Assignment-3 :: Fundamentals of Computational Mathematics

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1. What is the rank of the matrix A?

$A=\left[\begin{array}{cccc}1& 2& 3& 4\\ -1& 0& 1& 3\\ 0& 1& -2& 1\\ 5& 4& -2& -3\end{array}\right]$

A=matrix(c(1,-1,0,5,2,0,1,4,3,1,-2,-2,4,3,1,-3), nrow = 4)
A

##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]   -1    0    1    3
## [3,]    0    1   -2    1
## [4,]    5    4   -2   -3

print("Rank of matrix A")

## [1] "Rank of matrix A"

qr(A)$rank

## [1] 4

2. Given an mxn matrix where m > n, what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?

Maximum possible rank is n. Minimum possible rank is 1.

3. What is the rank of the matrix B?

$B=\left[\begin{array}{ccc}1& 2& 1\\ 3& 6& 3\\ 2& 4& 2\end{array}\right]$

B=matrix(c(1,3,2,2,6,4,1,3,2), nrow = 3)
B

##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2

print("Rank of matrix B")

## [1] "Rank of matrix B"

qr(B)$rank

## [1] 1

4. Compute the eigenvalues and eigenvectors of the matrix A. You’ll need to show your work. You’ll need to write out the characteristic polynomial and show your solution.

As per https://www.youtube.com/watch?v=IdsV0RaC9jM

$$\begin{gathered} A=\left[\begin{array}{ccc}1& 2& 3\\ 0& 4& 5\\ 0& 0& 6\end{array}\right] \\ \lambda I =\lambda \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}\lambda & 0& 0\\ 0& \lambda & 0\\ 0& 0& \lambda \end{array}\right] \\ A-\lambda I=\left[\begin{array}{ccc}1& 2& 3\\ 0& 4& 5\\ 0& 0& 6\end{array}\right]-\left[\begin{array}{ccc}\lambda & 0& 0\\ 0& \lambda & 0\\ 0& 0& \lambda \end{array}\right]=\left[\begin{array}{ccc}1-\lambda & 2& 3\\ 0& 4-\lambda & 5\\ 0& 0& 6-\lambda \end{array}\right] \\ det \left[\begin{array}{ccc}1-\lambda & 2& 3\\ 0& 4-\lambda & 5\\ 0& 0& 6-\lambda \end{array}\right]=(1-\lambda ) det( \left[\begin{array}{cc}4-\lambda & 5\\ 0& 6-\lambda \end{array}\right] ) \\ -0 det \left(\left[\begin{array}{cc}2& 3\\ 0& 6-\lambda \end{array}\right]\right) + \left(0\right) det \left(\left[\begin{array}{cc}2& 3\\ 4-\lambda & 5\end{array}\right]\right) \\ =(1-\lambda )(4-\lambda )(6-\lambda ) \\ so eigenvalues \lambda =1, \lambda =4, \lambda =6 \\ \mathit{F}\mathit{o}\mathit{r}\mathbf{ }\mathit{\lambda }\mathbf{=}\mathbf{1} \\ A-\lambda I=\left[\begin{array}{ccc}1-\lambda & 2& 3\\ 0& 4-\lambda & 5\\ 0& 0& 6-\lambda \end{array}\right]=\left[\begin{array}{ccc}0& 2& 3\\ 0& 3& 5\\ 0& 0& 5\end{array}\right] \\ First pivot is 0 that means x1 can take any value. Taking value=1 \\ 3x_{2 }+5x_{3 }=0 ( considring 2nd row ) \\ 5x_3=0 ( considring 3rd row ) \\ {x}_{\lambda =1}=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right] \\ \mathit{F}\mathit{o}\mathit{r}\mathbf{ }\mathit{\lambda }\mathbf{=}\mathbf{4} \\ A-\lambda I=\left[\begin{array}{ccc}1-\lambda & 2& 3\\ 0& 4-\lambda & 5\\ 0& 0& 6-\lambda \end{array}\right]=\left[\begin{array}{ccc}-3& 2& 3\\ 0& 0& 5\\ 0& 0& 2\end{array}\right] \\ -3x_1+2x_2+3x_3=0 \\ 2x_3=0 \\ i.e. x_3 =0 \\ considering x_2=1 \\ {x}_1=\frac{2}{3} \\ {x}_{\lambda =4}=\left[\begin{array}{c}\frac{2}{3}\\ 1\\ 0\end{array}\right] \\ \mathit{F}\mathit{o}\mathit{r}\mathbf{ }\mathit{\lambda }\mathbf{=}\mathbf{6} \\ A-\lambda I=\left[\begin{array}{ccc}1-\lambda & 2& 3\\ 0& 4-\lambda & 5\\ 0& 0& 6-\lambda \end{array}\right]=\left[\begin{array}{ccc}-5& 2& 3\\ 0& -2& 5\\ 0& 0& 0\end{array}\right] \\ -5x_1+2x_2+3x_3=0 \\ -2x_2+5x_3=0 \\ Considering x_3=1 \\ {x}_2=\frac{5}{2},x_1=\frac{8}{5} \\ so x_{\lambda =6}=\left[\begin{array}{c}\frac{8}{5}\\ \frac{5}{2}\\ 1\end{array}\right] \end{gathered}$$