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## New answers tagged eigenvalues-eigenvectors

1

Given an invertible symmetric matrix $A \in \mathbb{R}^{n \times n}$, an eigendecomposition of $A$ is $$A = Q \Lambda Q^T = \begin{bmatrix} Q_1 & Q_2\end{bmatrix} \begin{bmatrix} \Lambda_1 & O\\ O & \Lambda_2\end{bmatrix} \begin{bmatrix} Q_1^T\\ Q_2^T\end{bmatrix}$$ where the columns of orthogonal matrix $Q$ are the eigenvectors of $A$ and the ...

1

Let $$v_3=\begin{pmatrix}-1\\0\\0\end{pmatrix}.$$ Then $$(M-2I)v_3 = \begin{pmatrix}-1&1&1\\-1&1&1\\0&0&0\end{pmatrix}\begin{pmatrix}-1\\0\\0\end{pmatrix} = \begin{bmatrix}1\\1\\0\end{bmatrix}=v_1,$$ so $$(M-2I)^2v_3 = (M-2I)(M-2I)v_3=(M-2I)v_1=0,$$ and hence $v_3$ is a generalized eigenvector of $M$.

1

If $A$ is symmetric and negative definite (i.e., all its eigenvalues are negative), then $-A$ is symmetric and positive definite (i.e., all its eigenvalues are positive). From the symmetry of $-A$, we conclude that it is diagonalizable, that its eigenvalues are real, and that its eigenvectors are orthogonal. Hence, $-A$ has an eigendecomposition $-A = Q ... 3 Easily we see that$-A$is symmetric with positive eigenvalues and by spectral theorem it's diagonalizable in orthonormal basis: there is an orthogonal matrix$P$and$\lambda_1,\ldots,\lambda_n>0$such that $$-A=P DP^T$$ where$D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$. Let$\Delta=\operatorname{diag}(\sqrt\lambda_1,\ldots,\sqrt\lambda_n)$then ... 1$\DeclareMathOperator{\rank}{rank}$If I understand, you're asking: Let$A$be an$n \times n$(real/complex) matrix with$k < n$non-zero eigenvalues (counting multiplicity) and$m \geq 1$the multiplicity of$0$as an eigenvalue. Is$\det A = 0$? Is$\rank A = n - m = k$, the number of non-zero eigenvalues counted with multiplicity, or only ... 2 The “maximum number of linearly dependent rows” doesn't really make sense. Consider the matrix $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}$$ The three rows form a linearly dependent set, but no subset of rows is linearly dependent. What you can define is the maximum number of linearly independent rows, which ... 0 If$\lambda_1=0$, then$\dim(\mathrm{ker}(A))=\alpha$implies that the geometric multiplicity is$\alpha$. If$\lambda_1\neq 0$, then$A$is non-singular (for otherwise$0$would be an eigenvalue of$A$), hence$\alpha=0$. In this case you need to determine$\dim(\mathrm{ker}(A-\lambda_1I))$in order to find the geometric multiplicity of$\lambda_1$. 1 1) From$N^2=0$we have$\operatorname{Im}N\subset \ker N$and from the rank-nullity theorem we get$\dim\ker N=2$and$\dim\operatorname{Im}N=1$(assuming$N\ne0$). So since$\ker N= E_0(N)$the eigenspace associated to$0$so$N$has two linearly independent eigenvectors associated to$0$. 2) Now if$N^2\ne0$then there is$e\in E$such that$N^2e\ne0$. ... 0 In general, the symmetric eigenvalue problem is solved by reducing the matrix to tridiagonal form and then applying the QR algorithm with implicit shifts. Only the last stage is relevant here and the run time is$O(n^2)$as you only want the eigenvalues. I recommend starting with the routines implemented in LAPACK. A short list of relevant routines is ... 1 Given$\mathrm{u} \in \mathbb{R}^n \setminus \{0_n\}$, an eigendecomposition of the projection matrix$P := \dfrac{\mathrm{u} \mathrm{u}^T}{\mathrm{u}^T \mathrm{u}}is $$P = Q \Lambda Q^T = \begin{bmatrix} | & | & & |\\ \dfrac{\mathrm{u}}{\|\mathrm{u}\|} & \mathrm{q}_2 & \cdots & \mathrm{q}_n\\ | & | & & |\end{bmatrix} ... 2 Just express the identity matrix as a product \mathbf{VV^T} where \mathbf{V} is an orthonormal matrix with first column u/\|u\| and the remaining columns some basis of the perpendiculsr space of u. You'll get a matrix decomposition of H showing that the eigenvalues are (-1,1,..,1), eigenvectors are the same as columns of \mathbf{V}. 