1. Problem set 1

(1) What is the rank of the matrix A?

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

A<-matrix(c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3), byrow =TRUE,nrow = 4,ncol = 4)
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?

Since the pivot columns of A form a basis for ColA, the rank of A is just the number of pivot columns in A. Therefore the maximum rank in the case is rank(A)=n.

Since A is non-zero, there exists a minimum one non-zero column in echelon reduction form, Therefore the minmum rank is 1.

(3) What is the rank of matrix B?

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

B<-matrix(c(1,2,1,3,6,3,2,4,2), byrow =TRUE,nrow = 3,ncol = 3)
qr(B)$rank
## [1] 1

2. Problem set 2

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.

\[ A = \left[\begin{array}{cc} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6\\ \end{array}\right] \]

For eigenvalues:

\(p_A\) = det (A - x\(I_n\))

\[ p_A = det\left|\begin{array}{cc} 1-x & 2 & 3 \\ 0 & 4-x & 5 \\ 0 & 0 & 6-x\\ \end{array}\right| \]

\[ p_A = (1-x )\left|\begin{array}{cc} 4-x & 5 \\ 0 & 6-x \end{array}\right| = (1-x)(4-x)(6-x)= -(x-1)(x-4)(x-6) \]

\(\lambda_1\) = 1 , \(\lambda_2\) = 4, \(\lambda_3\) = 6

For eigenvectors:

\[ \lambda_1 = 1 , A - 1*I_n = \left|\begin{array}{cc} 0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 &-1\end{array}\right| -(RREF)-> \left|\begin{array}{cc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 &0\end{array}\right| \] \[ E_A(1) = N(A - 1*I_3 )= \left<\{\begin{array}{cc} 1\\ 0 \\ 0\end{array}\}\right> \]

\[ \lambda_2 = 4 , A - 4*I_3 = \left|\begin{array}{cc} -3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 &2\end{array}\right| -(RREF)-> \left|\begin{array}{cc} 1 & -2/3 & 0 \\ 0 & 0 & 1 \\ 0 & 0 &0\end{array}\right| \]

\[ E_A(4) = N(A - 4*I_3 )= \left<\{\begin{array}{cc} 2/3\\ 1 \\ 0\end{array}\}\right> \]

\[ \lambda_3 = 6 , A - 6*I_3 = \left|\begin{array}{cc} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 &0\end{array}\right| -(RREF)-> \left|\begin{array}{cc} 1 & 0 & -8/5 \\ 0 & 1 & -5/2 \\ 0 & 0 &0\end{array}\right| \]

\[ E_A(6) = N(A - 4*I_3 )= \left<\{\begin{array}{cc} 8/5\\ 5/2\\ 1\end{array}\}\right> \]

Please show your work using an R-markdown document. Please name your assignment submission with your ???rst initial and last name.