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User John Brevik gives as a hint: What is the characteristic polynomial of $$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ How could you have come up with this hint yourself? Well, as you say, the characteristic polynomial only cares about eigenvalues, so you need to find two matrices which are not similar but which have the same eigenvalues. ...

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For the record, a counterexample to 1 (which you correctly disproved) is $A=2I$: $A^4=16I=8A$. If $A=P^{-1}DP$, substitute that into $A^2-4A+8I$, you get: \begin{align*} A^2-4A+8I={}&(P^{-1}DP)^2-4P^{-1}DP+8I=(P^{-1}DP)(P^{-1}DP)-P^{-1}\cdot4D\cdot P+8I={} \\ {}={}&P^{-1}D(PP^{-1})DP-P^{-1}4DP+P^{-1}8IP=P^{-1}(D^2P-4DP+8P)={} \\ ...

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\begin{bmatrix} 1&1 \\ 0& 1 \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} consider these two matrices.(Jordan Canonical form is the answer).

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That's not true in general. let $A=\begin{pmatrix} 1&1\\0&\frac12 \end{pmatrix}$, $B=\begin{pmatrix}1&1\\0&2 \end{pmatrix}$, those 2 matrices are clearly diagonalizable since they have distinct eigenvalues, while $AB=\begin{pmatrix}1&3\\0&1\end{pmatrix}$ isn't.

