(1) What is the rank of the matrix A?
We will perform Gaussian elimination to find the rank of matrix \(A\). The steps are as follows:
\[ A = \begin{pmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{pmatrix} \]
\[ R3 = R3 - R2 \rightarrow \begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & \frac{7}{2} \\ 0 & 0 & -4 & -\frac{5}{2} \\ 0 & -6 & -17 & -23 \end{pmatrix} \]
\[ R4 = R4 + 6R2 \rightarrow \begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & \frac{7}{2} \\ 0 & 0 & -4 & -\frac{5}{2} \\ 0 & 0 & -5 & -5 \end{pmatrix} \]
\[ R3 = -\frac{1}{4} R3 \rightarrow \begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & \frac{7}{2} \\ 0 & 0 & 1 & \frac{5}{8} \\ 0 & 0 & -5 & -5 \end{pmatrix} \]
\[ R4 = R4 + 5R3 \rightarrow \begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & \frac{7}{2} \\ 0 & 0 & 1 & \frac{5}{8} \\ 0 & 0 & 0 & \frac{15}{8} \end{pmatrix} \]
Since the last row is not all zeros, we have a non-zero element in the bottom-right corner of the matrix. This indicates that the matrix has full rank, which is 4 for a 4x4 matrix. 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?
For an \(m \times n\) matrix where \(m > n\), the maximum rank it can have is limited by the number of columns. Hence, the maximum rank is \(n\). This is because the rank of a matrix is defined as the maximum number of linearly independent columns (or rows). Since there are only \(n\) columns, the rank cannot exceed this number.
Conversely, the minimum rank for a non-zero matrix is 1. This is because if the matrix is non-zero, there must be at least one non-zero column (or row). Therefore, there is at least one linearly independent vector, which guarantees the minimum rank of 1.
In summary:
Maximum Rank = \(n\) (the number of columns)
Minimum Rank = 1 (at least one non-zero row or column)
(3) What is the rank of matrix B?
Given the matrix \(B\):
\[ A = \begin{pmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{pmatrix} \]
We will perform Gaussian elimination to find the rank of matrix \(B\). The steps are as follows: 1. Swap the first and third rows to start with a non-zero pivot element in the first position: \[ R1 \leftrightarrow R3 \] \[ \begin{pmatrix} 2 & 4 & 2 \\ 3 & 6 & 3 \\ 1 & 2 & 1 \end{pmatrix} \] 2. Scale the first row by \(\frac{1}{2}\) to get a leading one: \[ R1 = \frac{1}{2} R1 \] \[ \begin{pmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 1 & 2 & 1 \end{pmatrix} \] 3. Subtract 3 times the first row from the second row and subtract the first row from the third row to create zeros below the pivot: \[ R2 = R2 - 3R1 \] \[ R3 = R3 - R1 \] \[ \begin{pmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \] After performing these operations, we find that the second and third rows are all zeros, indicating that they are not contributing to the rank of the matrix. Therefore, the rank of matrix \(B\) is 1, as there is only one non-zero row.
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.
