Question 1:
What is the rank of matrix \(A\)?
\[ A = \begin{pmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{pmatrix} \]
Solution:
Upon first inspection, it appears as though none of the rows of the matrix are multiples of each other, meaning that each row is linearly independent and that the rank of the matrix is 4 (the number of rows). To confirm this, we can start by converting the matrix into row-echelon form. This is done via Gaussian elimination in the steps below, which highlight each required row operation:
\[ \begin{align} A &\Rightarrow R_1 + R_2 &= \begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{pmatrix} \\ &\Rightarrow -5R_1 + R_4 &= \begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 0 & -6 & -17 & -23 \end{pmatrix} \\ &\Rightarrow -R_2 + R_3 &= \begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 0 & -6 & -17 & -23 \end{pmatrix} \\ &\Rightarrow 3R_2 + R_4 &= \begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 0 & -4 & -\frac{5}{2} \\ 0 & 0 & -5 & -2 \end{pmatrix} \\ &\Rightarrow -\frac{5}{4} + R_4 &= \begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 0 & -4 & -\frac{5}{2} \\ 0 & 0 & 0 & \frac{9}{8} \end{pmatrix} \end{align} \]
Thus, after the steps outlined above, our new row-reduced matrix is:
\[ A = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 0 & -4 & -\frac{5}{2} \\ 0 & 0 & 0 & \frac{9}{8} \end{pmatrix} \]
In order to determine the rank of this matrix, we can count the
number of non-zero rows (any row that has one or more entry not equal to
zero). For the matrix above, all of the rows fit this definition,
meaning that the rank of the matrix is 4. This is confirmed via R in the
code below using the Matrix package’s
RankMatrix() function:
A <- rbind(c(1, 2, 3, 4), c(-1, 0, 1, 3), c(0, 1, -2, 1), c(5, 4, -2, -3))
rankMatrix(A)[1]## [1] 4Question 2:
Given an \(m\)x\(n\) matrix where \(m > n\), what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?
Solution:
The rank of a matrix is defined as the number of linearly independent columns or rows of a matrix. Since the two definitions are synonymous, it must be the case that the maximum rank of a \(n\)x\(m\) matrix must be \(n\) if \(n<m\) and \(m\) if \(m<n\).
The minimum rank of a matrix - assuming the matrix is non-zero - must be one, as at least one of the row and columns must contain a non-zero element.
Question 3:
What is the rank of matrix \(B\)?
\[ \begin{pmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{pmatrix} \]
Solution:
We can see from initial inspection of the matrix that rows 2 and 3 are multiples of row 1, and are thus linearly dependent. Namely, \(R_2 = 3R_1\) and \(R_3 = 2R_1\). As such, we can conclude that this matrix has only one linearly independent row and that the rank of matrix \(B\) is equal to one. We can confirm this via reduction of the matrix into row echelon form:
\[ \begin{align} B &\Rightarrow -3R_1 + R_2 &= \begin{pmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 2 & 4 & 2 \end{pmatrix} \\ &\Rightarrow -2R_2 + R_3 &= \begin{pmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \end{align} \]
As we see from the above, the reduced matrix only has one non-zero column. Lastly, the cell below confirms our result again in R:
B <- rbind(c(1, 2, 1), c(3, 6, 3), c(2, 4, 2))
rankMatrix(B)[1]## [1] 1Question: Compute the eigenvalues and eigenvectors of the matrix \(A\).
