Lab 7. Eigenvalues, Eigenvectors, and Diagonalization: Independent Study

This lab accompanies Chapter 7: Eigenvalues, Eigenvectors, and Diagonalization.

The goal is to connect the story of eigenvectors with computation:

Eigenvectors: special directions that do not rotate under a linear transformation.
Eigenvalues: scaling factors along eigenvector directions.
Diagonalization: replacing a complicated transformation by a diagonal one after choosing the right basis.
Powers of matrices: using $A=PDP^{-1}$ to compute $A^k$ efficiently.
Applications: Markov chains, dynamical systems, dominant eigenvectors, and failure of diagonalization.

This is an independent-study lab. Each main question includes a worked solution and a similar practice question.

Python practice notebook

You may use the Jupyter notebook version for longer Python practice:

Interactive lab

Study guide and worked questions

Question 1. Check an eigenvector

Let

\[ A=\begin{bmatrix}2&1\\0&3\end{bmatrix}, \qquad v=\begin{bmatrix}1\\0\end{bmatrix}. \]

Determine whether $v$ is an eigenvector of $A$.

Solution

Compute

\[ Av=\begin{bmatrix}2&1\\0&3\end{bmatrix} \begin{bmatrix}1\\0\end{bmatrix} =\begin{bmatrix}2\\0\end{bmatrix}=2v. \]

Therefore $v$ is an eigenvector with eigenvalue $\lambda=2$.

Similar practice

For the same matrix $A$, decide whether

\[ w=\begin{bmatrix}1\\1\end{bmatrix} \]

is an eigenvector.

Answer

\[ Aw=\begin{bmatrix}3\\3\end{bmatrix}=3w. \]

So $w$ is an eigenvector with eigenvalue $3$.

Question 2. Compute eigenvalues and eigenspaces

Let

\[ A=\begin{bmatrix}4&1\\2&3\end{bmatrix}. \]

Find its eigenvalues and eigenspaces.

Solution

The characteristic polynomial is

\[ \det(A-\lambda I) =\det\begin{bmatrix}4-\lambda&1\\2&3-\lambda\end{bmatrix} =(4-\lambda)(3-\lambda)-2. \]

Thus

\[ \lambda^2-7\lambda+10=(\lambda-5)(\lambda-2), \]

so the eigenvalues are $5$ and $2$.

For $\lambda=5$,

\[ A-5I=\begin{bmatrix}-1&1\\2&-2\end{bmatrix}, \]

so $-x+y=0$ and $y=x$. Hence

\[ E_5=\operatorname{span}\left\{\begin{bmatrix}1\\1\end{bmatrix}\right\}. \]

For $\lambda=2$,

\[ A-2I=\begin{bmatrix}2&1\\2&1\end{bmatrix}, \]

so $2x+y=0$ and $y=-2x$. Hence

\[ E_2=\operatorname{span}\left\{\begin{bmatrix}1\\-2\end{bmatrix}\right\}. \]

Similar practice

Find the eigenvalues and eigenspaces of

\[ B=\begin{bmatrix}3&1\\0&2\end{bmatrix}. \]

Answer

The eigenvalues are $3$ and $2$. The eigenspaces are

\[ E_3=\operatorname{span}\left\{\begin{bmatrix}1\\0\end{bmatrix}\right\}, \qquad E_2=\operatorname{span}\left\{\begin{bmatrix}-1\\1\end{bmatrix}\right\}. \]

Question 3. Diagonalize a matrix

Let

\[ A=\begin{bmatrix}4&1\\2&3\end{bmatrix}. \]

Using the eigenvectors from Question 2, diagonalize $A$.

Solution

Use

\[ P=\begin{bmatrix}1&1\\1&-2\end{bmatrix}, \qquad D=\begin{bmatrix}5&0\\0&2\end{bmatrix}. \]

The columns of $P$ are eigenvectors corresponding to the diagonal entries of $D$. Therefore

\[ AP=PD, \]

and since the two eigenvectors are linearly independent,

\[ A=PDP^{-1}. \]

Similar practice

Diagonalize

\[ B=\begin{bmatrix}3&1\\0&2\end{bmatrix}. \]

Answer

One choice is

\[ P=\begin{bmatrix}1&-1\\0&1\end{bmatrix}, \qquad D=\begin{bmatrix}3&0\\0&2\end{bmatrix}. \]

Then $B=PDP^{-1}$.

