# Tag Info

1

The shortcut is a method to (more or less) quickly solve full rank $(n-1)\times n$ linear systems (i.e., all $n-1$ lines are linearly independent). It is IMO awkwardly presented by writing a the determinant of a matrix in which some entries are "unit vectors" (which makes no sense, matrix entries must be scalars). The proper way to present is it is as ...

0

Your method $(3)$ has nothing to do with eigenvectors per se. Instead it is a quick method for solving homogeneous systems of rank $2$ in three variables; but it can only be used in this case, namely rank $2$ and dimension $3$. Given such a system $$a_{i1}x_1 +a_{i2} x_2+a_{i3}x_3=0\qquad(1\leq i\leq m)\ ,\tag{1}$$ where usually $m=2$ or $m=3$, it is easy ...

4

It looks like you are computing the cross product of two rows of the matrix from the second method. This will give a vector that is orthogonal to those two rows. An eigenvector will be orthogonal to all rows, since it is in the null space of the matrix. So, if the two chosen vectors span the row-space, the third method gives an eigenvector.

1

I'll provide an example of computing an eigenvector for a different matrix. Consider the matrix below which has eigenvalues $\lambda_1=2$ and $\lambda_2=0$, $$A = \left[ \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right].$$ To find $v_1$ we write the equation $A v_1 = \lambda_1 v_1$. Here we represent $v_1$ as the column vector $(a,b)$, $$... 2 If you are trying to verify the semicircular law, the elements need to be chosen from a Gaussian distribution, which is why randn has no outliers. The outlier effect is a consequence of Perron-Frobenius theory; you are generating a non-negative matrix, so there is a single largest real eigenvalue, and all the others are strictly smaller. There is apparently ... 1 You are doing nothing wrong. Gaussian elimination does not preserve eigenvalues, and this method simply does not work. If you are lucky enough to start with an upper triangular matrix, then it is true that the diagonals are the eigenvalues. This is very convenient. Otherwise, you have to go and actually compute them. For two by two matrices, there is a ... 0 It means that you want : M = a*I =N*A*N^{-1} => A = a*N^{-1}*I*N = a*I = M So in this case there is only one solution. For the second case, all you have to do is find a diagonalizable matrix with the coefficient of M as eigenvalues : X_A(X)=X^2 -tr(A)*X +det(A)=(X-a_1)*(X-a_2)  =X^2 -(a_1+a_2)*X +a_1*a_2 For instance, we can take:$$N = ...

3

Since the family $(I,A,\ldots,A^{n-1})$ is linearly independent then there's no non trivial linear combination of these matrices that gives $0$ so the polynomial $P=x^n-1$ which annihilates $A$ has the minimal degree so it's the minimal polynomial and the characteristic polynomial so their roots: the $n$- roots of the unity are the eigenvalues of $A$ and ...

3

Hint: Since $A^n - I = 0$, the minimal polynomial of $A$ divides $x^n - 1$. Since $(I,\dots,A^{n-1})$ are linearly independent, the minimal polynomial must have degree $n$. We conclude that the minimal polynomial of $A$ is $q(x)= x^{n} - 1$.

2

Hint: for any square matrix $\;n\times n\;,\;\;Tr. A\;$ is the coefficient of $\;x^{n-1}\;$ in the characteristic polynomial of $\;A\;$ . Another hint in the same spirit as above: the trace of $\;A\;$ is the sum of all its eigenvalues, which are the roots of the charac. polynomial, and then...

2

You have the right idea. To prove multiplicity: it is a theorem that for any real polynomial $p(t)$, if $z$ is a root with multiplicity $p$, then so is $\overline{z}$. Apply this to the polynomial $\det(A - tI)$.

2

Write your eigenvectors into a matrix $V$: $$Q = \pmatrix{-2 & 2 & 1\\ 1 & 0 & 2 \\ 0 & 1 & -2}.$$ Then it holds $$AQ = QL$$ by definition of the eigenvectors. Hence $A = QLQ^{-1}$. In essence: the columns of $Q$ form a basis of the vector space $\mathbb R^3$ of eigenvectors of $A$. In order to obtain such a basis, you have to ...

