Difference between revisions of "031 Review Part 1, Problem 3"

From Grad Wiki
Jump to navigation Jump to search
 
(One intermediate revision by the same user not shown)
Line 4: Line 4:
 
!Solution:    
 
!Solution:    
 
|-
 
|-
−
|First, we switch to the limit to <math style="vertical-align: 0px">x</math> so that we can use L'Hopital's rule.
+
|The eigenvalues of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; are &nbsp;<math style="vertical-align: -4px"> 0, 1, -1, -e.</math>
 
|-
 
|-
−
|So, we have
+
|Hence, the eigenvalues of &nbsp;<math style="vertical-align: 0px">A</math> are distinct.
 
|-
 
|-
−
|
+
|Therefore, &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is diagonalizable and the statement is true.
−
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
−
\displaystyle{\lim_{x \rightarrow \infty}\frac{3-2x^2}{5x^2 + x +1}} & \overset{L'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{-4x}{10x+1}}\\
 
−
&&\\
 
−
& \overset{L'H}{=} & \displaystyle{\frac{-4}{10}}\\
 
−
&&\\
 
−
& = & \displaystyle{-\frac{2}{5}}.
 
−
\end{array}</math>
 
 
|}
 
|}
 +
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
−
|&nbsp;&nbsp; &nbsp; &nbsp; False
+
|&nbsp;&nbsp; &nbsp; &nbsp; TRUE
 
|}
 
|}
−
[[031_Review_Part_1|'''<u>Return to Sample Exam</u>''']]
+
[[031_Review_Part_1|'''<u>Return to Review Problems</u>''']]

Latest revision as of 12:02, 15 October 2017

True or false: If    is a    matrix with characteristic equation    then    is diagonalizable.

Solution:  
The eigenvalues of    are  
Hence, the eigenvalues of   are distinct.
Therefore,    is diagonalizable and the statement is true.


Final Answer:  
       TRUE

Return to Review Problems

Retrieved from "https://gradwiki.math.ucr.edu/index.php?title=031_Review_Part_1,_Problem_3&oldid=3651"