Given the matrix \(A\):
\[ A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{pmatrix} \]
To find the eigenvalues of \(A\), we compute the characteristic polynomial: \[ \det(A - \lambda I) = \det\begin{pmatrix} 1-\lambda & 2 & 3 \\ 0 & 4-\lambda & 5 \\ 0 & 0 & 6-\lambda \end{pmatrix} \] The determinant of a triangular matrix is the product of its diagonal entries, so: \[ \det(A - \lambda I) = (1-\lambda)(4-\lambda)(6-\lambda) \] Setting this equal to zero gives us the characteristic polynomial: \[ (1-\lambda)(4-\lambda)(6-\lambda) = 0 \] The roots of the characteristic polynomial are the eigenvalues of \(A\): \[ \lambda_1 = 1, \quad \lambda_2 = 4, \quad \lambda_3 = 6 \]
For \(\lambda_1 = 1\), we find the eigenvectors as follows:
Solving \((A - \lambda_1 I)v = 0\): \[ (A - I)v = 0 \] \[ \begin{pmatrix} 0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 5 \end{pmatrix} \begin{pmatrix} v_{11} \\ v_{12} \\ v_{13} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \] This system reduces to: \[ 2v_{12} + 3v_{13} = 0 \] \[ 3v_{12} + 5v_{13} = 0 \] Since the first two equations are multiples of each other, we can let \(v_{13} = t\) and solve for \(v_{12}\) in terms of \(t\). We find that \(v_{12} = -\frac{3}{2}t\) and \(v_{11}\) is free. Thus, an eigenvector is: \[ v_1 = \begin{pmatrix} s \\ -\frac{3}{2}t \\ t \end{pmatrix} \]
In the case of the eigenvector \(v_1\) for the eigenvalue \(\lambda_1 = 1\), we have the freedom to choose the value for \(s\). Setting \(s = 0\) is a valid choice and simplifies the eigenvector. Thus, assuming \(s = 0\) and \(v_{13} = 1\) (for simplicity), the eigenvector \(v_1\) becomes: \[ v_1 = \begin{pmatrix} 0 \\ -\frac{3}{2} \\ 1 \end{pmatrix} \] This is the simplest non-trivial eigenvector corresponding to \(\lambda_1\).
For \(\lambda_2 = 4\) , we find the eigenvectors as follows:Solving \((A - \lambda_2 I)v = 0\): \[ (A - 4I)v = 0 \] \[ \begin{pmatrix} -3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & 2 \end{pmatrix} \begin{pmatrix} v_{21} \\ v_{22} \\ v_{23} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \] Here, the last two rows tell us \(5v_{23} = 0\) and \(2v_{23} = 0\), which implies \(v_{23} = 0\). The first equation becomes \(-3v_{21} + 2v_{22} = 0\), so we can choose \(v_{22} = t\) and \(v_{21} = \frac{2}{3}t\). Therefore, an eigenvector is: \[ v_2 = \begin{pmatrix} \frac{2}{3}t \\ t \\ 0 \end{pmatrix} \]
For the eigenvalue \(\lambda_2 = 4\), we can choose the simplest non-trivial eigenvector by setting \(v_{23} = 0\) and solving for \(v_{21}\) and \(v_{22}\). Choosing \(v_{22} = 1\), \(v_{21} = \frac{2}{3}\) gives us: \[ v_2 = \begin{pmatrix} \frac{2}{3} \\ 1 \\ 0 \end{pmatrix} \] This is the simplest non-trivial eigenvector corresponding to \(\lambda_2\).
For \(\lambda_3 = 6\) , we find the eigenvectors as follows:
Solving \((A - \lambda_3 I)v = 0\): \[ (A - 6I)v = 0 \] \[ \begin{pmatrix} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} v_{31} \\ v_{32} \\ v_{33} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \] In this case, the last row is all zeros, meaning \(v_{33}\) is free. The second row tells us \(-2v_{32} + 5v_{33} = 0\), so \(v_{32} = \frac{5}{2}v_{33}\). There are no constraints on \(v_{31}\), so it is free.
For the eigenvalue \(\lambda_3 = 6\), the eigenvector \(v_3\) is found by solving the equation \((A - \lambda_3 I)v_3 = 0\). The last row of matrix \(A - \lambda_3 I\) is zero, so \(v_{33}\) can be any scalar. We can choose \(v_{33} = 1\) for simplicity. The second row of the matrix provides the equation \(-2v_{32} + 5 = 0\), giving us \(v_{32} = \frac{5}{2}\). Since there are no constraints on \(v_{31}\), it can be any scalar. Therefore, we can choose \(v_{31} = 0\) for simplicity. Thus, an eigenvector for \(\lambda_3\) is: \[ v_3 = \begin{pmatrix} 0 \\ \frac{5}{2} \\ 1 \end{pmatrix} \] \end{document}