Problem Set 1 1. What is the rank of the matrix \(A\)? \[\begin{equation} A= \begin{pmatrix} 1 & 2 & 3 & 4\\ -1 & 0 & 1 & 3\\ 0 & 1 & -2 & 1\\ 5 & 4 & -2 & -3 \end{pmatrix} \end{equation}\]
The rank of a matrix is the dimension of the span of the column vectors. Let \(\vec v_1, \vec v_2, \vec v_3, \vec v_4\) denote the column vectors of \(A\), i.e. \(A=\left(\vec v_1\,\vec v_2\,\vec v_3\,\vec v_4\right)\). To show that the vectors \(\vec v_1, \vec v_2, \vec v_3, \vec v_4\) is linearly independent, we must show that the only solution to the the equation \(a\vec v_1+b\vec v_2+c\vec v_3+d\vec v_4=\vec 0\) is \(a=b=c=d=0\).
Now, given a vector \(\vec x=\left(a,b,c,d\right)^\top\in R^4\), \[A\vec x=a\vec v_1+b\vec v_2+c\vec v_3+d\vec v_4.\] This means that \(range A= span {\vec v_1, \vec v_2, \vec v_3, \vec v_4}\).
The matrix \(A\), using Gaussian elimination, reduces to the \(4\times 4\) identity matrix. This means that the matrix is invertible. Alternatively, it is easy to check that \(\det A=-9\), so \(A\) is invertible. This means that \(A\) is injective, so the equation \(A\vec x=\vec 0\) has only the trivial solution, \(\vec x=\vec 0\).
An \(m\times n\) matrix represents a linear transformation \(L_A:R^n\to R^m\). Since \(rank A\) is the dimension of \(range {L_A}\). In general, the maximum dimension of the range of any such transformation would be \(\texttt{min}\left\{m,n\right\}\). In this case, \(\texttt{min}\left\{m,n\right\}=n\).
The minimum dimension of the range for a non-trivial linear transformation is 1 since there is at least one non-zero column vector.
What is the rank of the matrix \(B\)? \[\begin{equation} B= \begin{pmatrix} 1 & 2 & 1\\ 3 & 6 & 3\\ 2 & 4 & 2 \end{pmatrix} \end{equation}\]
Let \(\vec w_1, \vec w_2, \vec w_3\) denote the column vectors of \(B\). It is easy see that \(2\vec w_1=\vec w_2\), thus \(span{\vec w_1, \vec w_2, \vec w_3}=span {\vec w_1, \vec w_3}\). But \(\vec w_1=\vec w_3\), and so \(span{\vec w_1, \vec w_3}=span{\vec w_1}\), whose dimension is clearly 1.
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.
\[\begin{equation} A= \begin{pmatrix} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6 \end{pmatrix} \end{equation}\]
\[\begin{equation} A-\lambda I = \begin{pmatrix} 1-\lambda & 2 & 3\\ 0 & 4-\lambda & 5\\ 0 & 0 & 6-\lambda \end{pmatrix} \end{equation}\]
The \(\det{A-\lambda I}=0\) yields the polynomial \((1-\lambda)(4-\lambda)(6-\lambda)=0\). Thus the eigenvalues are 1, 4, and 6.
To find the eigenvector associated with the eigenvalue \(\lambda=1\), we solve \(A\vec x=\vec x\). This gives the following system of equations: \[\begin{align} x_1+2x_2+3x_3 &= x_1\\ 4x_2+5x_3 &= x_2\\ 6x_3 &= x_3 \end{align}\]
The second and third equations force \(x_2=x_3=0\). Then the first equation reduces to \(x_1=x_1\). This leaves the solution \(\vec x=k(1,0,0)^\top\), where \(k\in R\).
Solving for the eigenvector for \(\lambda=4\), we solve \(A\vec x=4\vec x\).
\[\begin{align} x_1+2x_2+3x_3 &= 4x_1\\ 4x_2+5x_3 &= 4x_2\\ 6x_3 &= 4x_3 \end{align}\]
The last equation yields \(x_3=0\). The remaining two equations reduce to \(x_2=x_2\) (a redundant equation) and \(x_1+2x_2=4x_1\), yielding the solutions \(\vec x=k(2,3,0)^\top\) for any \(k\in R\).
Finally, solving for \(\lambda=6\), we solve \(A\vec x=6\vec x\).
\[\begin{align} x_1+2x_2+3x_3 &= 6x_1\\ 4x_2+5x_3 &= 6x_2\\ 6x_3 &= 6x_3 \end{align}\]
The third equation is redundant, and we get the two remaining equations reuduce to \(2x_2=5x_3\) and \(2x_2+x_3=5x_1\). Substituting the first equation into the second, we get \(5x_1=6x_3\). This yields the solutions \(\vec x=k(12,25,10)^\top\) for any \(k\in R\).