Question 4. Use diagonalization to compute powers

Let $A=PDP^{-1}$, where

\[ P=\begin{bmatrix}1&1\\1&-2\end{bmatrix}, \qquad D=\begin{bmatrix}5&0\\0&2\end{bmatrix}. \]

Compute $A^{10}$ in terms of $P$ and $D$.

Solution

Because

\[ A=PDP^{-1}, \]

we have

\[ A^{10}=PD^{10}P^{-1}. \]

Since $D$ is diagonal,

\[ D^{10}=\begin{bmatrix}5^{10}&0\\0&2^{10}\end{bmatrix}. \]

Therefore

\[ A^{10} =P\begin{bmatrix}5^{10}&0\\0&2^{10}\end{bmatrix}P^{-1}. \]

The main computational benefit is that raising a diagonal matrix to a power only requires raising its diagonal entries to that power.

Similar practice

If $B=PDP^{-1}$ with $D=\operatorname{diag}(3,2)$, write a formula for $B^k$.

Answer

\[ B^k=P\begin{bmatrix}3^k&0\\0&2^k\end{bmatrix}P^{-1}. \]

Question 5. A matrix that is not diagonalizable

Let

\[ J=\begin{bmatrix}2&1\\0&2\end{bmatrix}. \]

Show that $J$ is not diagonalizable.

Solution

The only eigenvalue is $\lambda=2$. Compute

\[ J-2I=\begin{bmatrix}0&1\\0&0\end{bmatrix}. \]

The equation $(J-2I)x=0$ gives $x_2=0$, so

\[ E_2=\operatorname{span}\left\{\begin{bmatrix}1\\0\end{bmatrix}\right\}. \]

This eigenspace has dimension $1$, but a $2\times2$ diagonalizable matrix needs two linearly independent eigenvectors. Therefore $J$ is not diagonalizable.

Similar practice

\[ C=\begin{bmatrix}4&0\\0&4\end{bmatrix} \]

diagonalizable?

Answer

Yes. Although $4$ is a repeated eigenvalue, the eigenspace is all of $\mathbb R^2$. Thus $C$ has two linearly independent eigenvectors and is diagonalizable.

Question 6. Markov chain and long-run behavior

Let

\[ M=\begin{bmatrix}0.8&0.3\\0.2&0.7\end{bmatrix}. \]

This is a column-stochastic matrix. Find the steady-state vector $p$ satisfying

\[ Mp=p,\qquad p_1+p_2=1. \]

Solution

Solve

\[ (M-I)p=0. \]

This gives

\[ \begin{bmatrix}-0.2&0.3\\0.2&-0.3\end{bmatrix} \begin{bmatrix}p_1\\p_2\end{bmatrix}=0. \]

The equation $-0.2p_1+0.3p_2=0$ gives

\[ p_1=1.5p_2. \]

Together with $p_1+p_2=1$, we get $2.5p_2=1$, so $p_2=0.4$ and $p_1=0.6$. Therefore

\[ p=\begin{bmatrix}0.6\\0.4\end{bmatrix}. \]

Similar practice

Find the steady state of

\[ N=\begin{bmatrix}0.9&0.4\\0.1&0.6\end{bmatrix}. \]

Answer

Solve $Np=p$ and $p_1+p_2=1$. The equation is $-0.1p_1+0.4p_2=0$, so $p_1=4p_2$. Hence $5p_2=1$, and

\[ p=\begin{bmatrix}0.8\\0.2\end{bmatrix}. \]

AI companion activities

Use an AI tool as a tutor, not as a replacement for computation. For each prompt, first work by hand or in Python, then ask the AI to check your reasoning.

Explain in plain language why an eigenvector is a direction preserved by a linear transformation.
Given a $2\times2$ matrix with two distinct eigenvalues, ask the AI to help you find eigenvectors step by step.
Ask for a geometric explanation of why diagonalization makes powers of matrices easier.
Ask the AI to compare a diagonalizable matrix with a Jordan block.
Ask for a real-world example where a dominant eigenvector is useful.

Submission checklist

Before moving on, make sure you can do the following without looking at the solutions:

Compute eigenvalues from a characteristic polynomial.
Compute eigenspaces by solving $(A-\lambda I)x=0$.
Decide whether a matrix is diagonalizable.
Build $P$ and $D$ for a diagonalization.
Use $A^k=PD^kP^{-1}$.
Interpret a steady-state vector as an eigenvector with eigenvalue $1$.

