• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q14SE

Expert-verified
Found in: Page 267

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# Use Exercise 12 to find the eigenvalues of the matrices in Exercises 13 and 14.14. $$A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{5}}&{{\bf{ - 6}}}&{{\bf{ - 7}}}\\{\bf{2}}&{\bf{4}}&{\bf{5}}&{\bf{2}}\\{\bf{0}}&{\bf{0}}&{{\bf{ - 7}}}&{{\bf{ - 4}}}\\{\bf{0}}&{\bf{0}}&{\bf{3}}&{\bf{1}}\end{array}} \right)$$

The eigenvalues of $$A$$ are $$- 1$$, $$- 1$$, $$- 5$$, $$6$$.

See the step by step solution

## Step 1: Find the characteristic polynomial of $$U$$

Assume $$G = \left( {\begin{array}{*{20}{c}}U&X\\0&V\end{array}} \right)$$.

Then we have,

\begin{aligned}{c}A &= \left( {\begin{aligned}{*{20}{c}}1&5&{ - 6}&{ - 7}\\2&4&5&2\\0&0&{ - 7}&{ - 4}\\0&0&3&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}U&X\\0&V\end{aligned}} \right)\end{aligned}

On comparison we get,

U = \left( {\begin{aligned}{*{20}{c}}1&5\\2&4\end{aligned}} \right), V = \left( {\begin{aligned}{*{20}{c}}{ - 7}&{ - 4}\\3&1\end{aligned}} \right)

Now characteristic polynomial of $$U$$ are as shown below:

\begin{aligned}{c}\det \left( {\begin{aligned}{*{20}{c}}{1 - \lambda }&5\\2&{4 - \lambda }\end{aligned}} \right) &= \left( {1 - \lambda } \right)\left( {4 - \lambda } \right) - 10\\ &= {\lambda ^2} - 5\lambda - 6\\ &= \left( {\lambda + 1} \right)\left( {\lambda - 6} \right)\end{aligned}

## Step 2: Find the characteristic polynomial of $$V$$

\begin{aligned}{c}\det \left( {\begin{aligned}{*{20}{c}}{ - 7 - \lambda }&{ - 4}\\3&{1 - \lambda }\end{aligned}} \right) &= \left( { - 7 - \lambda } \right)\left( {1 - \lambda } \right) + 12\\ &= {\lambda ^2} + 6\lambda + 5\\ &= \left( {\lambda + 1} \right)\left( {\lambda + 5} \right)\end{aligned}

On multiplication we get the characteristic polynomial of $$A$$.

\begin{aligned}{c}A &= \left( {\lambda + 1} \right)\left( {\lambda - 6} \right)\left( {\lambda + 1} \right)\left( {\lambda + 5} \right)\\ &= {\left( {\lambda + 1} \right)^2}\left( {\lambda + 5} \right)\left( {\lambda - 6} \right)\end{aligned}

Thus, the eigenvalues of $$A$$ are $$- 1$$, $$- 1$$, $$- 5$$, $$6$$.