[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.5.3

Let $\scrM$ be a $p$-dimensional subspace of $\scrH$ and $\scrN$ its orthogonal complement. Choosing $j$ vectors from $\scrM$ and $k-j$ vectors from $\scrN$ and forming the linear span of the antisymmetric tensor products of all such vectors, we get different subspaces of $\wedge^k\scrH$; for example, one of those is $\vee^k\scrM$. Determine all the subspaces thus obtained and their dimensionalities. Do the same for $\vee^k\scrH$.

Solution. (1). Let $e_1,\cdots,e_p$ be the orthonormal basis of $\scrM$, and $e_{p+1},\cdots,e_k$ be the orthonormal basis of $\scrN$. Then for $0\leq j\leq k$, the subspace we consider has a basis $$\bex e_{i_1}\wedge \cdots \wedge e_{i_j}\wedge e_{i_{j+1}}\wedge\cdots \wedge e_{i_k}, \eex$$ where $$\bex 1\leq i_1<\cdots<i_j\leq p<p+1\leq i_{j+1}<\cdots<i_k\leq n. \eex$$ Thus its dimension is $$\bex \sex{p\atop j}\cdot \sex{n-p\atop k-j}. \eex$$ (2). Now we consider the subspace of $\vee^k\scrH$. In this case, it has a basis $$\bex e_{i_1}\vee \cdots \vee e_{i_j}\vee e_{i_{j+1}}\vee \cdots \vee e_{i_k}, \eex$$ where $$\bex 1\leq i_1\leq\cdots\leq i_j\leq p<p+1\leq i_{j+1}\leq\cdots\leq i_k\leq n. \eex$$ Thus its dimension is $$\bex \sex{p+j-1\atop j}\cdot \sex{n-p+k-j+1\atop k-j}. \eex$$

时间: 2025-01-14 21:21:26

[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.5.3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. 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

[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.2.7

The set of all invertible matrices is a dense open subset of the set of all $n\times n$ matrices. The set of all unitary matrices is a compact subset of all $n\times n$ matrices. These two sets are also groups under multiplication. They are called th

[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.2.5

Show that matrices with distinct eigenvalues are dense in the space of all $n\times n$ matrices. (Use the Schur triangularisation) Solution.  By the Schur triangularisation, for each matrix $A$, there exists a unitary $U$ such that $$\bex A=U\sex{\ba