1) Find the rank of matrix \(A = \left[\begin{array}{cc} 1 & 2 & 3 & 4\\ -1 & 0 & 1 & 3\\0 & 1 & -2 & 1\\5 & 4 & -2 & -3\end{array}\right]\)
See how many pivots there are:
Replace \(r_{2}\) with \(r_{2} + r_{1}\): \(\left[\begin{array}{cc} 1 & 2 & 3 & 4\\ 0 & 2 & 4 & 7\\0 & 1 & -2 & 1\\5 & 4 & -2 & -3\end{array}\right]\)
Replace \(r_{1}\) with \(r_{1} - r_{2}\): \(\left[\begin{array}{cc} 1 & 0 & -1 & -3\\ 0 & 2 & 4 & 7\\0 & 1 & -2 & 1\\5 & 4 & -2 & -3\end{array}\right]\)
Replace \(r_{3}\) with \(2r_{3}\) - \(r_{2}\): \(\left[\begin{array}{cc} 1 & 0 & -1 & -3\\ 0 & 2 & 4 & 7\\0 & 0 & -8 & -5\\5 & 4 & -2 & -3\end{array}\right]\)
Replace \(r_{4}\) with \(r_{4} - 5r_{1} - 2r_{2}\): \(\left[\begin{array}{cc} 1 & 0 & -1 & -3\\ 0 & 2 & 4 & 7\\0 & 0 & -8 & -5\\0 & 0 & -5 & -2\end{array}\right]\)
Replace \(r_{4}\) with \(r_{4} - \frac{5}{8}r_{3}\): \(\left[\begin{array}{cc} 1 & 0 & -1 & -3\\ 0 & 2 & 4 & 7\\0 & 0 & -8 & -5\\0 & 0 & 0 & \frac{9}{8}\end{array}\right]\)
Replace \(r_{2}\) with \(r_{3} + 2r_{2}\): \(\left[\begin{array}{cc} 1 & 0 & -1 & -3\\ 0 & 4 & 0 & 9\\0 & 0 & -8 & -5\\0 & 0 & 0 & \frac{9}{8}\end{array}\right]\)
Replace \(r_{1}\) with \(8r_{1} - r_{3}\): \(\left[\begin{array}{cc} 8 & 0 & 0 & -19\\ 0 & 4 & 0 & 9\\0 & 0 & -8 & -5\\0 & 0 & 0 & \frac{9}{8}\end{array}\right]\)
At this point, if not before, it’s clear that the rank of the matrix is 4.
We could zero out the 4th column of rows 1-3 by subtracting multiples of row 4 from each of them, and we will be left with 4 pivots.
2) Given an \(m \times n\) matrix where \(m > n\), what can be the maximum rank?
The max rank is \(n\), since the highest number of linearly independent columns is the total number of columns, when \(m \ge n\)
What can be the minimum rank, assuming that the matrix is non-zero?
The min rank is 1, as evidenced by arbitrarily choosing the matrix to be a single column such as \(\left[\begin{array}{cc} 1\\ 0\end{array}\right]\) or more generally a set of columns that are all scalar multiples of each other, such as \(\left[\begin{array}{cc} 1 & 2 & 3\\ 2 & 4 & 6\\3 & 6 & 9\\4 & 8 & 12\end{array}\right]\)
3) Find the rank of matrix \(B = \left[\begin{array}{cc} 1 & 2 & 1\\ 3 & 6 & 3\\2 & 4 & 2\end{array}\right]\)
A quick glance shows that the columns are all multiples of each other. Therefore the rank is 1.
Compute the eigenvalues and eigenvectors of the matrix \(A\)
\(A = \left[\begin{array}{cc}1 & 2 & 3\\0 & 4 & 5\\0 & 0 & 6 \end{array}\right]\)
Start by calculating the characteristic polynomial of \(A\)
Solve for all \(\lambda\) that make the determinant of \(\lambda\cdot I_{3} - A\) equal to 0.
That is, the determinant of \(\left(\left[\begin{array}{cc} \lambda & 0 & 0\\0 & \lambda & 0\\ 0 & 0 & \lambda\end{array}\right] - \left[\begin{array}{cc}1 & 2 & 3\\0 & 4 & 5\\0 & 0 & 6 \end{array}\right]\right) = 0\)
or \(\det\left[\begin{array}{cc}\lambda-1 & -2 & -3\\0 & \lambda-4 & -5\\0 & 0 & \lambda-6 \end{array}\right] = 0\)
If we call the matrix \(D\), so that we are calculating \(det(D) = 0\), the determinant is calculated as
\(D_{11}D_{22}D_{33}+D_{12}D_{23}D_{31}+D_{13}D_{21}D_{32}-D_{11}D_{23}D_{32}-D_{12}D_{21}D_{33}-D_{13}D_{22}D_{31}\)
Conveniently for us, the only non-zero term of that equation is the first one, which hands us our characteristic polynomial on a silver platter:
\((\lambda-1)\cdot(\lambda-4)\cdot(\lambda-6) = 0\)
The roots of that polynomial, \(\lambda=1, \lambda=4\) and \(\lambda=6\), are the eigenvalues of \(A\).
