Problem set 1

1) What is the rank of Matrix A?

A = \(\begin{bmatrix}1 & 2 & 3 & 4\\-1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1\\ 5 & 4 & -2 &-3 \end{bmatrix}\)

The row echelon form is

##      [,1] [,2] [,3]   [,4]
## [1,]    1    2    3  4.000
## [2,]    0    2    4  7.000
## [3,]    0    0   -4 -2.500
## [4,]    0    0    0  1.125

The reduced row echelon form is \(\begin{bmatrix}1&2&3&4\\0&1&2&3.5\\0&0&1&6.25\\0&0&0&1\end{bmatrix}\) There are 4 pivot points, therefore the rank of matrix A is 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 rank is determined by the number of pivot columns or nonzero rows in reduced row echelon form, the maximum rank would be restricted to the smallest number of rows or columns.

3) What is the rank of matrix B?

B = \(\begin{bmatrix}1 &2 & 1 \\ 3 & 6 & 3 \\ 2 &4 & 2\end{bmatrix}\)

\(R_3 = R_3 -2R_1 \rightarrow\) \(\begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 0 & 0 & 0\end{bmatrix}\)

\(R_2 = R_2 - 3R_1 \rightarrow\) \(\begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\)

The rank of Matrix B is 1.

Problem Set 2

Compute the eigenvectors and eigenvalues of matrix A

A = \(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6\end{bmatrix}\)

characteristic polynomial = det \(\begin{bmatrix} 1 - \lambda & 2 & 3 \\ 0 & 4 - \lambda & 5 \\ 0 & 0 & 6 - \lambda\end{bmatrix}\)

=(1 - \(\lambda\))[(4 - \(\lambda\))(6-\(\lambda\))] + 2[(0)(6-\(\lambda\)) -(5)(0)] - 3[(0)(0)-(4-\(\lambda\))(0)]

=(1-\(\lambda\))[24 - 10\(\lambda\) + \(\lambda^2\)] + 2[0-0] -3[0-0]

=(1-\(\lambda\))[24 - 10\(\lambda\) +\(\lambda^2\)]

=24 - 10\(\lambda\) +\(\lambda^2\) -24\(\lambda\) + 10\(\lambda^2\) - \(\lambda^3\)

=-\(\lambda^3\) + 11\(\lambda^2\) - 34\(\lambda\) + 24

Using synthetic division and 4 as a possible root, we have: (\(\lambda\)-4)(\(\lambda\)-6)(\(\lambda\)-1)

our Eigenvalues are \(\lambda_1\) = 6, \(\lambda_2\) = 1 and \(\lambda_3\) = 4

Eigenvector for \(\lambda_1\) = 6

A\(v_1\) = \(\lambda_1 v_1\)

\(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6\end{bmatrix}\) \(v_1\) = 6\(v_1\)

\(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6\end{bmatrix}\) \(\begin{bmatrix}x \\ y \\ z\end{bmatrix}\) = 6\(\begin{bmatrix}x \\ y \\ z\end{bmatrix}\)

x + 2y + 3z = 6x

4y + 5z = 6y

6z = 6z

2y + 3z = 5x

5z = 2y

z = z

Let z = 1 to create the simplest eigenvector that satisfies the eigenvalue, then y = 5/2 and x = 8/5

The eigenvector for \(\lambda_1\) = 6 is \(\begin{bmatrix}8/5 \\ 5/2 \\ 1\end{bmatrix}\)

Eigenvector for \(\lambda_2\) = 1

A\(v_2\) = \(\lambda_2 v_2\)

\(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6\end{bmatrix}\) \(v_2\) = 1\(v_2\)

\(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6\end{bmatrix}\) \(\begin{bmatrix}x \\ y \\ z\end{bmatrix}\) = 6\(\begin{bmatrix}x \\ y \\ z\end{bmatrix}\)

x + 2y + 3z = 1x

4y + 5z = 1y

6z = 1z

The eigenvector for \(\lambda_2\) = 1 is \(\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}\)

Eigenvector for \(\lambda_3\) = 4

A\(v_3\) = \(\lambda_3 v_3\)

\(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6\end{bmatrix}\) \(v_3\) = 4\(v_3\)

\(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6\end{bmatrix}\) \(\begin{bmatrix}x \\ y \\ z\end{bmatrix}\) = 4\(\begin{bmatrix}x \\ y \\ z\end{bmatrix}\)

x + 2y + 3z = 4x

4y + 5z = 4y

6z = 4z

The eigenvector for \(\lambda_3\) = 4 is \(\begin{bmatrix}2/3\\ 1 \\ 0\end{bmatrix}\)