[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]PrI.6.1

Given a basis $U=(u_1,\cdots,u_n)$ not necessarily orthonormal, in $\scrH$, how would you compute the biorthogonal basis $\sex{v_1,\cdots,v_n}$? Find a formula that expresses $\sef{v_j,x}$ for each $x\in\scrH$ and $j=1,\cdots,k$ in terms of Gram matrices.

Soluton. Let $V=(v_1,\cdots,v_k)$, then $$\bex V^*U=I_n\lra U^*V=I_n. \eex$$ We may just set $v_i$ to be the solution of the linear system $U^*x=e_i$, where $e_i=(\underbrace{0,\cdots,1}_{i},\cdots, 0)^T$. Suppose now $$\bex x=\sum_{j=1}^n x_jv_j\in \scrH, \eex$$ then $$\bex \sef{v_i,x}=\sum_{j=1}^n \sef{v_i,v_j}x_j,\quad i=1,\cdots,n. \eex$$ And hence $$\bex \sex{\ba{cc} \sef{v_1,x}\\ \vdots\\ \sef{v_n,x} \ea}=\sex{\sef{v_i,v_j}}\sex{\ba{cc} x_1\\\vdots\\ x_n \ea}. \eex$$

时间： 2024-08-30 04:02:25

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[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]Contents

[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.1.3 Use the QR decomposition to prove Hadamard's inequality: if $X=(x_1,\cdots,x_n)$, then $$\bex |\det X|\leq \prod_{j=1}^n \sen{x_j}. \eex$$ Equality holds here if and only if the $x_j

[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. Solution.

[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.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 y

[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\wedg

[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.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,\c

[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. \ee

[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