenvalues of$Q^T Q + \epsilon V$. For small$\epsilon$you can approximate eigenvalues and eigenvectors using perturbation theory, it is a standard tool in quantum mechanics (https://en.wikipedia.org/wiki/Perturbation_theory_(... 2 Since$|a+ib|^2=\det\begin{pmatrix} a & & - b \\ b & & a \end{pmatrix},\begin{align}\det X\cdot\overline{\det X}&=\sum_{\sigma,\,\sigma^\prime\in S_n}\varepsilon_\sigma\varepsilon_{\sigma^\prime}\prod_{j=1}^n\prod_{k=1}^nX_{j\sigma(j)}\overline{X_{k\sigma^\prime(k)}}\\&=\sum_{\sigma,\,\sigma^\prime}\varepsilon_{\sigma\circ\sigma^\... 1 Here's a proof that holds over any field \mathbb F for a skew symmetric (or indeed a regular symmetric) matrix. M\mathbf x = \mathbf 0 =-M^T\mathbf x =M^T\mathbf x\implies \mathbf x^T M= \mathbf 0^T having rank r, rank-nullity tells us \dim\ker\big(M\big) = n-r. Build a basis for the nullspace. \big\{\mathbf x_1, ..., \mathbf x_{n-r}\big\}. Now ... 0 Hint: Consider the standard bases of\mathcal M_2(\mathbf R) and \mathcal M_{2\times 3}(\mathbf R)::E_1=\begin{bmatrix}1&0 \\0&0\end{bmatrix},\enspace E_2=\begin{bmatrix}0&1 \\ 0&0\end{bmatrix},\enspace E_3=\begin{bmatrix}0&0 \\1&0\end{bmatrix},\enspace E_4=\begin{bmatrix}0&0 \\0&1\end{bmatrix} e_1=\begin{bmatrix}1&... 0 Start by showing that the four2\times 2$matrices$A_1, A_2, A_3, A_4$you know the images under$f$of form a basis of$\mathbb R^{2\times 2}$. Then any$A\in\mathbb R^{2\times 2}$can uniquely be expressed as$A = \lambda_1 A_1 + \dots + \lambda_4 A_4$. Using linearity,$f(A)=0is equivalent to $$\lambda_1 \begin{pmatrix} 1 & 2 & 3\\ 1 &... 0 The input matrices form a basis of \Bbb R^{2\times2}. Write v\in\Bbb R^{2\times2} as a_1\begin{bmatrix}1&1\\1&1\end{bmatrix}+a_2\begin{bmatrix}0&1\\1&1\end{bmatrix}+a_3\begin{bmatrix}0&0\\1&1\end{bmatrix}+a_4\begin{bmatrix}0&0\\0&1\end{bmatrix}. Then v\in\ker f\iff f(v)=0_{\Bbb R^{2\times3}} i.e.$$a_1\begin{bmatrix}1&... 0 Hint: $$f(x)=f\left(\alpha\begin{pmatrix} 1 & 1\\ 1 & 1\\ \end{pmatrix}+\beta\begin{pmatrix} 0 & 1\\ 1 & 1\\ \end{pmatrix}+\gamma\begin{pmatrix} 0 & 0\\ 1 & 1\\ \end{pmatrix}+\delta\begin{pmatrix} 0 & 0\\ 0 & 1\\ \end{pmatrix}\right)=$$ $$=\alpha f\begin{pmatrix} 1 & 1\\ 1 &... 1 For any matrices X and Y that have compatible sizes (the number of columns of X equals to that of the rows of Y), and any scalar \lambda, we have$$ \lambda(XY)=(\lambda X)Y. $$(You can even do X(\lambda Y) and get the same result.) This is why we usually omit the paratheses and write \lambda XY. 1 To find the kernel of a linear map f:V\to W is the same as solving the equation f(x)=0\;. What you know in your problem are the values of f(\alpha_k) where \alpha =(\alpha_k:k=1,2,3,4) is a basis of V. You can then write out the matrix representation of f with respect to the basis \alpha of V and the standard basis of W. Then you can ... 2 the companion matrix for polynomial x^5 - 1 is$$ \left( \begin{array}{ccccc} 0&1&0&0&0 \\ 0&0&1&0&0 \\ 0&0&0&1&0 \\ 0&0&0&0&1 \\ 1&0&0&0&0 \\ \end{array} \right) $$or, if preferred, the transpose of this. 2 A straightforward example would be a matrix that rotates on two of the dimensions by some multiple of 72^\circ - as noted in comments here by @BenGrossmann this is a Givens rotation matrix. ... and I now noticed Anthony Saint-Criq's permutation suggestion & subsequent discussion in main comments, that's even easier (as well as being numerically stable)... 4 A quicker way to see that G is not cyclic: note that$$ \pmatrix{2 & 0\\0 & 1} \pmatrix{1 & 1\\0&1} = \pmatrix{2 & 2\\0 & 1}, \\ \pmatrix{1 & 1\\0 & 1} \pmatrix{2 & 0\\0 & 1} = \pmatrix{2 & 1\\0 & 1}. $$That is, we have found elements g,h \in G with hg \neq gh. Because G is not abelian, it cannot be ... 1 \def\p#1#2{\frac{\partial #1}{\partial #2}}Let \,(\alpha,\beta)\, be fourth-order tensors with components$$\eqalign{ \alpha_{ijk\ell} &= \delta_{ik}\,\delta_{j\ell} \\ \beta_{ijk\ell} &= \delta_{i\ell}\,\delta_{jk} \\ }$$and properties with respect to the matrices (F,G,H)$$\eqalign{ \alpha:H &= H:\alpha = H \\ \beta:F &= F:\beta = F^... 3\begin{pmatrix} a &b \\0 &1 \end{pmatrix}\begin{pmatrix} c &d \\0 &1 \end{pmatrix}=\begin{pmatrix} ac &ad+b \\0 &1 \end{pmatrix}\Rightarrow\begin{pmatrix} a &b \\0 &1 \end{pmatrix}^n=\begin{pmatrix} a^n &b(a^n+a^{n-1}+...+1)\\0 &1 \end{pmatrix}$which means$a \neq 1$if there exists an element ... 0 With$11$unknowns and$8$data points, there isn't enough information given to decide what values$a,\ldots, k$have.$\$\begin{pmatrix}1&1&1&1&0&0&0&0&0&0&0\\0&0&0&0&1&1&1&0&0&0&0\\0&0&0&0&0&0&0&1&1&1&0\\0&0&0&0&0&0&...

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