Exercise Sheet 4 Solutions
1.
\[\begin{split}
A = \begin{pmatrix} 3 & 2 \\ -2 & -1 \end{pmatrix}
\end{split}\]
Characteristic Polynomial:
We compute the characteristic polynomial:
\[\begin{split}
p(\lambda) = \det(A - \lambda I)
= \det \begin{pmatrix} 3 - \lambda & 2 \\ -2 & -1 - \lambda \end{pmatrix}
\end{split}\]
\[
= (3 - \lambda)(-1 - \lambda) - (-2)(2)
= \lambda^2 - 2\lambda + 1 = (\lambda - 1)^2
\]
Eigenvalues:
Solving the characteristic polynomial:
\[
(\lambda - 1)^2 = 0 \Rightarrow \lambda_1 = \lambda_2 = 1
\]
Eigenvectors:
We solve:
\[\begin{split}
(A - \lambda I)\vec{x} = 0 \Rightarrow
\begin{pmatrix} 2 & 2 \\ -2 & -2 \end{pmatrix} \vec{x} = \vec{0}
\end{split}\]
This reduces to:
\[\begin{split}
x_1 + x_2 = 0 \Rightarrow \vec{x} = \alpha \begin{pmatrix} 1 \\ -1 \end{pmatrix}
\end{split}\]
So, the eigenvector is any scalar multiple of:
\[\begin{split}
\vec{x} = \begin{pmatrix} 1 \\ -1 \end{pmatrix}
\end{split}\]
Now, we solve them for B:
\[\begin{split}
B = \begin{pmatrix}
0 & 1 & 0 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{pmatrix}
\end{split}\]
Characteristic Polynomial:
We compute the characteristic polynomial:
\[\begin{split}
p(\lambda) = \det(B - \lambda I) =
\det \begin{pmatrix}
-\lambda & 1 & 0 \\
1 & -\lambda & 1 \\
1 & 1 & -\lambda
\end{pmatrix}
\end{split}\]
Expanding along the first row:
\[\begin{split}
= -\lambda \cdot \det \begin{pmatrix}
-\lambda & 1 \\
1 & -\lambda
\end{pmatrix}
- 1 \cdot \det \begin{pmatrix}
1 & 1 \\
1 & -\lambda
\end{pmatrix}
+ 0
\end{split}\]
\[
= -\lambda(\lambda^2 - 1) - (-\lambda - 1)
= -\lambda^3 + \lambda + \lambda + 1
= -\lambda^3 + 2\lambda + 1
\]
So the characteristic polynomial is:
\[
p(\lambda) = -\lambda^3 + 2\lambda + 1
\]
Eigenvalues:
Solving the characteristic polynomial:
\[
-\lambda^3 + 2\lambda + 1 = 0
\]
\[
-\left( \lambda + 1 \right)\left( \lambda - \frac{1 + \sqrt{5}}{2} \right)\left( \lambda - \frac{1 - \sqrt{5}}{2} \right) = 0
\]
so:
\[
\lambda_1 \approx -0.62, \quad
\lambda_2 = -1, \quad
\lambda_3 \approx 1.62
\]
Eigenvectors:
To find the eigenvectors, we solve:
\[
(B - \lambda I)\vec{x} = 0
\]
For $\( \lambda_1 \approx -0.62 \)$:
\[\begin{split}
\vec{v}_1 = \begin{pmatrix}
1 \\
-0.62 \\
-0.62
\end{pmatrix}
\end{split}\]
For $\( \lambda_2 = -1 \)$:
\[\begin{split}
\vec{v}_2 = \begin{pmatrix}
-1 \\
1 \\
0
\end{pmatrix}
\end{split}\]
For $\( \lambda_3 \approx 1.62 \)$:
\[\begin{split}
\vec{v}_3 = \begin{pmatrix}
1 \\
1.62 \\
1.62
\end{pmatrix}
\end{split}\]
\[\blacksquare\]
2.
We start by expressing the trace of $\( ABC \)$:
\[
\mathrm{tr}(ABC) = \sum_{i=1}^{n} (ABC)_{ii}
\]
Computing $\( (ABC)_{ii} \)$:
We want to compute the entry in the $\( i \)\(-th row and \)\( i \)\(-th column of the matrix product \)\( ABC \)$.