2 This should be more seen as a hint: Consider (x_1,\ldots,x_n)=X \in \mathbb{R}^{n\times n}, where the x_i denote the column vectors of X. Now consider \mathbb{R}^{n\times n} as a "normal" vectorspace, i.e. with column vectors, then you have$$ X=\pmatrix{x_1\\ \vdots \\ x_n}. $$Now what is meant by AX in this representation? Since the columns of ... 1 Matrices of rank~1 are easy to understand due to their large eigenspace for \lambda=0. Let A be square of size~n and have trace~c. Then its characteristic polynomial is X^{n-1}(X-c), so the algebraic multiplicity of the eigenvalue~0 is either n-1 when c\neq0, or n when c=0. The minimal polynomial is X(X-c) provided n>1 (when ... 0 Important properties of Kronecker product operation:$$(A\otimes B)(C\otimes D)=AC\otimes BD,A\otimes(B+C)=A\otimes B+A\otimes C.Now if L(G)X_i=\lambda_i X_i and L(F)Y_j=\mu_j Y_j, then \begin{align*}L(G\times H)(X_i\otimes Y_j)&=(L(G)\otimes I_m+I_n\otimes L(H))(X_i\otimes Y_j)\\&=L(G)X_i\otimes I_mY_j+I_nX_i\otimes ... 2 There can be no definite answer because it has been left unspecified what good results are and what the origin of the undesired negative eigenvalues is. Anyway, the following might have to do with it: Adding \lambda I to C also distorts the originally positive eigenvalues, whereas B leaves those unchanged. B is auto-adaptive: The errors in B ... 2 Presumably, the "true" matrix that you're measuring really is only positive semidefinite with a large kernel. Since you're doing physical measurements, the chance of getting any number exactly is nil, so all the zero eigenvalues are coming out as either tiny positive or tiny negative numbers. If you flip the negative eigenvalues to positive ones, you're ... 2 As hinted in Carmichael's comment, we have that the following are equivalent: (1) (I_n-BA) is invertible. (2) 1 is not an eigenvalue of BA. (3) 1 is not an eigenvalue of AB. (4) (I_m-AB) is invertible. You've proven that (2) \iff (3), so just note that(I-M)v \iff Mv=Iv=v$$for any matrix M. 2 For part (2): Let 𝐼_{𝑚}−𝐴𝐵 be invertible and let’s consider the following matrix expression (𝐼_{𝑛}−𝐵𝐴)𝐵(𝐼_{𝑚}−𝐴𝐵)^{−1}𝐴 which you can verify simplifies to BA. If that is true, then what is (𝐼_{𝑛}−𝐵𝐴)[𝐵(𝐼_{𝑚}−𝐴𝐵)^{−1}𝐴+I_{n}]? You can easily construct a similar argument if you first let 𝐼_{𝑛}−𝐵𝐴 be invertible. 1 The space of endomorphisms of E ( dimE= n), is isomorphic to the space of square matrices of order n; but this is not canonical isomorphism, it depends on the basis of E chosen, the expression of eigenvectors also depends on the basis. The invariant element which does not depend on the basis are eingenvalues, eingenspace also the dimension of the ... 2 A self-adjoint operator S : X \to X (where X is an inner product space) is an operator such that for all x,y \in X, we have$$\langle Sx,y \rangle = \langle x,Sy\rangle.$$This is a generalization of a real, symmetric matrix. One important property of such operators is that the eigenvalues of a self-adjoint operator are necessarily real. Indeed, if ... 3 The first two diagonal matrices~D commute with every matrix, so S^{-1}DS=SS^{-1}D=D for any S, so these two are each only similar to themselves. So concentrate on the final matrix~B. You can in principle compute the set \{\, S^{-1}BS\mid s\in GL(2,\Bbb R)\,\} explicitly by using the formula for the inverse and matrix multiplication. You can maybe ... 1 As has been noted in the comments, the first two matrices commute with all S, so they're similar only to themselves. For c), note that any 2\times2 matrix S can be written as the product of a shear, a rotation and a scaling:$$ S=\pmatrix{p\\&r}\pmatrix{1\\q&1}\pmatrix{\cos\phi&\sin\phi\\-\sin\phi&\cos\phi} $$(see this answer). ... 