1

The eigenvalues $\lambda_1=2+i\sqrt{3}$ and $\lambda_2=2-i\sqrt{3}$ are correct. Now you have to solve $$(A-\lambda_1I_2)x=0$$ which is the system having matrix $$\begin{bmatrix} 1-\lambda_1 & 2 \\ -2 & 3-\lambda_1 \end{bmatrix} = \begin{bmatrix} -1-i\sqrt{3} & 2 \\ -2 & 1-i\sqrt{3} \end{bmatrix}$$ Since you know this matrix has rank ...

0

I think I know where you made a mistake; when you calculated $A-(2+\sqrt{3}*i)I$ , you wrote that the term on the up-left was $(\sqrt{3}*i-1)$ whereas it was $(-\sqrt{3}*i-1)$, you could have tried anything from there it would have been confusing at best :). I'll show you a practical way of dealing with this: Calculate $B=A-(2+\sqrt{3}*i)I$ B = $\left( ... 1 To add a few details: to get an automorphism you'd better be able to undo the map$T$, and to be able to map the image of the square back to the torus$T$had better map integer lattice points to integer lattice points. If$T$is also linear, as we have here, then these conditions are equivalent to requiring$T\in GL_2(\mathbb{Z})$, that is,$T$is ... 2 The fundamental difference is that the Jacobi method attempts to reduce the matrix to diagonal form, and successive rotations undo previously set zeros, but the off-diagonal elements successively get smaller and smaller (thus it is an "iterative" method). The sequence of Givens rotations tries to do something easier: It reduces the matrix to tridiagonal ... 1 There's no natural ordering on the eigenvalues. The coefficients of the characteristic polynomials$P_i$certainly converge to the coefficients of$P$, but does that mean that the "roots converge"? It's probably better to put it this way: the coefficients of a polynomial are related to the elementary symmetric functions in its roots. So yes, the ... 2 Note that$A$is diagonalizable since$A$is$3\times 3and has three eigenvalues. Hence $$A=P\Lambda P^{-1}$$ where \begin{align*} P&= \begin{bmatrix} 0 & 1 & 0 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix} & \Lambda &= \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} ... 0 Result from MATLAB, using symbolic manipulation on MuPad (Source: Guodong Xie). 1 $$\lambda I-A=\begin{pmatrix}\lambda-a&-b\\-b&\lambda+a\end{pmatrix}$$ Substituting in the above and forming tha corresponding homogeneous system (and pay attention to the fact that only one equation is needed (why?)) when: $$\lambda=\pm\sqrt{a^2+b^2}\;\;\implies\;\;\;(\pm\sqrt{a^2+b^2}-a)x-by=0\iff x=\frac b{\pm\sqrt{a^2+b^2}-a}y$$ Thus, we get ... 0 Note that the sum of the eigenvalues is 0. If all eigenvalues are integers, then all coefficients of the characteristic polynomial are integers. Setd=a^2+b^2+c^2$, an integer. Suppose$n$is an integer eigenvalue. Dividing$\lambda-n$into the char poly gives $$\lambda^2+n\lambda+(n^2-d).$$ This has roots $$\frac{-n\pm \sqrt{4d-3n^2}}{2}.$$ These are ... 3 Set$a = 0$, then we have $$\pmatrix{0 & 0 & b \\ 0 & 0 & c \\ b & c & 0}.$$ The eigenvalues are defined by$ -\lambda(\lambda^2 - c^2) + b^2\lambda = 0 $, so we have$\lambda = 0$or$\lambda = \pm\sqrt{b^2 + c^2}$. Choose Pythagorean triple values such as$b = 3$and$c = 4$. Then we have$\lambda = 0$and$\lambda = \pm5$. ... 0 I think this comes down to understanding the characteristic polynomial. Suppose that$PQ = 0$, where$P$and$Q$are relatively prime polynomials of$T$. Then$V = PV \oplus QV$. This is basically the Chinese Remainder Theorem, but we can see it directly: if$AP + BQ = 1$, then$v = APv+BPv$, and if$w \in PV \cap QV$, then$Pw = Qw = 0$, so$w = (AP+BQ)w ...

0

Assume that all the eigenvalues of $T$ are in the field of scalars of $V$ ( true if $k=\mathbb{C}$. Then $V$ is a direct sum of the generalized eigenspaces $V_{\lambda}$. That is, every vector can be written uniquely as $$v = \sum_{\lambda} v_{\lambda}$$ where $(T-\lambda)^{n_{\lambda}}v_{\lambda} = 0$. Note that any vector for which more than one ...

