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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

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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