\[ A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \\ \end{pmatrix} \]
Solution:
To find the eigenvalues \(\lambda\) of \(A\) we must find the roots of its characteristic polynomial. The characteristic polynomial \(p_A(x)\) of matrix \(A\) is defined as:
\[ \begin{equation} p_A(x) = \text{det}(A-xI_n) \end{equation} \]
where \(I_n\) is the identity matrix of size \(n\) equal to the size of \(A\). Thus to solve for the eigenvalues, we must solve the following equation:
\[ 0 = \text{det}(A-\lambda I_n) \]
Plugging in, we have:
\[ 0 = \begin{pmatrix} 1 - \lambda & 2 & 3 \\ 0 & 4- \lambda & 5 \\ 0 & 0 & 6 - \lambda \end{pmatrix} \]
Using the definition of the determinant of a 3x3 matrix:
\[ 0 = (1-\lambda)\cdot\text{det} \begin{pmatrix} 4 - \lambda & 5 \\ 0 & 6 - \lambda \end{pmatrix} -2\cdot\text{det} \begin{pmatrix} 0 & 5 \\ 0 & 6 - \lambda \end{pmatrix} +3\cdot\text{det} \begin{pmatrix} 0 & 4-\lambda \\ 0 & 0 \end{pmatrix} \]
Using the definition of the determinant of a 2x2 matrix:
\[ \begin{align} 0 &= (1-\lambda)\cdot((4-\lambda)(6-\lambda) - 0) - 2(0 - 0) + 3(0 - 0) \\ 0 &= (1-\lambda)\cdot(4-\lambda)\cdot(6-\lambda) \\ \lambda &= 1, 4, 6 \end{align} \]
Given that many of the calculations in the above steps evaluate to 0, it is easy to express our characteristic polynomial above in already factored form, namely \(p_A(x) = (1-x)(4-x)(6-x)\). Thus, the eigenvalues of \(A\) are 1, 4, and 6. This is confirmed in the R cell below:
A <- rbind(c(1, 2, 3), c(0, 4, 5), c(0, 0, 6))
eigen(A)$values## [1] 6 4 1To determine the eigenvectors \(\mathbf{x}\) of \(A\), we must solve the following equation for each of our three eigenvalues:
\[ \begin{equation} (A - \lambda I)\mathbf{x} = 0 \end{equation} \]
For \(\lambda = 1\):
Plugging in we have that : \[ A - \lambda I = A - I = \begin{pmatrix} 0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 5 \end{pmatrix} \]
Via Gaussian elimination we have:
\[ \begin{align} A - I &\Rightarrow \frac{1}{2}R_1 &= \begin{pmatrix} 0 & 1 & \frac{3}{2} \\ 0 & 3 & 5 \\ 0 & 0 & 5 \end{pmatrix} \\ &\Rightarrow -2R_1 + R_2 &= \begin{pmatrix} 0 & 1 & \frac{3}{2} \\ 0 & 0 & \frac{1}{2} \\ 0 & 0 & 0 \end{pmatrix} \\ &\Rightarrow 2R_2 &= \begin{pmatrix} 0 & 1 & \frac{3}{2} \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \\ &\Rightarrow -\frac{3}{2}R_2 + R_1 &= \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \end{align} \]
Using this row reduced matrix, the our equation becomes:
\[ \begin{align} (A - \lambda I)\mathbf{x} &= 0 \\ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \end{align} \]
The above provides us with the following three equations: \(x_2 = 0\), \(x_3 = 0\), and \(x_1 = x_1\). Thus, setting \(x_1\) equal to 1, we have that the eigenvector associated with \(\lambda=1\) is \(\begin{pmatrix} 1\\0\\0\end{pmatrix}\). Technically, since \(x_1\) can be any value, any scalar multiple of this eigenvector is also an eigenvector. This set of vectors, referred to as the eigenspace, are of the form \(a\begin{pmatrix} 1\\0\\0\end{pmatrix}\) such that \(a \in \mathbb{R}\). Vector spaces can be more easily denoted as \(\text{Vect}\{\begin{pmatrix} 1\\0\\0\end{pmatrix}\}\)
For \(\lambda = 4\):