# Lab 7. Eigenvalues, Eigenvectors, and Diagonalization: Independent Study {.unnumbered} This lab accompanies **Chapter 7: Eigenvalues, Eigenvectors, and Diagonalization**. The goal is to connect the story of eigenvectors with computation: 1. **Eigenvectors:** special directions that do not rotate under a linear transformation. 2. **Eigenvalues:** scaling factors along eigenvector directions. 3. **Diagonalization:** replacing a complicated transformation by a diagonal one after choosing the right basis. 4. **Powers of matrices:** using $A=PDP^{-1}$ to compute $A^k$ efficiently. 5. **Applications:** Markov chains, dynamical systems, dominant eigenvectors, and failure of diagonalization. This is an **independent-study lab**. Each main question includes a worked solution and a similar practice question. ## Python practice notebook You may use the Jupyter notebook version for longer Python practice: - [Download Lab 7 independent-study notebook](lab-07-diagonalization-eigenvalues-independent-study.ipynb) - [Open Lab 7 in Google Colab](https://colab.research.google.com/github/wanghemath/MATH5110/blob/main/Labs/lab-07-diagonalization-eigenvalues-independent-study.ipynb) ## Interactive lab ```{=html} <iframe src="lab-07-interactive.html" width="100%" height="3600" style="border:1px solid #ddd; border-radius:12px;"> </iframe> ``` ## Study guide and worked questions ### Question 1. Check an eigenvector Let $$ A=\begin{bmatrix}2&1\\0&3\end{bmatrix}, \qquad v=\begin{bmatrix}1\\0\end{bmatrix}. $$ Determine whether $v$ is an eigenvector of $A$. #### Solution Compute $$ Av=\begin{bmatrix}2&1\\0&3\end{bmatrix} \begin{bmatrix}1\\0\end{bmatrix} =\begin{bmatrix}2\\0\end{bmatrix}=2v. $$ Therefore $v$ is an eigenvector with eigenvalue $\lambda=2$. #### Similar practice For the same matrix $A$, decide whether $$ w=\begin{bmatrix}1\\1\end{bmatrix} $$ is an eigenvector. #### Answer $$ Aw=\begin{bmatrix}3\\3\end{bmatrix}=3w. $$ So $w$ is an eigenvector with eigenvalue $3$. ### Question 2. Compute eigenvalues and eigenspaces Let $$ A=\begin{bmatrix}4&1\\2&3\end{bmatrix}. $$ Find its eigenvalues and eigenspaces. #### Solution The characteristic polynomial is $$ \det(A-\lambda I) =\det\begin{bmatrix}4-\lambda&1\\2&3-\lambda\end{bmatrix} =(4-\lambda)(3-\lambda)-2. $$ Thus $$ \lambda^2-7\lambda+10=(\lambda-5)(\lambda-2), $$ so the eigenvalues are $5$ and $2$. For $\lambda=5$, $$ A-5I=\begin{bmatrix}-1&1\\2&-2\end{bmatrix}, $$ so $-x+y=0$ and $y=x$. Hence $$ E_5=\operatorname{span}\left\{\begin{bmatrix}1\\1\end{bmatrix}\right\}. $$ For $\lambda=2$, $$ A-2I=\begin{bmatrix}2&1\\2&1\end{bmatrix}, $$ so $2x+y=0$ and $y=-2x$. Hence $$ E_2=\operatorname{span}\left\{\begin{bmatrix}1\\-2\end{bmatrix}\right\}. $$ #### Similar practice Find the eigenvalues and eigenspaces of $$ B=\begin{bmatrix}3&1\\0&2\end{bmatrix}. $$ #### Answer The eigenvalues are $3$ and $2$. The eigenspaces are $$ E_3=\operatorname{span}\left\{\begin{bmatrix}1\\0\end{bmatrix}\right\}, \qquad E_2=\operatorname{span}\left\{\begin{bmatrix}-1\\1\end{bmatrix}\right\}. $$ ### Question 3. Diagonalize a matrix Let $$ A=\begin{bmatrix}4&1\\2&3\end{bmatrix}. $$ Using the eigenvectors from Question 2, diagonalize $A$. #### Solution Use $$ P=\begin{bmatrix}1&1\\1&-2\end{bmatrix}, \qquad D=\begin{bmatrix}5&0\\0&2\end{bmatrix}. $$ The columns of $P$ are eigenvectors corresponding to the diagonal entries of $D$. Therefore $$ AP=PD, $$ and since the two eigenvectors are linearly independent, $$ A=PDP^{-1}. $$ #### Similar practice Diagonalize $$ B=\begin{bmatrix}3&1\\0&2\end{bmatrix}. $$ #### Answer One choice is $$ P=\begin{bmatrix}1&-1\\0&1\end{bmatrix}, \qquad D=\begin{bmatrix}3&0\\0&2\end{bmatrix}. $$ Then $B=PDP^{-1}$. ### Question 4. Use diagonalization to compute powers Let $A=PDP^{-1}$, where $$ P=\begin{bmatrix}1&1\\1&-2\end{bmatrix}, \qquad D=\begin{bmatrix}5&0\\0&2\end{bmatrix}. $$ Compute $A^{10}$ in terms of $P$ and $D$. #### Solution Because $$ A=PDP^{-1}, $$ we have $$ A^{10}=PD^{10}P^{-1}. $$ Since $D$ is diagonal, $$ D^{10}=\begin{bmatrix}5^{10}&0\\0&2^{10}\end{bmatrix}. $$ Therefore $$ A^{10} =P\begin{bmatrix}5^{10}&0\\0&2^{10}\end{bmatrix}P^{-1}. $$ The main computational benefit is that raising a diagonal matrix to a power only requires raising its diagonal entries to that power. #### Similar practice If $B=PDP^{-1}$ with $D=\operatorname{diag}(3,2)$, write a formula for $B^k$. #### Answer $$ B^k=P\begin{bmatrix}3^k&0\\0&2^k\end{bmatrix}P^{-1}. $$ ### Question 5. A matrix that is not diagonalizable Let $$ J=\begin{bmatrix}2&1\\0&2\end{bmatrix}. $$ Show that $J$ is not diagonalizable. #### Solution The only eigenvalue is $\lambda=2$. Compute $$ J-2I=\begin{bmatrix}0&1\\0&0\end{bmatrix}. $$ The equation $(J-2I)x=0$ gives $x_2=0$, so $$ E_2=\operatorname{span}\left\{\begin{bmatrix}1\\0\end{bmatrix}\right\}. $$ This eigenspace has dimension $1$, but a $2\times2$ diagonalizable matrix needs two linearly independent eigenvectors. Therefore $J$ is not diagonalizable. #### Similar practice Is $$ C=\begin{bmatrix}4&0\\0&4\end{bmatrix} $$ diagonalizable? #### Answer Yes. Although $4$ is a repeated eigenvalue, the eigenspace is all of $\mathbb R^2$. Thus $C$ has two linearly independent eigenvectors and is diagonalizable. ### Question 6. Markov chain and long-run behavior Let $$ M=\begin{bmatrix}0.8&0.3\\0.2&0.7\end{bmatrix}. $$ This is a column-stochastic matrix. Find the steady-state vector $p$ satisfying $$ Mp=p,\qquad p_1+p_2=1. $$ #### Solution Solve $$ (M-I)p=0. $$ This gives $$ \begin{bmatrix}-0.2&0.3\\0.2&-0.3\end{bmatrix} \begin{bmatrix}p_1\\p_2\end{bmatrix}=0. $$ The equation $-0.2p_1+0.3p_2=0$ gives $$ p_1=1.5p_2. $$ Together with $p_1+p_2=1$, we get $2.5p_2=1$, so $p_2=0.4$ and $p_1=0.6$. Therefore $$ p=\begin{bmatrix}0.6\\0.4\end{bmatrix}. $$ #### Similar practice Find the steady state of $$ N=\begin{bmatrix}0.9&0.4\\0.1&0.6\end{bmatrix}. $$ #### Answer Solve $Np=p$ and $p_1+p_2=1$. The equation is $-0.1p_1+0.4p_2=0$, so $p_1=4p_2$. Hence $5p_2=1$, and $$ p=\begin{bmatrix}0.8\\0.2\end{bmatrix}. $$ ## AI companion activities Use an AI tool as a tutor, not as a replacement for computation. For each prompt, first work by hand or in Python, then ask the AI to check your reasoning. 1. Explain in plain language why an eigenvector is a direction preserved by a linear transformation. 2. Given a $2\times2$ matrix with two distinct eigenvalues, ask the AI to help you find eigenvectors step by step. 3. Ask for a geometric explanation of why diagonalization makes powers of matrices easier. 4. Ask the AI to compare a diagonalizable matrix with a Jordan block. 5. Ask for a real-world example where a dominant eigenvector is useful. ## Submission checklist Before moving on, make sure you can do the following without looking at the solutions: - Compute eigenvalues from a characteristic polynomial. - Compute eigenspaces by solving $(A-\lambda I)x=0$. - Decide whether a matrix is diagonalizable. - Build $P$ and $D$ for a diagonalization. - Use $A^k=PD^kP^{-1}$. - Interpret a steady-state vector as an eigenvector with eigenvalue $1$.