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22M:100:001 (MATH:3600:0001) Quiz 2 \hb
April 3, 2013

1.) Determine which of the following sets of vectors are linearly dependent versus linearly
independent.  Circle the correct answer


[10]~ 1i.)  $\{\left(\matrix{1 \cr 2}\right), \left(\matrix{-1 \cr 3}\right), \left(\matrix{4 \cr 5}\right) \}$

\hfil a.) linearly dependent \hfil \hfil \hfil

The span of these three vectors is 2-dimensional.  Any collection of more than two vectors 
spanning a 2-dimensional space must be linearly dependent



\v

[10]~ 1ii.)  $\{\left(\matrix{1 \cr 2 \cr 3}\right), \left(\matrix{-1 \cr -2 \cr -3}\right) \}$

\hfil a.) linearly dependent \hfil \hfil  \hfil
\v

The second vector is a multiple of the first vector.


[10]~ 1iii.)  $\{\left(\matrix{1 \cr 2 \cr 4}\right), \left(\matrix{-1 \cr 3 \cr 4}\right) \}$

\hfil \hfil \hfil b.)  linearly independent \hfil

There are only two vectors and the second vector is NOT a multiple of the first vector.


\v

[10]~ 1iv.)  $\{\left(\matrix{1 \cr 2 \cr 4}\right), \left(\matrix{1 \cr 3 \cr 2 }\right), 
\left(\matrix{2 \cr 5 \cr 6 }\right)\}$

\hfil a.) linearly dependent \hfil \hfil \hfil


$\left(\matrix{1 \cr 2 \cr 4}\right)+ \left(\matrix{1 \cr 3 \cr 2 }\right),
= \left(\matrix{2 \cr 5 \cr 6 }\right)$
or $\left(\matrix{1 & 1 & 2 \cr 2 & 3 & 5\cr 4 & 2 & 6}\right) \sim
 \left(\matrix{1 & 1 & 2 \cr 0 & 1 & 1\cr 0 & -2 & 2}\right)$

\v

[10]~ 1v.)  $\{\left(\matrix{1 \cr 2 \cr 0}\right), \left(\matrix{1 \cr 3 \cr 0 }\right), \left(\matrix{2 \cr 3 \cr 6 }\right)\}$

\hfil  \hfil \hfil b.)  linearly independent \hfil

Note the first two vectors span a 2-dimensional space that does not contain the third vector.  
Hence these 3 vectors are linearly independent.

Alternatively 
$\left(\matrix{1& 1 & 2 \cr 2 & 3 & 3 \cr 0 & 0 & 6}\right) \sim$
$\left(\matrix{1& 1 & 2 \cr 0 & 1 & -1 \cr 0 & 0 & 6}\right) $


\vfill
\eject
 2.)  If
$A\left(\matrix{1 \cr 2}\right) = \left(\matrix{5 \cr 5}\right)$,
$A\left(\matrix{2 \cr 1}\right) = \left(\matrix{4 \cr 4}\right)$,
$A\left(\matrix{2 \cr 2}\right) = \left(\matrix{6 \cr 6}\right)$,
$A\left(\matrix{2 \cr 3}\right) = \left(\matrix{8 \cr 8}\right)$,
$A\left(\matrix{3 \cr 2}\right) = \left(\matrix{7 \cr 7}\right)$.

\w

[10]~ 2a.)  An eigenvalue of $A$ is $\underline{~~ 3~~}$ 

[15]~ 2b.) 4 eigenvectors corresponding to this eigenvalue are $\underline{\left(\matrix{2 \cr 
2}\right), \left(\matrix{1 \cr 1}\right), \left(\matrix{-1 \cr -1}\right), \left(\matrix{\pi 
\cr 
\pi}\right)}$ 

Any non-zero scalar multiple of $\left(\matrix{2 \cr 2}\right)$ is an eigenvector of $A$ with 
eigenvalue 3.


FYI:  
$A\left(\matrix{1 \cr 0}\right) =$ $A\left[\left(\matrix{2 \cr 2}\right) - \left(\matrix{1 
\cr 
2}\right)\right] =$  $A\left\left(\matrix{2 \cr 2}\right) - A\left(\matrix{1  
\cr
2}\right) =  \left(\matrix{6 \cr 6}\right)- \left(\matrix{5 \cr 5}\right) = 
\left(\matrix{1 \cr 1}\right)$


$A\left(\matrix{0 \cr 1}\right) =$ $A\left[\left(\matrix{2 \cr 2}\right) - \left(\matrix{2 
\cr 
1}\right)\right] =$  $A\left\left(\matrix{2 \cr 2}\right) - A\left(\matrix{2  
\cr 1}\right) =  \left(\matrix{6 \cr 6}\right)- \left(\matrix{4 \cr 4}\right) = 
\left(\matrix{2 
\cr 2}\right)$

Thus $A = \left(\matrix{1 & 2 \cr 1 & 2 }\right)$.  Since the columns (and similarly the rows) 
are not linearly independent, 0 is also an eigenvalue of $A$.  

Note  $A = \left(\matrix{1 & 2 \cr 1 & 2 }\right) \left(\matrix{2 \cr -1}\right) = 
\left(\matrix{0 \cr 0}\right) = 0 \left(\matrix{2 \cr -1}\right)$

Thus an alternate answer is


[10]~ 2a.)  An eigenvalue of $A$ is $\underline{~~ 0~~}$

[15]~ 2b.) 4 eigenvectors corresponding to this eigenvalue are $\underline{\left(\matrix{2 \cr
-1}\right), \left(\matrix{-2 \cr 1}\right), \left(\matrix{-4 \cr 2}\right), \left(\matrix{2\pi
\cr
-\pi}\right)}$

Any non-zero scalar multiple of $\left(\matrix{2 \cr -1}\right)$ is an eigenvector of $A$ with 
eigenvalue 0.




[15]~ 3a.)  Find the eigenvalues  of
 $A = \left(\matrix{1 & ~1\cr 2 & -1}\right)$ \hb
[10]~ 3b.)  Find one eigenvector corresponding to each eigenvalue.

 $|A - rI| = \left|\matrix{1 - r & ~1\cr 2 & -1 - r}\right|$ 
$= (1-r)(-1-r) - 2 = r^2 - 3 = 0$.  Thus $r = \pm \sqrt{3}$

 $A - rI$
$ = \left(\matrix{1 - ( \pm \sqrt{3}) & ~1\cr 2 & -1 -  (\pm \sqrt{3})}\right)$ 
$ = \left(\matrix{1  \mp \sqrt{3} & ~1\cr 2 & -1 \mp \sqrt{3}}\right)$ 

Note $ \left(\matrix{1  \mp \sqrt{3} & ~1\cr 2 & -1 \mp \sqrt{3}}\right)$$\left(\matrix{ 1\cr 
-1 \pm \sqrt{3}}\right)$ = $\left(\matrix{ 0\cr 0}\right)$

Thus a nonzero solution to $(A - rI){\bf x} = {\bf 0}$ is  
$\left(\matrix{ 1\cr -1 \pm \sqrt{3}}\right)$ 


An e. value of $A$ is $\underline{~~\sqrt{3}~~}$ \& 
an e. vector corresponding to this e. value is $\underline{~~\left(\matrix{ 1\cr -1 + 
\sqrt{3}}\right)~~}$ 

An e. value of $A$ is $\underline{~~-\sqrt{3}~~}$ \&
an e. vector corresponding to this e. value is $\underline{~~\left(\matrix{ 1\cr -1 - 
\sqrt{3}}\right)~~}$ 





\end