\[ A - \lambda I = A - 4I = \begin{pmatrix} -3 & 2 & 3 \\ 0 & 0 & 1 \\ 0 & 0 & 2 \end{pmatrix} \]
Via Gaussian elimination we have:
\[ \begin{align} A - I &\Rightarrow -\frac{1}{3}R_1 &= \begin{pmatrix} 1 & -\frac{2}{3} & -1 \\ 0 & 0 & 5 \\ 0 & 0 & 2 \end{pmatrix} \\ &\Rightarrow \frac{1}{5}R_2 &= \begin{pmatrix} 1 & -\frac{2}{3} & -1 \\ 0 & 0 & 1 \\ 0 & 0 & 2 \end{pmatrix} \\ &\Rightarrow \frac{1}{2}R_3 &= \begin{pmatrix} 1 & -\frac{2}{3} & -1 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{pmatrix} \\ &\Rightarrow \frac{1}{2}R_3 &= \begin{pmatrix} 1 & -\frac{2}{3} & -1 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{pmatrix} \\ &\Rightarrow R_1 + R_2 &= \begin{pmatrix} 1 & -\frac{2}{3} & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{pmatrix} \\ &\Rightarrow -R_2 + R_3 &= \begin{pmatrix} 1 & -\frac{2}{3} & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \end{align} \]
Our equation to solve becomes:
\[ \begin{align} (A - \lambda I)\mathbf{x} &= 0 \\ \begin{pmatrix} 1 & -\frac{2}{3} & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \end{align} \]
This gives the following equations: \[ \begin{align} x_1 - \frac{2}{3}x_2 = 0 \Rightarrow x_1 &= \frac{2}{3}x_2 \\ x_2 & = x_2 \\ x_3 &= 0 \end{align} \]
Thus, when \(x_2 = 1\) our eigenvector associated with \(\lambda = 4\) is \(\begin{pmatrix}\frac{2}{3}\\1\\1\end{pmatrix}\). Written without the fraction, this eigenvector becomes \(\begin{pmatrix}2\\3\\0\end{pmatrix}\). The eigenspace associated with \(\lambda = 6\) is then \(\text{Vect}\{\begin{pmatrix} 16\\25\\10\end{pmatrix}\}\).
Lastly, For \(\lambda = 6\):
\[ A - \lambda I = A - 6I = \begin{pmatrix} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{pmatrix} \]
Via Gaussian elimination we have:
\[ \begin{align} A - I &\Rightarrow -\frac{1}{5}R_1 &= \begin{pmatrix} 1 & -\frac{2}{5} & -\frac{3}{5} \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{pmatrix} \\ &\Rightarrow -\frac{1}{2}R_2 &= \begin{pmatrix} 1 & -\frac{2}{5} & -\frac{3}{5} \\ 0 & 1 & -\frac{5}{2} \\ 0 & 0 & 0 \end{pmatrix} \\ &\Rightarrow \frac{2}{5}R_2 + R_1 &= \begin{pmatrix} 1 & 0 & -\frac{8}{5} \\ 0 & 1 & -\frac{5}{2} \\ 0 & 0 & 0 \end{pmatrix} \\ \end{align} \]
Our equation to solve becomes:
\[ \begin{align} (A - \lambda I)\mathbf{x} &= 0 \\ \begin{pmatrix} 1 & 0 & -\frac{8}{5} \\ 0 & 1 & -\frac{5}{2} \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \end{align} \]
This gives the following equations: \[ \begin{align} x_1 - \frac{8}{5}x_3 = 0 \Rightarrow x_1 &= \frac{8}{5}x_3 \\ x_2 - \frac{5}{2}x_3 = 0 \Rightarrow x_1 &= \frac{8}{5}x_3 \\ x_3 &= x_3 \end{align} \]
Thus, when \(x_3 = 1\) our eigenvector associated with \(\lambda = 6\) is \(\begin{pmatrix}\frac{8}{5}\\\frac{5}{2}\\1\end{pmatrix}\). Written without the fractions, this eigenvector becomes \(\begin{pmatrix}16\\25\\10\end{pmatrix}\). The eigenspace associated with \(\lambda = 6\) is then \(\text{Vect}\{\begin{pmatrix} 16\\25\\10\end{pmatrix}\}\).
Thus, in conclusion, the eigenvalues of \(A\) are 1, 4, and 6, with associated eigenvectors: \(\begin{pmatrix}1\\0\\0\end{pmatrix}\), \(\begin{pmatrix}2\\3\\0\end{pmatrix}\), and \(\begin{pmatrix}16\\25\\10\end{pmatrix}\), respectively.