Q23E Find the basis of all 2x2 lower ... [FREE SOLUTION]

91影视

Linear Algebra With Applications

Otto Bretscher

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 442 pages

ISBN: 9780321796974

Chapter 4: Q23E (page 176)

Find the basis of all 2x2 lower triangular matrix, and determine its dimension.

Short Answer

Expert verified

The dimension of a 2x2 lower triangular matrix is 3 which is spanned by

$S p a n \{[\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}], [\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}], [\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}]\}$ .

Step by step solution

Determine the matrix.

Consider the matrix $A = [\begin{matrix} a & 0 \\ b & c \end{matrix}]$ where a, b and c are real.

Simplify the equation $A = [\begin{matrix} a & 0 \\ b & c \end{matrix}]$ as follows.

$A = [\begin{matrix} a & 0 \\ b & c \end{matrix}] A = [\begin{matrix} a & 0 \\ b & 0 \end{matrix}] + [\begin{matrix} 0 & 0 \\ b & 0 \end{matrix}] + [\begin{matrix} 0 & 0 \\ 0 & c \end{matrix}] A = a [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}] + b [\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}] + [\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}]$

Find the basis of each required space

In the matrix, $[\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}], [\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}] and [\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}]$ and are linear independent.

Therefore, the matrix A is spanned by $S p a n \{[\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}], [\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}], [\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}]\}$ .

Hence, the dimension of A is 3 and spanned by $S p a n \{[\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}], [\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}], [\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}]\}$ .

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91影视

Find the basis of all 2x2 lower triangular matrix, and determine its dimension.

Short Answer

Step by step solution

Determine the matrix.

Find the basis of each required space

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