Difference between revisions of "031 Review Part 1, Problem 8"

From Grad Wiki
Jump to navigation Jump to search
(Created page with "<span class="exam">True or false: If all the entries of a  <math style="vertical-align: 0px">7\times 7</math>  matrix  <math style="vertical-align: 0px">A</math...")
 
 
(2 intermediate revisions by the same user not shown)
Line 1: Line 1:
−
<span class="exam">True or false: If all the entries of a &nbsp;<math style="vertical-align: 0px">7\times 7</math>&nbsp; matrix &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; are &nbsp;<math style="vertical-align: -4px">7,</math>&nbsp; then &nbsp;<math style="vertical-align: 0px">\text{det }A</math>&nbsp; must be &nbsp;<math style="vertical-align: 0px">7^7.</math>
+
<span class="exam">True or false: Let &nbsp;<math style="vertical-align: 0px">W</math>&nbsp; be a subspace of &nbsp;<math style="vertical-align: 0px">\mathbb{R}^4</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">\vec{v}</math>&nbsp; be a vector in &nbsp;<math style="vertical-align: 0px">\mathbb{R}^4.</math>&nbsp; If &nbsp;<math style="vertical-align: 0px">\vec{v}\in W</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">\vec{v}\in W^\perp,</math>&nbsp; then &nbsp;<math style="vertical-align: 0px">\vec{v}=\vec{0}.</math>
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Solution: &nbsp;  
 
!Solution: &nbsp;  
 
|-
 
|-
−
|First, we switch to the limit to <math style="vertical-align: 0px">x</math> so that we can use L'Hopital's rule.
+
|Since &nbsp;<math style="vertical-align: -4px">\vec{v}\in W^\perp,</math>&nbsp; we know &nbsp;<math style="vertical-align: 0px">\vec{v}</math>&nbsp; is orthogonal to every vector in &nbsp;<math style="vertical-align: 0px">W.</math>
 
|-
 
|-
−
|So, we have
+
|In particular, since &nbsp;<math style="vertical-align: -4px">\vec{v}\in W,</math>&nbsp; we have that &nbsp;<math style="vertical-align: 0px">\vec{v}</math>&nbsp; is orthogonal to &nbsp;<math style="vertical-align: 0px">\vec{v}.</math>
 +
|-
 +
|Hence,
 
|-
 
|-
 
|
 
|
−
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
+
::<math>\vec{v}\cdot \vec{v}=0.</math>
−
\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}}\\
+
|-
−
&&\\
+
|But, this tells us that &nbsp;<math style="vertical-align: 0px">\vec{v}=\vec{0}.</math>
−
& \overset{L'H}{=} & \displaystyle{\frac{-4}{10}}\\
+
|-
−
&&\\
+
|Therefore, the statement is true.
−
& = & \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 12:21, 15 October 2017

True or false: Let    be a subspace of    and    be a vector in    If    and    then  

Solution:  
Since    we know    is orthogonal to every vector in  
In particular, since    we have that    is orthogonal to  
Hence,
But, this tells us that  
Therefore, 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_8&oldid=3657"