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The minimal polynomial of $A$ divides $x^4 - 4x^2 = x^2(x-2)(x+2)$ and the characteristic polynomial and the minimal polynomial have the same irreducible factors (possibly with different multiplicities), so a priori, you can only tell that the possible eigenvalues are $0,\pm 2$ and not necessarily all must occur (for example, the matrix $A = cI$ where $c \in ... 3 Hint Consider the eigenvalues of$A \in SO(3, \Bbb R)$. In particular, (1) their product is$\det A = 1$, (2) they all have modulus$1$, and (3) any nonreal eigenvalues come in complex conjugate pairs. Now, if$A \neq I$, what can you say about the$1$-eigenspace of$A$? 2 Answer to your last question: because there are nontrivial Jordan canonical forms. 2 Your matrix, as it is written, is singular: it must be that zero is one of its eigenvalues. In fact, only zero is an eigenvalue of algebraic order two, and the homogeneous system to obtain its eigenspace is $$-\frac12x+\frac12y=0\implies x=y\implies V_{\lambda=0}=Span\left\{\binom11\right\}$$ and it's dimension is one, thus there is no other eigenvector ... 2 So, it's a known result that symmetric matrices (over$\mathbb{R}$), have an orthonormal basis of eigenvectors. We have two eigenvectors. In 3D, a two perpendicular vectors span a plane, so the third vector perpendicular to the first two will be the vector perpendicular to this plane, and therefore unique (up to a constant factor). It can be shown that ... 2 The matrix$A$is real and symmetric, so it must be diagonalisable. Therefore the minimal polynomial (supposedly called$m_A$) has simple roots. The matrix$A+2I$has rank$1$and trace$6$, so its characteristic polynomial is$X^2(X-6)$, and the characteristic polynomial of$A$(supposedly called$f_A$) is obtained from it by substituting$X+2$for$X$: ... 1 You could just expand everything out in series, for example the first term goes like $$( \bar A ^T \bar A ) = ( A^TA+E_1^T A +A^TE_1 + E_1^TE_1)$$ $$( \bar B ^T \bar B ) ^{-1} = ( B^TB+E_2^T B +B^TE_2 + E_2^TE_2)^{-1}=(B^TB)^{-1}( I+E_2^T B(B^TB)^{-1} +B^TE_2(B^TB)^{-1} +E_2^TE_2(B^TB)^{-1} )^{-1}$$ Then use $$(I + C)^{-1} \approx I - C$$ you may use ... 1 The quick answer to your question: note that the only diagonalizable matrix whose eigenvalues are all$0$is the zero-matrix, and that a rank$1$matrix can have at most one non-zero eigenvalue. Another approach: Note that any rank$1$matrix can be written in the form$uv^T$for column vectors$u,v$, and that a rank-$1$matrix will be symmetric if and ... 1 If$\chi_A(A)$is$0$on each vector of a basis,$\chi_A(A)$is$0$on the whole space. 1 Try to do what the link you posted says: $$1=x^2+6xy+y^2=(x\;y)\begin{pmatrix}1&3\\3&1\end{pmatrix}\binom xy=\binom xy^tA\binom xy$$ Diagonalize orthogonally the matrix$\;A\;$(it's possible because it is symmetric): $$\begin{vmatrix}x-1&-3\\-3&x-1\end{vmatrix}=x^2-2x-8=(x-4)(x+2)$$ Now eigenvectors: ... 1 To find the generalised eigenvector you simply have to solve for $$\begin{bmatrix}\frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}1 \\ 1\end{bmatrix}$$ If$A$is your given matrix, this indeed means a solution$e_2$satisfies $$(A+I)e_2=e_1,\quad\text{whence}\quad A ... 1 You should solve \left[\begin{array}{cc}0.5 & 0 \\ 2 & 0.5\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right] = 0.5\left[\begin{array}{c}x \\ y\end{array}\right] That gives 0.5 x = 0.5x 2x + 0.5y = 0.5y So the only restriction is x = 0 . You have only 1 eigenvector, and you are not getting 2 because the matrix is ... 1 In general it is not sufficient to check the characteristic polynomial to make sure that two matrices are similar. In order to be similar, there needs to exist an invertible matrix P such that A = P^{-1}B P. If two matrices are similar and one of them is diagonalizible (say, B=Q^{-1}DQ), then A is automatically diagonalizible, too (by means of ... 1 In general, if p(x) is a polynomial an A a diagonalisable matrix, then so is p(A). First, note that if A=U^{-1}DU, then A^n=U^{-1}D^nU, and hence$$ p(A)=\sum_{k=0}^n c_kA^k=\sum_{k=0}^n c_kU^{-1}D^kU= U^{-1}\left(\sum_{k=0}^n c_k D^k\right)U. $$Clearly, \sum_{k=0}^n c_k D^k is also a diagonal matrix. 1 Recall that \lambda is an eigenvalue of A if \det(A-\lambda I) = 0. This means A - \lambda I is singular and therefore$$(A - \lambda I)v = 0$$has a non-trivial solution v \neq 0 called an eigenvector of A with respect to \lambda. With this, for any t \in \mathbb{F}, we have$$(A-\lambda I) (tv) = t (A-\lambda I) v = 0$$which shows tv ... 1 The formula$$X=k_1[-1/2,0,1]+k_2[1/2,1,0]$$tells you, that for each cobination of k_1,k_2, the resulting vector satisfies Ax=\lambda_1x where \lambda_1=2. Therefore the eigenvectors for \lambda_1 are [-1/2,0,1] and [1/2,1,0] The previous step$$[k_2/2-k_1/2,k_2,k_1]= k_1[-1/2,0,1]+k_2[1/2,1,0]$$is basically a factorization. ... 1 Recall that the set of all eigenvectors of matrix A corresponding to eigenvalue \lambda=2$$ V_2=\{ [\frac{k_2}{2} - \frac{k_1}{2}, k_2, k_1] \mid k_1,k_2 \in \mathbb R\}$$forms a vector subspace of \mathbb R^3. Since every element of V_2 can be expressed as a linear combination$$ [\frac{k_2}{2} - \frac{k_1}{2}, k_2, k_1] = k_1 ... 1 Normal means precisely that$AA^\ast =A^\ast A$. Then $$\langle Av,Av\rangle=\langle v,A^\ast Av\rangle=\langle v,AA^\ast v\rangle=\overline{\langle AA^\ast v,v\rangle}=\overline{\langle A^\ast v,A^\ast v\rangle}$$ This gives the first result. Can you deduce the second from the first? 1 A graph with least eigenvalue at least$-1$is a disjoint union of cliques. (Proof: the least eigenvalue of$K_{1,2}$is$-\sqrt2$, interlacing.) The graphs with least eigenvalue at least$-2$were characterized by Cameron, Goethals and Seidel. They are line graphs, so-called generalized line graphs, and a finite set of graphs associated to$E_6$,$E_7\$, ...

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