2

Picking up from where you left off. From $\text{rank}(A-5I)=3$ and from $\text{rank}(A+2I)+\text{rank}(A+3I)+\text{rank}(A-5I)=9$ you get $$\text{rank}(A+2I)+\text{rank}(A+3I)=6.$$ Now prove that $A+2I$ is invertible. What does that tell you about the rank of $A+2I$? Infer that $\text{rank}(A+3I)=2$. What does that tell you about the geometric ...

0

Hint: The rank of $A - \lambda I$ is $n$ minus (the number of Jordan blocks of $A$ that are associated with $\lambda$).

0

Hint: with such a characteristic polynomial, the matrix is congruent to either $$\left( \begin{array}{ccc} -3 & 1 & 0 &0\\ 0 & -3 & 0 &0\\ 0 & 0 & 1 &0\\ 0 & 0 & 0 &5\end{array} \right)$$ or $$\left( \begin{array}{ccc} -3 & 0 & 0 &0\\ 0 & -3 & 0 &0\\ 0 & 0 & 1 &0\\ 0 ... 2$$\lambda_1 + \lambda_2 = \textrm{tr}\ A\\ \lambda_1 \lambda_2 = \det A$$Suppose \lambda_1 = -\frac{5}{2}. Then can \lambda_2 satisfy those relationships? 6$$det(A-\lambda I)=0(a-\lambda)(d-\lambda)-bc=0ad-a\lambda-d\lambda +\lambda² -bc=0 \lambda² -(a+d)\lambda+(ad-bc)=0 $$Making analysis under discriminant and you will see... (a+d)²-4(ad-bc)>0 it is \Delta >0 <=> 2 real eigenvalues (a+d)²-4(ad-bc)=0 it is \Delta =0 <=> 1 real eigenvalues ... 1 For an arbitrary trasformation T:V \to V: We say that T is diagonalizable if and only if V can be written as the direct sum of the eigenspaces of T, which is equivalent to finding a basis of V consisting of eigenvectors. Not all transformations are diagonalizable. If V is a vector space over an algebraically closed field, then we can always ... 2 Observe that I, B, B^2, \ldots, B^{n-1} are linearly independent (because their nonzero entries lie on entirely different diagonals). Hence no nonzero degree-(n-1) polynomial over \mathbb R can annilhilate B. In other words, x^n-1 is the minimal polynomial. You may continue from here. Edit. On second thought, perhaps the quickest way to prove the ... 0 If you delete the first row and last column from an irreducible n\times n tridiagonal matrix T, the resulting submatrix is triangular with non-zero diagonal entries. Hence it is invertible, and it follows that \mathbb{rank}(T-\lambda I) is always at least n-1. 0 If you use Lagrange multiplier technique, then you get a (maybe) long list of candidates for global minima. Then you can evaluate the function to minimize over these set of candidates. However, this method is not very efficient in general. It may work for special cases, where something about the structure of the problem is known. 1 If the matrices A,B\in M_n are nxn, then yes. Since those eigenvectors have the same complex modulus, you can rotate matrix B to be in the same plane as A. That rotation matrix is nonsingular, therefore you can combine it with the matrix that diagonalizes A to get a nonsigular matrix that makes it similar to B. For instance, for A,B ... 1 if A is diagonalizable, it's possible I think. You define the opposite of the projection on the subspace formed by your two eigenvalues with the cos and sin, and you change the base to be in the diagonalisation base: E = P*D*P^{−1} , A = P*D1∗P^{−1} , A*E = P*D3∗P^{−1} , with the opposite eigenvalues ( the eigenvectors have been changed into their ... 0 Your proof is basically correct. Since 0 is not an eigenvector of T (if this is not obvious to you, you should think about proving it), every eigenvector of T is also an eigenvector of T^{-1}. There exists by hypothesis a basis consisting of eigenvectors for T, and that basis is also a basis of eigenvectors for~T^{-1}. It depends on how ... 0 Yep. Concretely, if T = UDU^{-1}, then T^{-1} = UD^{-1}U^{-1}. 1 It is easy to see that$$ Ce_j=e_j $$for all j, and hence C is the identity matrix. 1 i think the question is asking for the exercise of "intuitive" perceptions. e.g. a projection clearly has eigen-values 0,1 (for vectors perpendicular to or (resp) lying in the invariant subspace). symmetric and anti-symmetric matrices show that the transpose has eigenvalues \pm 1 2 Sometimes its easier to work by inspection, noting any obvious properties a matrix may have, than to proceed with the full formalism of the characteristic polynomial, etc.; I think such is the case with the present matrix. Setting A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta ... 1 Starting from your last line (assuming it's right),$$-\lambda^3+\lambda^2(1+2\cos(\theta))-\lambda(\cos^2(\theta)+2\cos(\theta)+\sin^2(\theta))+\cos^2(\theta)+\sin^2(\theta)=0$$Since \sin^2\theta+\cos^2\theta=1,$$-\lambda^3+\lambda^2(1+2\cos\theta)-\lambda(1+2\cos\theta)+1=0$$You can factor this as:$$ (1-\lambda) (\lambda^2 -2\lambda \cos\theta +1) = ...