First, compute the product $\( AB \)$, which gives:
\[
(AB)_{il} = \sum_{j=1}^{m} A_{ij} B_{jl}
\]
Then compute $\( ABC = (AB)C \)\(. The \)\( (i, i) \)\(-th element of \)\( ABC \)$ is:
\[
(ABC)_{ii} = \sum_{l=1}^{k} (AB)_{il} C_{li}
\]
Substitute the expression for $\( (AB)_{il} \)$:
\[
(ABC)_{ii} = \sum_{l=1}^{k} \left( \sum_{j=1}^{m} A_{ij} B_{jl} \right) C_{li}
\]
Rearranging the summation order:
\[
(ABC)_{ii} = \sum_{j=1}^{m} \sum_{l=1}^{k} A_{ij} B_{jl} C_{li}
\]
So, the trace becomes:
\[
\mathrm{tr}(ABC) = \sum_{i=1}^{n} \sum_{j=1}^{m} \sum_{k=1}^{k} A_{ij} B_{jk} C_{ki}
\]
Computing the trace of $\( BCA \)$:
\[
\mathrm{tr}(BCA) = \sum_{j=1}^{m} (BCA)_{jj}
\]
Expand the product:
First compute $\( BC \)$:
\[
(BC)_{ji} = \sum_{k=1}^{k} B_{jk} C_{ki}
\]
Then compute:
\[
(BCA)_{jj} = \sum_{i=1}^{n} (BC)_{ji} A_{ij} = \sum_{i=1}^{n} \sum_{k=1}^{k} B_{jk} C_{ki} A_{ij}
\]
So the trace is:
\[
\mathrm{tr}(BCA) = \sum_{j=1}^{m} \sum_{k=1}^{k} \sum_{i=1}^{n} B_{jk} C_{ki} A_{ij}
\]
Final Step:
Since scalar multiplication is commutative:
\[
\sum_{i=1}^{n} \sum_{j=1}^{m} \sum_{k=1}^{k} A_{ij} B_{jk} C_{ki}
= \sum_{j=1}^{m} \sum_{k=1}^{k} \sum_{i=1}^{n} B_{jk} C_{ki} A_{ij}
\]
Therefore:
\[
\mathrm{tr}(ABC) = \mathrm{tr}(BCA)
\]
\[\blacksquare\]
3.
Eigenvalues are the same
Let $\( p(\lambda) \)\( be the characteristic polynomial of \)\( A \)$. By definition:
\[
p(\lambda) = \det(A - \lambda I)
\]
Now consider the transpose:
\[
\det(A^\top - \lambda I)
= \det((A - \lambda I)^\top)
= \det(A - \lambda I)
= p(\lambda)
\]
So both $\( A \)\( and \)\( A^\top \)$ have the same characteristic polynomial, which means they have the same eigenvalues, including their algebraic multiplicities.
Eigenvectors may differ
Although $\( A \)\( and \)\( A^\top \)$ have the same eigenvalues, their eigenvectors need not be the same.
To see this, consider a concrete example:
Let
\[\begin{split}
A = \begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}
\end{split}\]
Then:
\[\begin{split}
A^\top = \begin{pmatrix}
1 & 0 \\
1 & 1
\end{pmatrix}
\end{split}\]
The characteristic polynomial of both is:
\[
\det(A - \lambda I) = (1 - \lambda)^2
\]
So they both have a repeated eigenvalue $\( \lambda = 1 \)$.
Now compute eigenvectors:
\[\begin{split}
(A - I)\vec{x} = \begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix} \vec{x} = 0
\Rightarrow x_2 = 0 \Rightarrow \vec{x} = \begin{pmatrix}
1 \\
0
\end{pmatrix}
\end{split}\]
\[\begin{split}
(A^\top - I)\vec{x} = \begin{pmatrix}
0 & 0 \\
1 & 0
\end{pmatrix} \vec{x} = 0
\Rightarrow x_1 = 0 \Rightarrow \vec{x} = \begin{pmatrix}
0 \\
1
\end{pmatrix}
\end{split}\]
Thus, the eigenvectors are different, even though the eigenvalues are the same.
So:
\[ \begin{align}\begin{aligned} A $$ and $$ A^\top $$ always have the **same eigenvalues**, including multiplicities.
- However, they may have **different eigenvectors**, especially when the matrix is **not symmetric**.\\$$\blacksquare\end{aligned}\end{align} \]
4.
(a) The rank of $\( B \)$
We observe that each row (and column) of $\( B \)$ is a linear combination of the others:
So all rows lie in the span of the first row.
Let’s do row reduction to confirm:
\[\begin{split}
\begin{pmatrix}
1 & 2 & 3 \\
2 & 4 & 6 \\
3 & 6 & 9
\end{pmatrix}
\rightarrow
\begin{pmatrix}
1 & 2 & 3 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
\end{split}\]
Only one non-zero row remains after Gaussian elimination.
Therefore, the rank of $\( B \)$ is:
\[
\mathrm{rank}(B) = 1
\]
(b) Are the columns of $\( B \)$ linearly independent?
Recall that the number of linearly independent columns is equal to the rank of the matrix.
Since:
\[
\mathrm{rank}(B) = 1 < 3
\]
This means that the columns are linearly dependent.
In fact:
\[\begin{split}
\text{Col}_2 = 2 \cdot \text{Col}_1 \\
\text{Col}_3 = 3 \cdot \text{Col}_1
\end{split}\]
\[\blacksquare\]
