Math Insight

Finding eigenvalues and eigenvectors of 2x2 matrices

Math 2241, Spring 2023
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Due date: Feb. 8, 2023, 11:59 p.m.
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Total points: 1
  1. Sometimes, when we multiply a matrix $A$ by a vector, we get the same result as multiplying the vector by a scalar $\lambda$: $$A\vc{x}=\lambda \vc{x}$$ A vector $\vc{x}$ that satisfies this equation for some value of $\lambda$ is called an eigenvector of $A$, and the value of $\lambda$ is called the corresponding eigenvalue. No matter what $A$ is, there is always one vector $\vc{x}$ for which this equation is true, regardless of what $\lambda$ is. What is this vector?
    Because this vector is always a solution, we do not consider it an eigenvector.

    Eigenvalues and eigenvectors tell us a lot about the effect of the matrix. Let's figure out how to find them.

    1. We will restrict our attention to $2\times 2$ matrices. We're looking for solutions to the equation $$A\vc{x}=\lambda \vc{x}.$$ We know how to solve $B\vc{y}=\vc{z}$ if we know $\vc{z}$, so the first step is to convert the equation to that format. Start by subtracting $\lambda \vc{x}$ from both sides, and enter the new equation below.


      (Online, enter $\lambda$ as lambda or with the symbol λ. For the former option, to write $\lambda x$, you need a space or an asterisk, as in lambda x or lambda*x. Just enter the vector $\vc{x}$ as x; don't worry about making it a boldface $\vc{x}$ or adding an arrow like $\vec{x}$.)

      Next, we want to factor out the $\vc{x}$ so that it's multiplied by only one thing. We have to be careful about this, though. What does $A-\lambda$ mean when $A$ is a matrix and $\lambda$ is a scalar?
      . In order for this to make sense, we need to convert $\lambda \vc{x}$ into a matrix times $\vc{x}$. The matrix $$I=\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$$ is called the identity matrix. The identity matrix $I$ acts like $1$, as you can verify by multiplying $I$ times $\vc{x}$:



      $\displaystyle \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix} = $



      Anywhere we have a vector $\vc{x}$, we can replace it with $I\vc{x}$. With this change, we have $$A\vc{x}-\lambda I \vc{x} = \vc{0}$$ What does it mean to multiply a matrix by a scalar?
      That means that


      $\displaystyle \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} = $





      Now we can factor out the $\vc{x}$ to get $(A-\lambda I)\vc{x}=\vc{0}$. Suppose $\displaystyle A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Then


      $\displaystyle A-\lambda I = $





      Need help?

    2. We're looking for solutions to $(A-\lambda I)\vc{x}=\vc{0}$. We observed at the beginning of this question that one specific vector was always a solution to this equation for any choice of $A$ and $\lambda$. What was this vector?

      We want another vector to be a solution. Recall that the determinant of a matrix can tell us how many solutions the system has. The system has exactly one solution if $\det (A-\lambda I) \neq$
      , and either no solutions or infinitely many solutions if $\det (A-\lambda I) =$
      . Since we know the system has at least one solution, the latter condition will guarantee that there is another solution.

      This condition $\det (A-\lambda I) = ＿$ for the eigenvalues is such an important equation in linear algebra that we give it a name, the characteristic equation.

      Our first step is to find the values of $\lambda$ that solve the characteristic equation. Let's work through an example: $$A=\begin{bmatrix} -1 & 1 \\ 4 & -1 \end{bmatrix}$$ Then



      $\displaystyle A-\lambda I = $




      Since we have a $2 \times 2$ matrix, the characteristic equation, $\det (A-\lambda I )= 0$ will be a quadratic equation for $\lambda$. Write the quadratic here:

      $=0$

      We can find the roots of the characteristic equation by either factoring or using the quadratic formula. Write down the roots in increasing order:
      ,
      These roots are the eigenvalues of $A$.