3 Let B be an n\times n-matrix and let k:=\dim N(B). Then the row-echelon form has all zeroes in its last k rows, so B^{\top} has all zeroes in its last k columns, meaning that B^{\top}e_i=0 for the last k basis vectors e_{n-k+1},\ldots,e_n. Hence \dim N(B^{\top})\geq\dim N(B) holds for all square matrices B. Then$$\dim N(B)\leq\dim ... 3 This is because of the Rank nullity theorem: as matrix and its transpose have the same rank, we have $$\DeclareMathOperator\rk{rank} \rk(A-\lambda I)=\rk{}{}^\mathrm t\mkern-1.5mu(A-\lambda I)\iff \dim \ker(A-\lambda I)=\dim\ker{}{}^\mathrm t\mkern-1.5mu(A-\lambda I).$$ 3 Every matrix of rank 1 is similar to one of the Jordan forms $$\left(\begin{array}{cc|ccc}0&1&\\&0&\\ \hline && 0 \\ &&&\ddots \\ &&&&0 \end{array}\right) \qquad\text{or}\qquad \left(\begin{array}{c|ccc}\lambda&\\\hline &0\\&&\ddots\\&&&0\end{array}\right)$$ for some ... 2 The algebraic multiplicity of0$, being at least the geometric one, is at least$n-1$in your case, more generally$n - r$with$r$the rank. Contrary to what you say it is not necessarily equal to this value. Now, it is always at most$n$, the dimension. Thus, in the characteristic polynomial it is either$n-1$or$n$. Both can happen. For the$n-1... 2 There can't be a closed form expression here (for any meaning of "closed form" that is weaker than roots of sextic polynomials). For example, try $$H = \pmatrix{3 & 1 & 0 & 0 & 0 & 1\cr 1 & 3 & 1 & 0 & 0 & 0\cr 0 & 1 & 3 & 1 & 0 & 0\cr 0 & 0 & 1 & 3 ... 1 (AB)^{-1} = B^{-1} A^{-1}\\ A = PDP^{-1}\\ A^{-1} = PD^{-1}P^{-1} 1 Here's a hint for the second question. By the definition of inverses, we have the following:$$BC \cdot (BC)^{-1}=I$$In order to get from BC to I, you want to cancel out the C on the right and then cancel out the B. How would this be written in terms of B^{-1} and C^{-1}? 4 Hint If A is invertible and {\bf w} = A {\bf v}, then {\bf v} = A^{-1} {\bf w}. How does this specialize if \bf v is an eigenvector of A? 4 To expand upon the answer by loup blanc, the minimal polynomial of~A is of degree~n, namely it is X^n-a_1X^{n-1}-a_2X^{n-2}-\cdots-a_n. If some eigenvalue~\lambda had geometric multiplicity{}>1 then there would to the contrary exist a monic polynomial of degree less than~n that annihilates~A (for instance the one obtained by dividing its ... -1 You can use the following property of a matrix: Given some n by n matrix, A, the trace of the matrix is equivalent to the sum of its eigenvalues. The sum of the eigenvalues of the matrix is: trace (A) = a_{1} + 0(n-1) = a_{1} = \lambda Using the above theorem, we can then clearly see the following must hold true: 1.) \left | \lambda \right | ... 6 The answer is no. Your matrix A is an avatar of a companion matrix. The geometric multiplicity of an eigenvalue of such a matrix is always 1; 2 Let \lambda \in \mathbb{C} with |\lambda|=1. Then \lambda = e^{it} for t \in \mathbb{R}. Now consider the matrix$$A=\begin{pmatrix} \lambda & 0 \\ 0 & E_{n-1} \end{pmatrix} \in \mathbb{C}^{n \times n}.$$What are the eigenvalues of A? And is this unitary? 2 It's just that$$P^{-1}AP-\lambda I=P^{-1}(A-\lambda I)P. $$3 Actually A and P^{-1}AP share the same characteristic polynomial, hence they have the same eigenvalues. Note that$$\begin{align*} \det(P^{-1}AP-\lambda I_n) & = \det(P^{-1}(A-\lambda I_n)P))\\ & = \det(P^{-1})\det(A-\lambda I_n)\det(P)\\ & = \det(A-\lambda I_n). \end{align*} $$3 Let \lambda=a+ib, with \lvert \lambda\rvert^2=a^2+b^2=1. The matrix$$ U=\left(\begin{array}{rr} a & -b \\ b & a\end{array}\right)\in\mathbb R^{2\times2} $$is unitary, and its eigenvalues are \lambda,\overline{\lambda}. 