Exercise 08 solutions

📚 Exercises from the book

🔹 7.1.15

Find (a) the characteristic equation and (b) the eigenvalues (and corresponding eigenvectors) of the matrix \(A\):

Solution

First, to obtain characteristic equation, we can write:

which means that the eigenvalues are \(\lambda_1=0\) and \(\lambda_2=7\)

Now, to find eigenvectors, we can simply plugging the found eigenvalues to \((\lambda I - A)x = 0\). Therefore:

For \(\lambda_1=0,\left[\begin{array}{cc}\lambda_1-6 & 3 \\ 2 & \lambda_1-1\end{array}\right]\left[\begin{array}{l}x_1 \\ x_2\end{array}\right]=\left[\begin{array}{l}0 \\ 0\end{array}\right] \Rightarrow\left[\begin{array}{rr}2 & -1 \\ 0 & 0\end{array}\right]\left[\begin{array}{l}x_1 \\ x_2\end{array}\right]=\left[\begin{array}{l}0 \\ 0\end{array}\right]\).

The solution set is \(\{(t, 2 t): t \in R\}\). So, an eigenvector corresponding to \(\lambda_1=0\) is \((1,2)\).

For \(\lambda_2=7,\left[\begin{array}{cc}\lambda_2-6 & 3 \\ 2 & \lambda_2-1\end{array}\right]\left[\begin{array}{l}x_1 \\ x_2\end{array}\right]=\left[\begin{array}{l}0 \\ 0\end{array}\right] \Rightarrow\left[\begin{array}{ll}1 & 3 \\ 0 & 0\end{array}\right]\left[\begin{array}{l}x_1 \\ x_2\end{array}\right]=\left[\begin{array}{l}0 \\ 0\end{array}\right]\).

The solution set is \(\{(-3 t, t): t \in R\}\). So, an eigenvector corresponding to \(\lambda_2=7\) is \((-3,1)\).

🔹 7.2.15

If possible, diagonalize matrix \(A\) from the previous exercise.

Solution

Since \(A\) is \(2 \times 2\) matrix, and in the previous exercise, we have found 2 distinct eigenvalues, we know that the matrix \(A\) is diagonalizable. To diagonalize, we can use the following formulu:

where \(D\) should be a diagonal matrix with eigenvalues on its main diagonal and \(P\) should be a matrix whose columns are formed by eigenvectors of \(A\). Therefore:

And plugging \(P\) and its invers to the formula yields:

⛑ 7.1.57

For an invertible matrix \(A\), prove that \(A\) and \(A^{-1}\) have the same eigenvectors. How are the eigenvalues of \(A\) related to the eigenvalues of \(A^{-1}\)?

Solution

First, we know that:

If we now multiply both sides by \(A^{-1}\):

If we now divide the equation by \(\lambda\), then:

As a result, we can see that the eigenvector \(x\) is the same, while the eigenvalues are different. More specifically, eigenvalues of the \(A^{-1}\) are inverses of the eigenvalues of the original matrix \(A\).

⛑ 7.2.43

Prove that if matrix \(A\) is diagonalizable with \(n\) real eigenvalues \(\lambda_1, \lambda_2, \ldots, \lambda_n\), then \(det(A)=\lambda_1 \lambda_2 \cdots \lambda_n\).

Solution

Since \(A\) diagonalizable, we can write it as \(D = P^{-1}AP\) where \(D\) represents the exact same transformation as \(A\), but its expressed in terms of eigenbasis instead of standard basis. This implies that determinant of these two transformations are indeed identical, i.e., \(det(A) = det(D)\).

We know that \(D\) is a diagonal matrix which has on its main diagonal the eigenvalues of matrix \(A\). Recall from lecture 3 that if we have a diagonal matrix, the determinant can simply be computed as the product of the entries on the main diagonal, therefore:

🔖 Exercises from Rasmus’ notes

🔸 Exercise 1

Consider the sequence given by the equations

i.e., each number is given as the average of the two preceding numbers. Construct a matrix \(A\) such that

Compute the eigenvectors and eigenvalues of \(A\). Compute the limit of \(G_k\) as \(k \rightarrow \infty\).

Solution

First, we want to find \(A\). We first realize that (based on the provided information):

Next, we rewrite this into the desired form:

Therefore, to conclude:

Next, we will compute the eigenvalues and eigenvectors of \(A\). We start by obtaining the characteristic polynomial of \(A\):

The polynomial has roots \(\lambda_1=1\) and \(\lambda_2=-\frac{1}{2}\). If we plugin each respective eigenvalue to \((\lambda I - A)x = 0\), use Gauss-Jordan elimination, we should obtain:

• for \(\lambda_1 = 1\): \(\{(t, t): t \in R\}\). So, an eigenvector corresponding to \(\lambda_1=1\) is for instance \((1,1)\).

• for \(\lambda_2 = -\frac{1}{2}\): \(\{(t, -2t): t \in R\}\). So, an eigenvector corresponding to \(\lambda_2=-\frac{1}{2}\) is for instance \((1,-2)\).

Finally, we can proceed to the last and most important task: determine to which number \(G_k\) converges as \(k \rightarrow \infty\). First of all, we realize that:

As we know from Rasmus’ note, the next step is to write vector \([1, 0]^T\) as linear combination of eigenbasis vectors, i.e., as a linear combination of eigenvectors we just found:

Just a side note, I skipped the computation of the coefficients that scale the basis vectors such that we get the vector \([1, 0]^T\) - again this is just solving system of linear equations using Gauss-Jordan elimination.

Now, we use this new expression to obtain better suited equation for computing \(G_k\):

We realize that \(A^kv_1 = \lambda^kv_1\) and similar for the other eigenvector and eigenvalue. We can use this to then obtain:

Since we are interested just in \(G_k\), we write:

Clearly, if we compute large power of a small number, in our case \(-\frac{1}{2}\), then we get a very small number approaching zero as we increase the power. This implies that:

Therefore, we conclude that:

🔸 Exercise 2

See the problem statement and solution here.

🔍 About the page

• Last updated: 27/10/2022
• Unless othewise stated, exercises come from the book: Elementary Linear Algebra, International Metric Edition, Ron Larson
• Purpose: This page was created as part of a preparation for my teaching assistant session in the course Linear algebra and optimization managed by Rasmus Ejlers Møgelberg. Please note that unless othewise stated, the solutions on this page are my own, thus it should not be considered as official solutions provided as part of the course.