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$$det(A-\lambda I)=0$$ $$(a-\lambda)(d-\lambda)-bc=0$$ $$ad-a\lambda-d\lambda +\lambda² -bc=0$$ $$\lambda² -(a+d)\lambda+(ad-bc)=0$$ 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 ...

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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 $\lambda = \pm5$. ...

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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 ...

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Note that $A$ is diagonalizable since $A$ is $3\times 3$ and 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} ...

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$$\lambda_1 + \lambda_2 = \textrm{tr}\ A\\ \lambda_1 \lambda_2 = \det A$$ Suppose $\lambda_1 = -\frac{5}{2}$. Then can $\lambda_2$ satisfy those relationships?

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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 ...

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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 ...

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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 ...

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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 $A,B$ ...

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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 $\pm 1$

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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 ...

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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 ...

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It is easy to see that $$Ce_j=e_j$$ for all $j$, and hence $C$ is the identity matrix.

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$$\lambda I-A=\begin{pmatrix}\lambda-a&-b\\-b&\lambda+a\end{pmatrix}$$ 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 ...

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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 ...

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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 ...

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