0

If $\lambda$ is an eigenvalue of a $2\times2$ matrix $A$, an eigenvector $v$ is a nonzero solution of $(A-\lambda I)v=0$. In your case, if $v=\begin{bmatrix}x_1\\x_2\end{bmatrix}$, we get the linear system $$\begin{bmatrix} -\lambda & -b \\ a & -\lambda \end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix} =0$$ If we assume $b\ne0$ (or the matrix ...

2

The eigenvectors will be the vectors which, by definition, verify the relation: $(J-\lambda·I)·w=0$ As we have $2$ eigenvalues $(\lambda_1,\lambda_2)$ we are going to have $2$ eigenvectors $(u,v)$: $(J-\lambda_1·I)·u=\left( \begin{array}{ccc} -\sqrt{ab}i & -b \\ a & -\sqrt{ab}i \end{array} \right)\left( \begin{array}{ccc} u_1 \\ u_2 \end{array} ... 1 Make an ansatz$v=\begin{pmatrix}1\\y\end{pmatrix}$and solve$Jv=\lambda v$. You immediately get$-by=\pm\sqrt{ab}i$(and$\pm\sqrt{ab}iy=a$), so$y=\mp\sqrt{\frac ab}i$. 3 There's two cases: if$a=-b$then the given matrix is$aI_2$and then any non zero vector$v$is an eigenvector for$J$if$a\ne-b$then the eigenvectors associated to$a$are$\lambda (1,0)^T$where$\lambda\ne0$and the eigenvectors associated to$-b$are$\lambda(0,1)^T$where$\lambda\ne0$. 0 One way to answer this question is to remember the definition of an eigenvector. An eigenvector$v$of a matrix$A$is a vector which satisfied$Av=\lambda v$for some scalar$\lambda$. So we can guess and check. My guesses would be$\left(\begin{array}{c} 1\\0\end{array}\right)$and$\left(\begin{array}{c} 0\\1\end{array}\right)$. And a simple check ... 2 if$a=b=0$Then you have $$\begin{pmatrix}a&0\\0&-b\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}.$$ This means that every basis$(v_1,v_2)$of$\Bbb{R}^2$is a basis of eigenvectors. One can verify this by seeing that the dimension of the kernel of$A$is$2$and the fact that multiplying the zero matrix by$v_1$and$v_2$we get the ... 0 Here's a bit more general/topological perspective for why this is true: The set of all invertible$n$-by-$n$matrices is an open set, so for any invertible matrix there exists a small neighborhood of invertible matrices surrounding it. In particular, there is a small neighborhood of invertible matrices surrounding the identity, so that$$I + ... 0 Let$T_\alpha=\alpha I -T:\mathbb{R}^n \to \mathbb{R}^n$Assume$T_\alpha$is not an isomorphism, so$ker(T_a) \ne \{ \vec{0}\}$Thus the set$S_\alpha:=\{\vec{v}\in \mathbb{R}^n: T(\vec{v})=\alpha \vec{v}\}= ker(T_\alpha) \subset \mathbb{R}^n$has$\dim(S_\alpha)\ge 1$For$\alpha \ne \beta$, if$T_\beta$is not an isomorphism, then$S_\beta \cap ...

2

We know by the Jordan block form of $T$ that $\alpha I-T$ is similar to an upper-triangular matrix with values on the diagonal equal to $\alpha-\lambda_i$ with $\lambda_i$ the eigenvalues of $T$. So long as $|\alpha|>|\lambda_i|$ for all $i$, the determinant is non-zero, hence the transformation $\alpha I-T$ is invertible.

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