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Hot answers tagged spectral-theory

3

Without using the Cayely-Hamilton theorem: For all $\lambda\ne0$ we see that $\frac1\lambda A$ is also nilpotent hence the matrix $I-\frac1\lambda A$ is invertible hence $$\chi_A(\lambda)=\det(\lambda I-A)\ne0$$ hence for all $\lambda\ne0$, $\lambda$ is not an eigenvalue of $A$ but since $\operatorname{sp}(A)\ne\emptyset$ then $0$ is the only eigenvalue of ...

2

$1$ is an eigenvalue of multiplicity $1$ of a stochastic matrix as long as your graph edges give a path between every pair of vertices $(v,w)$,and the matrix summation will not converge if $1$ is an eigenvalue. Actually, the matrix summation wouldn't converge for any stochastic matrix. You can see this by noting that every power of a stochastic matrix is ...

2

It's false: $A=\left(\begin{array}{cc}1 & 0 \\0 & 2 \\\end{array}\right)$, $B=\left(\begin{array}{cc}3 & 0 \\0 & 1 \\\end{array}\right)$ $ρ(A)<ρ(B)$, but $\left\|A\left(\begin{array}{c}0\\1\\\end{array}\right)\right\|>\left\|B\left(\begin{array}{c}0\\1\\\end{array}\right)\right\|$

2

Example 3 is correct, for $\lambda \notin S^1$, $z \mapsto \frac{1}{z-\lambda}$ is a continuous function on the unit circle, and hence $p(z) - \lambda$ is invertible. On the other hand, for $\lambda\in S^1$, $p(z)-\lambda$ has a zero on $S^1$ and is therefore not invertible, thus $\sigma_C(p) = S^1$. Example 4, however, is not correct. The spectrum of an ...

1

Attempt 1 looks perfect. Attempt 2: The vector $x$ is arbitrary, hence you obtain $(T^*-\bar\lambda I)y=0$. For this conclusion you do not need that range of $T-\lambda I$ is dense. For this attempt, also the backward conclusion is valid: $\bar\lambda\in \sigma_p(T^*)$, then exists $y\ne 0$, $y\in N(T^*-\bar \lambda I)$. This implies $y \in R(T-\lambda ... 1 Presumably$A$and$B$are real. Your requirement is possible if and only if$1$and$-1$are not eigenvalues of$A$. Suppose$1$(or$-1$) is an eigenvalue of$A$and$x$is a corresponding eigenvector. Then your requirement implies that$B\succ A^TBA$, which is impossible because$x^TBx=x^TA^TBAx$. Suppose$\pm1$are not eigenvalues of$A\$. By a change ...

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