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Let \(U\) and \(V\) denote the spaces of symmetric and skew-symmetric \(n \times n\) matrices. Show that \(\operatorname{dim} U+\operatorname{dim} V=n^{2}\)

Short Answer

Expert verified
The dimensions add up to \(n^2\), confirming the spaces span all \(n \times n\) matrices.

Step by step solution

01

Understand Definitions

A symmetric matrix is one where the elements are equal across the diagonal, i.e., element at position (i,j) is equal to element (j,i). A skew-symmetric matrix has elements where the element at position (i,j) is negative of element (j,i), and all diagonal elements must be zero.
02

Dimension of Symmetric Matrix Space

To find the dimension of symmetric matrices, note that for an element in position (i,j), with i not equal to j, both (i,j) and (j,i) need to be specified, but they are equal. Therefore, you only need to specify such elements once for i < j, which amounts to choosing elements from the upper triangle inclusive of the diagonal. This gives n diagonal elements and \(\frac{n(n-1)}{2}\) elements above the diagonal. The dimension of the space of symmetric matrices is \(\operatorname{dim} U = \frac{n(n+1)}{2}\).
03

Dimension of Skew-Symmetric Matrix Space

In skew-symmetric matrices, the diagonal elements are strictly zero. For non-diagonal elements (i,j) where i < j, knowing one of (i,j) or (j,i) determines the other (since they are negatives of each other). Hence, only elements above the diagonal need specifying, equal to \(\frac{n(n-1)}{2}\). Therefore, \(\operatorname{dim} V = \frac{n(n-1)}{2}\).
04

Add Dimensions

Sum the dimensions of the spaces of symmetric and skew-symmetric matrices. \[ \operatorname{dim} U + \operatorname{dim} V = \frac{n(n+1)}{2} + \frac{n(n-1)}{2} = \frac{n^2 + n + n^2 - n}{2} = \frac{2n^2}{2} = n^2 \] Thus, \(\operatorname{dim} U + \operatorname{dim} V = n^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Symmetric Matrices
Symmetric matrices are a fundamental concept in linear algebra, particularly known for their balanced structure. A matrix is termed symmetric if it is equal to its transpose. This means that the element at the ith row and jth column (\((i,j)\)) is identical to the element at the jth row and ith column (\((j,i)\)). In simpler terms, the left-right diagonal divides the matrix into two mirrored halves. Symmetric matrices appear frequently because of their inherent properties:
  • The eigenvalues of a symmetric matrix are always real numbers, which makes computations more intuitive.
  • Simplifying calculations further, symmetric matrices can be diagonalized using orthogonal matrices.
  • Due to their repetitive element pattern, only about half of the total elements in the matrix need independent specifying.
A significant contribution to understanding symmetric matrices is determining the dimension of their matrix space. For any symmetric \(n \times n\) matrix, the total number of independent entries can be calculated as the sum of diagonal elements and those above the diagonal. This results in a dimension of \(\frac{n(n+1)}{2}\).
Skew-Symmetric Matrices
Skew-symmetric matrices are another intriguing class within linear algebra. A matrix is skew-symmetric when its transpose is equal to the negative of the original matrix. For this to hold, the elements on the diagonal must always be zero, because symmetry across the diagonal with zero negation is zero itself. In essence, the off-diagonal elements (i,j) are the negatives of their corresponding (j,i) counterparts. Key features include:
  • All eigenvalues of a skew-symmetric matrix are either zero or purely imaginary numbers.
  • The determinant of any odd-dimensional skew-symmetric matrix is zero, making them singular matrices automatically.
  • Skew-symmetric matrices only require specifying entries above or below the diagonal, significantly reducing the number of elements to specify.
In calculating the dimension of the skew-symmetric matrix space, we observe that we only need the entries above the main diagonal. Given that diagonal elements are zero, the dimension can be identified as \(\frac{n(n-1)}{2}\).
Matrix Spaces Dimension
When exploring the dimensions of matrix spaces, especially symmetric and skew-symmetric types, each characteristic of these matrix families contributes to the overall scope of linear algebra. The basic premise in understanding the dimensionality of matrix spaces relies on the concept of specifying independent elements. These dimensions give life to vector spaces where these matrices reside.
The space of symmetric matrices focuses on \(\frac{n(n+1)}{2}\) unique values, supporting baselines where half of them mirror onto another. Meanwhile, skew-symmetric matrices occupy \(\frac{n(n-1)}{2}\) dimensions, considering the constraint of zero-diagonals and negatives across the leftover fields.
By adding the dimensions of both the symmetric and skew-symmetric matrices, we achieve:
  • The expression for the dimension of symmetric matrices:
  • The expression for the dimension of skew-symmetric matrices:
  • The combined dimension: \(\operatorname{dim} U + \operatorname{dim} V = n^2\). This tells us that every possible arrangement of an \(n \times n\) matrix can be expressed as a sum of symmetric and skew-symmetric portions.
Understanding these key concepts allows us to appreciate the structure and symmetry in matrices, enabling us to analyze and factorize matrices effectively in advanced mathematical applications.