    3. To find the eigenvectors, we need to find solutions to $(A-\lambda I)\vc{x}=\vc{0}$ for each value of $\lambda$. We'll start with the smaller of the two eigenvalues: $\lambda=＿$. Write down what $A-\lambda I$ is for this eigenvalue (online, this will be done for you). $$- ＿ + AI=\left[\begin{matrix}- ＿ - 1 & 1\\4 & - ＿ - 1\end{matrix}\right]$$ We will call this matrix $B$. Observe that the second row is double the first row. For $2 \times 2$ matrices, the rows will always be scalar multiples of each other if you found the eigenvalue correctly. We can find all solutions to the equation $B\vc{x}=\vc{0}$ just by looking at the first row. Let $\vc{x}=\begin{bmatrix}x \\ y \end{bmatrix}$. We must have $2x+y=0$. Any choice of $x$ will allow us to find a value of $y$ that will work. For example, we can plug in $x=1$ and find that $y=$
      . One eigenvector is therefore


      $\displaystyle \vc{x}= $



      We could have chosen a different value for $x$ other than $x=1$. For example, if we had chosen $x=2$, we would simply have gotten twice the value of $y$, or $y=$
      , giving us the eigenvector $\begin{bmatrix} 2\\＿\end{bmatrix}$. This eigenvector is twice the length of the original eigenvector $\begin{bmatrix} ＿ \\ ＿ \end{bmatrix}$, but is considered the same vector.

      In fact, any scalar multiple of $\begin{bmatrix} ＿ \\ ＿ \end{bmatrix}$ (except after multiplying by zero) is considered the eigenvector for $\lambda=＿$ . (This fact corresponds to the fact that we could flip or scale a vector when determining eigenvectors graphically.)

    4. Next, we'll find the eigenvector for the larger of the two eigenvalues: $\lambda=＿$. Write down what $A-\lambda I$ is for this eigenvalue (online, this will be done for you). $$- ＿ + AI=\left[\begin{matrix}- ＿ - 1 & 1\\4 & - ＿ - 1\end{matrix}\right]$$ We will call this matrix $C$. Observe that the second row is again a scalar multiple of the first row. We can find all solutions to the equation $C\vc{x}=\vc{0}$ just by looking at the first row. Let $\vc{x}=\begin{bmatrix}x \\ y \end{bmatrix}$. We must have $-2x+y=0$. Any choice of $x$ will allow us to find a value of $y$ that will work. For example, we can plug in $x=1$ and find that $y=$
      . One eigenvector is therefore


      $\displaystyle \vc{x}= $



      All other eigenvectors for $\lambda=＿$ are scalar multiples of $\begin{bmatrix} ＿ \\ ＿ \end{bmatrix}$.
    5. Plot the eigenvectors you found on this applet. (Any scalar multiple of the eigenvectors you found is fine.)
      Feedback from applet
      eigenvectors:

  2. Let's find the eigenvalues and eigenvectors of another matrix: $$A=\begin{bmatrix} 1 & -4 \\ 2 & -5 \end{bmatrix}$$
    1. The first step is to find the characteristic equation $\det (A-\lambda I)=0$. Calculate $A-\lambda I $.


      $\displaystyle A-\lambda I = $





      The characteristic equation $\det (A-\lambda I )=0$ will be a quadratic equation in $\lambda$. Write the quadratic here:

      =0

      We can find the roots of this characteristic equation by either factoring or using the quadratic formula. Write down the roots in increasing order:
      ,
      These roots are the eigenvalues of $A$.

    2. To find the eigenvectors, we need to find solutions to $(A-\lambda I)\vc{x}=\vc{0}$ for each value of $\lambda$. We'll start with the smaller of the two eigenvalues: $\lambda=＿$. Write down what $A-\lambda I$ is for this eigenvalue (online, this will be done for you). $$- ＿ + AI=\left[\begin{matrix}- ＿ + 1 & -4\\2 & - ＿ - 5\end{matrix}\right]$$ We will call this matrix $B$. Observe that the second row is half the first row, so we know that the eigenvalue we found was correct. We can find all solutions to the equation $B\vc{x}=\vc{0}$ just by looking at the first row. Let $\vc{x}=\begin{bmatrix} x \\ y \end{bmatrix}$. We must have $4x-4y=0$. Any choice of $x$ will allow us to find a value of $y$ that will work. For example, we can plug in $x=1$ and find that $y=$
      . One eigenvector is therefore