1 Normally the singular values of a matrix A are defined as the (positive) square roots of the eigenvalues of A^*A. 2 First off, it is well known any eigenvalue~\lambda of a unitary matrix~U has |\lambda|=1; this is so because of v\in\Bbb C^n is a corresponding eigenvector then 0\neq v^*v=v^*U^*Uv=(\lambda v)^*(\lambda v)=\lambda\overline\lambda\,(v^*v), so 1=\lambda\overline\lambda=|\lambda|^2. To conversely show that for each \lambda\in\Bbb C with ... 4 Suppose you have a non-trivial solution of$$ T^*Tf = \int_{t}^{1}\int_{0}^{s}f(y)dy ds = \lambda f(t) $$Then \lambda \ne 0 because the above would give f=0 after differentiating a couple of times. For \lambda \ne 0, any solution of the above must satisfy$$ \lambda f'' = -f \\ f(1)=0,\;\; f'(0)=0. $$Any ... 0 When \mathbf{B} is invertible (which isn't the case here),$$\mathbf{A}x=\lambda\mathbf{B} x$$can be rewritten$$\mathbf{B}^{-1}\mathbf{A}x=\lambda x,$$which is an ordinary Eigenproblem. That should be enough for you to understand the note. 1 Your '' instinct'' is correct only if we want Ax=\lambda Bx for all x. But, if we search some x for which the equation is true, than we have to solve$$ \begin{bmatrix} 3&1\\ 6&-2 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} -1&1\\ 0&0 \end{bmatrix} \begin{bmatrix} \lambda x\\ \lambda y \end{bmatrix} $$in ... 1 Since Ax=\lambda Bx, we have Ax-\lambda Bx=0 or (A-\lambda B)x=0 When we solve for eigenvalues, we have Ax=\lambda x so Ax-\lambda x=0 \rightarrow (A-\lambda I)x=0. Thus, \lambda is an eigenvalue iff A-\lambda I is singular or det(A-\lambda I)=0. Thus, in this case, it is correct that \lambda will be an "eigenvalue" if A-\lambda B is ... 1 The characteristic equation is:$$P(\mu)=\mu^4+(\omega^2-2\lambda)\mu^2+\lambda^2$$Where \mu are your eigenvectors. Note that u=\mu^2 results in an quadratic equation, for which you can simply find its roots.$$P(u)=u^2+(\omega^2-2\lambda)u+\lambda^2u_{1/2}=\frac{-(\omega^2-2\lambda)\pm\sqrt{(\omega^2-2\lambda)^2-4\lambda^2}}{2}$$Now use ... 1 Consider$$\left\{\begin{array}{c} −\Delta u = f & \text{on }\Omega \\ u(x)=0, \partial\Omega \end{array}\right.,$$where \Omega \subset \mathbb{R}^{N} is open, bounded and f \in L^{2}(\Omega) See that u \in H^{1}_{0}(\Omega) is a weak solution for the problem above if (u,v)_{H^{1}_{0}(\Omega)} = \displaystyle\int_{\Omega} \nabla u \nabla v = ... 0 Note: I realized eigenplan is the space formed by the eigenvectors for a repeated eigenvalue . So the title you ask a bout the existence of a eigenplan, but in the text you speak a bout invariant hyperplane . Then an invariant hyperplane is not necessarily a eigenplan, the converse is yes. The answer to the question given in the text, is already ... 1 Suppose we are given a matrix A \in \mathbb{R}^{n \times n}. If A = O_n, the one eigenvalue of A is 0, with multiplicity n. If the first n-1 rows or columns of A are either zero or a multiple of the n-th row or column, then we have a rank-1 matrix that can be written in the form$$A = \mathrm{u} \mathrm{v}^T$where$\mathrm{u}, ...

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Hint: $D$ is a rank-$1$ matrix (why?) and therefore it has an eigenspace of dimension $n-1$ (why?). Therefore, if you can find one vector $x$ such that $Dx=\lambda x$ which is not in that big eigenspace, then you have found all of the eigenvalues (why?). You shouldn't have to work too hard to come up with this $x$. (Note: Your guesses are largely correct, ...

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Your characteristic polynomial should be of the form $w^2-2\cos\theta \; w + 1$. Solve it to get your eigenvalues, $w$.

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Actually the two eigenvalues of your rotation matrix are complex (and not real), and more specifically they are $e^{i\phi}$ and $e^{-i\phi}$. Try to compute again (and carefully) the characteristic polynomial to find this result.

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