Elementary Linear Algebra 7th Edition

Published by Cengage Learning

Chapter 4 - Vector Spaces - 4.5 Basis and Dimension - 4.5 Exercises - Page 187: 25

Answer

$S$ does not span $M_{2,2}$.

Work Step by Step

The set $S=\left\{\left[\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right],\left[\begin{array}{ll}{0} & {1} \\ {1} & {0}\end{array}\right]\right\}$ is not a basis for $M_{2,2}$ because $S$ does not span $M_{2,2}$. For example, consider the matrix $\left[\begin{array}{ll}{1} & {0} \\ {0} & {0}\end{array}\right]$ as a linear combination of the elements of $S$ as follows $$\left[\begin{array}{ll}{1} & {0} \\ {0} & {0}\end{array}\right]=a\left[\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right]+b\left[\begin{array}{ll}{0} & {1} \\ {1} & {0}\end{array}\right].$$ We have the following system of equations \begin{align*} a&=1\\ a&=0\\ b&=0. \end{align*} It is clear that the above system is in-consistent. Hence, $S$ does not span $M_{2,2}$.

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