[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.2.6

If $\sen{A}<1$, then $I-A$ is invertible, and $$\bex (I-A)^{-1}=I+A+A^2+\cdots, \eex$$ aa convergent power series. This is called the Neumann series.

Solution. Since $\sen{A}<1$, $$\bex \sum_{n=0}^\infty \sen{A}^n=\frac{1}{1-\sen{A}}<\infty. \eex$$ Due to the completeness of the matrix space, $\dps{\sum_{n=0}^\infty A_n}$ converges. Since $$\bex (I-A)(I+\cdots+A^{n-1})=I-A^n, \eex$$ we may take limit to get $$\bex (I-A)\cdot \sum_{n=0}^\infty A^n=I. \eex$$

时间： 2024-08-20 12:59:48

[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.2.6的相关文章

[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.4.1

Let $x,y,z$ be linearly independent vectors in $\scrH$. Find a necessary and sufficient condition that a vector $w$ mush satisfy in order that the bilinear functional $$\bex F(u,v)=\sef{x,u}\sef{y,v}+\sef{z,u}\sef{w,v} \eex$$ is elementary. Solutio

[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.3.7

For every matrix $A$, the matrix $$\bex \sex{\ba{cc} I&A\\ 0&I \ea} \eex$$ is invertible and its inverse is $$\bex \sex{\ba{cc} I&-A\\ 0&I \ea}. \eex$$ Use this to show that if $A,B$ are any two $n\times n$ matrices, then $$\bex \sex{\ba{c

[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.4.6

Let $A$ and $B$ be two matrices (not necessarily of the same size). Relative to the lexicographically ordered basis on the space of tensors, the matrix for $A\otimes B$ can be written in block form as follows: if $A=(a_{ij})$, then $$\bex A\otimes B=

[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.5.10

Every $k\times k$ positive matrix $A=(a_{ij})$ can be realised as a Gram matrix, i.e., vectors $x_j$, $1\leq j\leq k$, can be found so that $a_{ij}=\sef{x_i,x_j}$ for all $i,j$. Solution. By Exercise I.2.2, $A=B^*B$ for some $B$. Let $$\bex B=(x_1,

[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.5.5

Show that the inner product $$\bex \sef{x_1\vee \cdots \vee x_k,y_1\vee \cdots\vee y_k} \eex$$ is equal to the permanent of the $k\times k$ matrix $\sex{\sef{x_i,y_j}}$. Solution. $$\beex \bea &\quad \sef{x_1\vee \cdots \vee x_k,y_1\vee \cdots \vee

[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.1.2

Let $X$ be nay basis of $\scrH$ and let $Y$ be the basis biorthogonal to it. Using matrix multiplication, $X$ gives a linear transformation from $\bbC^n$ to $\scrH$. The inverse of this is given by $Y^*$. In the special case when $X$ is orthonormal (

[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.2.8

For any matrix $A$ the series $$\bex \exp A=I+A+\frac{A^2}{2!}+\cdots+\frac{A^n}{n!}+\cdots \eex$$ converges. This is called the exponential of $A$. The matrix $A$ is always invertible and $$\bex (\exp A)^{-1}=\exp(-A). \eex$$ Conversely, every inver

[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.4.4

(1). There is a natural isomorphism between the spaces $\scrH\otimes \scrH^*$ and $\scrL(\scrH,\scrK)$ in which the elementary tensor $k\otimes h^*$corresponds to the linear map that takes a vector $u$ of $\scrH$ to $\sef{h,u}k$. This linear transfor

[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.5.1

Show that the inner product $$\bex \sef{x_1\wedge \cdots \wedge x_k,y_1\wedge \cdots\wedge y_k} \eex$$ is equal to the determinant of the $k\times k$ matrix $\sex{\sef{x_i,y_j}}$. Solution. $$\beex \bea &\quad \sef{x_1\wedge\cdots \wedge x_k,y_1\we