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Most popular questions from this chapter

Consider the linear transformations \(V \stackrel{T}{\rightarrow} W \stackrel{R}{\rightarrow} U\). a. Show that ker \(T \subseteq \operatorname{ker} R T\). b. Show that \(\operatorname{im} R T \subseteq \operatorname{im} R\).

If \(T: \mathbf{M}_{n n} \rightarrow \mathbb{R}\) is any linear transformation satisfying \(T(A B)=T(B A)\) for all \(A\) and \(B\) in \(\mathbf{M}_{n n}\) show that there exists a number \(k\) such that \(T(A)=k \operatorname{tr} A\) for all A. (See Lemma 5.5.1.) [Hint: Let \(E_{i j}\) denote the \(n \times n\) matrix with 1 in the \((i, j)\) position and zeros elsewhere. Show that \(E_{i k} E_{l j}=\left\\{\begin{array}{cl}0 & \text { if } k \neq l \\\ E_{i j} & \text { if } k=l\end{array}\right.\). Use this to show that \(T\left(E_{i j}\right)=0\) if \(i \neq j\) and \(T\left(E_{11}\right)=T\left(E_{22}\right)=\cdots=T\left(E_{n n}\right) .\) Put \(k=T\left(E_{11}\right)\) and use the fact that \(\left\\{E_{i j} \mid 1 \leq i, j \leq n\right\\}\) is a basis of \(\mathbf{M}_{n n}\).

In each case, show that \(T\) is not a linear transformation. a. \(T: \mathbf{M}_{n n} \rightarrow \mathbb{R} ; T(A)=\operatorname{det} A\) b. \(T: \mathbf{M}_{n m} \rightarrow \mathbb{R} ; T(A)=\operatorname{rank} A\) c. \(T: \mathbb{R} \rightarrow \mathbb{R} ; T(x)=x^{2}\) d. \(T: V \rightarrow V ; T(\mathbf{v})=\mathbf{v}+\mathbf{u}\) where \(\mathbf{u} \neq \mathbf{0}\) is a fixed vector in \(V(T\) is called the translation by \(\mathbf{u})\)

Let \(T: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}\) be a linear operator of rank \(1,\) where \(\mathbb{R}^{n}\) is written as rows. Show that there exist numbers \(a_{1}, a_{2}, \ldots, a_{n}\) and \(b_{1}, b_{2}, \ldots, b_{n}\) such that \(T(X)=X A\) for all rows \(X\) in \(\mathbb{R}^{n},\) where $$ A=\left[\begin{array}{cccc} a_{1} b_{1} & a_{1} b_{2} & \cdots & a_{1} b_{n} \\ a_{2} b_{1} & a_{2} b_{2} & \cdots & a_{2} b_{n} \\ \vdots & \vdots & & \vdots \\ a_{n} b_{1} & a_{n} b_{2} & \cdots & a_{n} b_{n} \end{array}\right] $$ \(\left[\right.\) Hint \(: \operatorname{im} T=\mathbb{R} \mathbf{w}\) for \(\mathbf{w}=\left(b_{1}, \ldots, b_{n}\right)\) in \(\left.\mathbb{R}^{n} .\right]\)

Let \(T: V \rightarrow W\) be a linear transformation and let \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\) denote vectors in \(V\). a. If \(\left\\{T\left(\mathbf{v}_{1}\right), \ldots, T\left(\mathbf{v}_{n}\right)\right\\}\) is linearly independent, show that \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\right\\}\) is also independent. b. Find \(T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) for which the converse of part (a) is false.

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