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Chapter 1: Problem 16

The set of all upper triangular \(n \times n\) matrices is a subspace \(\mathrm{W}\) of \(\mathrm{M}_{n \times n}(F)\) (see Exercise 12 of Section 1.3). Find a basis for W. What is the dimension of W?

Short Answer

Expert verified
A basis for the space of upper triangular \(n \times n\) matrices is given by the set of standard matrices \(\{ E_{ij} \mid i \le j \}\), and the dimension of this subspace is \(\frac{n(n + 1)}{2}\).

Step by step solution

01

Understand Upper Triangular Matrices

An upper triangular matrix is a special kind of square matrix in which all the elements below the main diagonal are zero. More formally, an \(n \times n\) matrix A is upper triangular if \(a_{ij} = 0\) for \(i > j\).
02

Define an Upper Triangular Matrix

Let's represent an arbitrary upper triangular matrix A in W as follows: \( A = \begin{bmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\ 0 & a_{22} & a_{23} & \cdots & a_{2n} \\ 0 & 0 & a_{33} & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & a_{nn} \\ \end{bmatrix} \)
03

Find a Basis for W

Now, let's consider the standard basis matrices \(E_{ij}\) for \(\mathrm{M}_{n \times n}(F)\), where the matrix has 1 at the (i, j)-th position and 0 elsewhere. Notice that for \(i \le j\), the matrix \(E_{ij}\) is an upper triangular matrix. Thus, the set of matrices \(\{ E_{ij} \mid i \le j \}\) spans W, and since they are linearly independent, they form a basis for W.
04

Determine the Dimension of W

Finally, we need to count the number of matrices in the basis to determine the dimension of W. Since the basis consists of matrices with \(i \le j\), there are \(n + (n - 1) + \cdots + 1 = \frac{n(n + 1)}{2}\) such matrices in the basis. Therefore, the dimension of W is \(\frac{n(n + 1)}{2}\). In summary, a basis for the space of upper triangular \(n \times n\) matrices is given by the set of standard matrices \(\{ E_{ij} \mid i \le j \}\), and the dimension of this subspace is \(\frac{n(n + 1)}{2}\).

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