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4

In fact, for $\lambda=2$ all three columns become equal, hence $2$ is a double eigenvalue. Then the third eigenvalue can be found from the fact thet the trace $3+4+3$ of the original matrix is the sum of eigenvalues.

4

The matrix $$T=\pmatrix{0&1\\0&0}$$ has a one-dimensional eigenvector space spanned by $[1,0]^T$, but any 2-vector is an eigenvector of $T^2=0$.

3

What about using the $\infty$-norm? That is $$\|A\|_\infty = \sup_{x: \|x\|_\infty=1} \|Ax\|_\infty.$$ Take a vector $x$. Then $$\|Px\|_\infty \le \max_{i}\left|\sum_j p_{ij} x_j\right| \le \max_{i}\sum_j p_{ij} (\max_k |x_k|) \le\|x\|_\infty.$$ Denote $z:=Px$. Then $$\|P^T\Xi^2 z\|_\infty = \max_i \left|\sum_j p_{ji}\xi_j^2 z_j\right| \le\max_i ... 2 Hint Take T the rotation of angle \frac{2\pi}n. 2 Let \lambda a eigenvalue of A and x \neq 0 respective eigenvector, then Ax = \lambda x \Leftrightarrow A^{-1}A x= \lambda A^{-1} x \Leftrightarrow x = \lambda A x \Leftrightarrow x = \lambda^2 x \Leftrightarrow (1-\lambda^2)x = 0 then \lambda =\pm 1 2 By induction: the inductive step assuming that Rank(T^m)=Rank(T^{m+1}) for m\ge n. It's easy to prove that$$Im(T^{m+2})\subset Im(T^{m+1})=Im(T^m)$$Now let y\in Im(T^{m+1}) and x such that y=T^{m+1}(x)=T(T^mx)  but T^m x\in Im(T^m)=Im(T^{m+1}) so there's z such that$$T^m x=T^{m+1}z$$hence$$y=T(T^{m+1}z)=T^{m+2}(z)\in Im(T^{m+2})$$so ... 2 I think that there are two true statements here: (1) If A is an n \times n real symmetric matrix, and A_k denotes its k \times k upper left corner, then the number of negative eigenvalues of A is the number of sign changes in the sequence (1, \det A_1, \det A_2, \ldots, \det A_n). (2) If \det (A+z \mathrm{Id}) = a_0 + a_1 z + \cdots + ... 1 Note that if A has eigenvector x associated with eigenvalue \lambda, then kx is also an eigenvector for any non-zero k \in \Bbb C. So, every matrix, orthogonal or otherwise, has a set of eigenvectors of identical length. 1 Your reasoning starts out fine, up to "In turn, this suggests that either$$ |C|,|D|, \textrm{ or } |C| \textrm{ and } |D| = 0.'' $$After that, I'm not sure what you're trying to do, but it is something circular/unnecessary. To start over from your last correct assertion, you now know that$$ \det[(B + 3I)(B-2I)] = 0 $$which tells you that either$$ ...

1

I don't understand what you can show, except the assumption... Hint: let $x$ be in the image of $T^{n+1}$ Can you show that it is in the image of $T^{n+2}$ by using that elements in the image of $T^n$ are in the image of $T^{n+1}$?

1

$\lambda=2$ is a double eigenvalue, since the rank of the matrix $$A-2 I = \left(\begin{array}{ccc} 1& 1 & 1\\ 2 & 2 & 2 \\ 1 & 1 & 1 \end{array}\right)$$ is equal to one - it has only one linearly independent row. Then Trace$(A)=10$. Thus the third eigenvalue is equal to $10-2-2=6$. Therefore $$\det (A-\lambda ... 1 Applying C_1'=C_1-C_2$$\left| {\matrix{ {3 - \lambda } & 1 & 1 \cr 2 & {4 - \lambda } & 2 \cr 1 & 1 & {3 - \lambda } \cr } } \right|=\left| {\matrix{ {3 - \lambda-1 } & 1 & 1 \cr 2-( 4 - \lambda)& {4 - \lambda } & 2 \cr 1-1 & 1 & {3 - \lambda } \cr } } \right|$$... 1 In the first case you get 4 distinct eigenvalues: \pm 1, \pm i. So T is diagonalizable over \mathbb{C}: there are four eigenspaces in \mathbb{C}^4, each of (complex) dimension 1. Hence the eigenvectors span \mathbb{C}^4. Hence T is diagonalizable over \mathbb{C}. (It is not diagonalizable over \mathbb{R}, because it has non-real eigenvalues.) ... 1 Let \lambda_{\max}<1. Then \rho(A)=\lambda_{\max}=1-\tau, \tau\in(0,1) and \|A^k\|< (1-\tau/2)^k for k sufficiently large. Thus the Neumann series$$ \sum_{k=0}^\infty A^k = (I-A)^{-1} $$converges. As A is non-negative, A^k is non-negative, and by the series representation (I-A)^{-1} is non-negative. If \lambda_{\max}=1 then I-A ... 1 Hint: The definition of the characteristic polynomial is$$f_A(\lambda) = \det(A-\lambda I).$$1 Developping$$ f_A(\lambda) =\begin{vmatrix} -1-\lambda & 2 & 2\\ 2 & 2-\lambda & -1\\ 2 & -1 & 2-\lambda\\ \end{vmatrix} $$gives as coefficient of -\lambda the number \left|\begin{smallmatrix}2&-1\\-1&2\end{smallmatrix}\right| + \left|\begin{smallmatrix}-1&2\\2&2\end{smallmatrix}\right| + ... 1 This Wikipedia article contains a sketch of a proof. It has three steps. If a normal matrix is upper triangular, then it's diagonal. (Proof: show the upper left corner is the only nonzero entry in that row/column using a matrix-norm argument; then use induction.) Details of proof: write A as Q T Q^{-1} for some unitary matrix Q, where T is upper ... 1 When X is a unitary space, and A:\>X\to X is a normal operator then one has$$\|Ax-\lambda x\|^2=\|A^*x-\bar\lambda x\|^2\qquad\forall x\in X,\ \forall\lambda\in{\mathbb C}\ .$$It follows that Ax=\lambda x implies A^*x=\bar\lambda x; whence A and A^* have the same eigenvectors. By the fundamental theorem of algebra A has an eigenvalue ... 1 You can use the Rayleigh quotient (see here: http://en.wikipedia.org/wiki/Rayleigh_quotient). This works as follows. Let A be a real symmetric matrix with minimal eigenvalue \mu_\min and maximal eigenvalue \mu_\max. Then we have$$ \mu_\min = \min_{v^Tv=1}v^TAv  \mu_\max = \max_{v^Tv=1}v^TAv $$This is easy to see if we transform A into ... 1 In general, to prove that a sub-set E is a sub-space, you have to show (1) that if v,w\in E, then v+w\in E, and (2) if v\in E, then \alpha v\in E for \alpha \in \mathbb{C} (or whatever field you are working over). Write v:=(1,-1) and let A and B be elements of E with eigenvalues \lambda and \mu respectively. Then,$$ ...

1

Use induction. Clearly, for $n=1$, $Tv = \lambda v$. Now, assume for $n=k$ that $T^k v = \lambda^k v$. Then, $T^{k+1} v = T (T^k v) = T(\lambda^k v) = \lambda^k T v = \lambda^k (\lambda v) = \lambda^{k+1} v$. Thus, by induction, we see $T^n v = \lambda^n v$.

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