Now we calculate the eigenvectors \(\overrightarrow{v}\) for each of those 3 eigenvalues.
\((\lambda_{1}\cdot I_{3} - A)\cdot\overrightarrow{v} = \overrightarrow{0}\), meaning \(\left(\left[\begin{array}{cc} 1 & 0 & 0\\0 & 1 & 0\\ 0 & 0 & 1\end{array}\right] - \left[\begin{array}{cc}1 & 2 & 3\\0 & 4 & 5\\0 & 0 & 6 \end{array}\right]\right)\cdot\left[\begin{array}{cc} v_{1}\\v_{2}\\v_{3}\end{array}\right] = \left[\begin{array}{cc} 0\\0\\0\end{array}\right]\)
which is \(\left[\begin{array}{cc} 0 & -2 & -3\\0 & -3 & -5\\ 0 & 0 & -5\end{array}\right]\cdot\left[\begin{array}{cc} v_{1}\\v_{2}\\v_{3}\end{array}\right] = \left[\begin{array}{cc} 0\\0\\0\end{array}\right]\)
row-reducing to \(\left[\begin{array}{cc} 0 & 0 & 0\\0 & -3 & 0\\ 0 & 0 & -5\end{array}\right]\cdot\left[\begin{array}{cc} v_{1}\\v_{2}\\v_{3}\end{array}\right] = \left[\begin{array}{cc} 0\\0\\0\end{array}\right]\)
This shows us that \(v_{2} = v_{3} = 0\), while \(v_{1}\) can equal anything, so that the one eigenvector corresponding to \(\lambda_{1}=1\) is \(\overrightarrow{v} =\left[\begin{array}{cc} 1\\ 0\\0\end{array}\right]\)
On to \(\lambda_{2}\)….
\((\lambda_{2}\cdot I_{3} - A)\cdot\overrightarrow{v} = \overrightarrow{0}\), meaning \(\left(\left[\begin{array}{cc} 4 & 0 & 0\\0 & 4 & 0\\ 0 & 0 & 4\end{array}\right] - \left[\begin{array}{cc}1 & 2 & 3\\0 & 4 & 5\\0 & 0 & 6 \end{array}\right]\right)\cdot\left[\begin{array}{cc} v_{1}\\v_{2}\\v_{3}\end{array}\right] = \left[\begin{array}{cc} 0\\0\\0\end{array}\right]\)
which is \(\left[\begin{array}{cc} 3 & -2 & -3\\0 & 0 & -5\\ 0 & 0 & -2\end{array}\right]\cdot\left[\begin{array}{cc} v_{1}\\v_{2}\\v_{3}\end{array}\right] = \left[\begin{array}{cc} 0\\0\\0\end{array}\right]\)
Since the 2nd and 3rd rows both tell us that \(v_{3}=0\), we are just left with the 1st row, which tells us that \(3v_{1} - 2v_{2} = 0\)
That means the one eigenvector corresponding to \(\lambda_{2}=4\) is \(\overrightarrow{v} =\left[\begin{array}{cc} \frac{2}{3}\\ 1\\0\end{array}\right]\)
And finally, \(\lambda_{3}\)….
\((\lambda_{3}\cdot I_{3} - A)\cdot\overrightarrow{v} = \overrightarrow{0}\), meaning \(\left(\left[\begin{array}{cc} 6 & 0 & 0\\0 & 6 & 0\\ 0 & 0 & 6\end{array}\right] - \left[\begin{array}{cc}1 & 2 & 3\\0 & 4 & 5\\0 & 0 & 6 \end{array}\right]\right)\cdot\left[\begin{array}{cc} v_{1}\\v_{2}\\v_{3}\end{array}\right] = \left[\begin{array}{cc} 0\\0\\0\end{array}\right]\)
which is \(\left[\begin{array}{cc} 5 & -2 & -3\\0 & 2 & -5\\ 0 & 0 & 0\end{array}\right]\cdot\left[\begin{array}{cc} v_{1}\\v_{2}\\v_{3}\end{array}\right] = \left[\begin{array}{cc} 0\\0\\0\end{array}\right]\)
row-reducing to \(\left[\begin{array}{cc} 5 & 0 & -8\\0 & 2 & -5\\ 0 & 0 & 0\end{array}\right]\cdot\left[\begin{array}{cc} v_{1}\\v_{2}\\v_{3}\end{array}\right] = \left[\begin{array}{cc} 0\\0\\0\end{array}\right]\)
This leaves us with two equations, \(v_{1} = \frac{8}{5}v_{3}\) and \(v_{2} = \frac{5}{2}v_{3}\) so that
the one eigenvector corresponding to \(\lambda_{3}=6\) is \(\overrightarrow{v} =\left[\begin{array}{cc} \frac{8}{5}\\ \frac{5}{2}\\1\end{array}\right]\)