      $\displaystyle \vc{x}= $



      All other eigenvectors for $\lambda=＿$ are scalar multiples of $\begin{bmatrix} ＿ \\ ＿ \end{bmatrix}$.
    3. Next, we'll find the eigenvector for the larger of the two eigenvalues: $\lambda=＿$. Write down what $A-\lambda I$ is for this eigenvalue (online, this will be done for you). $$- ＿ + AI=\left[\begin{matrix}- ＿ + 1 & -4\\2 & - ＿ - 5\end{matrix}\right]$$ We will call this matrix $C$. Observe that the second row is again a scalar multiple of the first row. We can find all solutions to the equation $C\vc{x}=\vc{0}$ just by looking at the first row. Let $\vc{x}=\begin{bmatrix}x \\ y \end{bmatrix}$. We must have $2x-4y=0$. We could choose any value of $x$ and find the corresponding value of $y$, or we could choose $y$ and find the corresponding value of $x$. What if we choose $y=1$? We find that $x=$
      . One eigenvector is therefore


      $\displaystyle \vc{x}= $



      All other eigenvectors for $\lambda=＿$ are scalar multiples of $\begin{bmatrix} ＿ \\ ＿ \end{bmatrix}$.

  3. Find the eigenvalues and eigenvectors of the matrix $$A=\begin{bmatrix} 6 & -4 \\ 3 & -1 \end{bmatrix}$$
    1. Step 1. Find the eigenvalues.


      $\displaystyle A-\lambda I = $





      What is characteristic equation?

      =0
      Find the eigenvalues and write them down (order does not matter): $\lambda_1=$
      , $\lambda_2=$

    2. Step 2: Find an eigenvector corresponding to the first eigenvalue you entered, $\lambda_1=＿$



      $\displaystyle A-\lambda_1 I = $





      Write down the equation corresponding to the first row of $(A-\lambda_1 I)\vc{x}=\vc{0}$, where $\vc{x}=\begin{bmatrix} x \\ y \end{bmatrix}$ and $\vc{0}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}$.

      Pick a value for $x$:
      . What is the corresponding value of $y$?
      Write down the eigenvector that you have found:

    3. Step 3: Find an eigenvector corresponding to the second eigenvalue you entered, $\lambda_2=＿$



      $\displaystyle A-\lambda_2 I = $





      Write down the equation corresponding to the first row of $(A-\lambda_2 I)\vc{x}=\vc{0}$, where $\vc{x}=\begin{bmatrix} x \\ y \end{bmatrix}$ and $\vc{0}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}$.

      Pick a value for $x$:
      . What is the corresponding value of $y$?
      Write down the eigenvector that you have found:

  4. Find the eigenvalues and eigenvectors of the matrix $$B=\begin{bmatrix} -3 & 2 \\ 4 & 4 \end{bmatrix}$$


    $\displaystyle B-\lambda I = $





    What is the characteristic equation $\det (B-\lambda I )=0$?


    Find the eigenvalues and write them down (order does not matter): $\lambda_1=$
    , $\lambda_2=$

    Find an eigenvector corresponding to the first eigenvalue you entered, $\lambda_1=＿$



    $\displaystyle B-\lambda_1 I = $





    Write down the equation corresponding to the first row of $(B-\lambda_1 I)\vc{x}=\vc{0}$, where $\vc{x}=\begin{bmatrix} x \\ y \end{bmatrix}$ and $\vc{0}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}$.

    Pick a value for $x$:
    . What is the corresponding value of $y$?
    Write down the eigenvector that you have found:

    Find an eigenvector corresponding to the second eigenvalue you entered, $\lambda_2=＿$



    $\displaystyle B-\lambda_2 I = $





    Write down the equation corresponding to the first row of $(B-\lambda_2 I)\vc{x}=\vc{0}$, where $\vc{x}=\begin{bmatrix} x \\ y \end{bmatrix}$ and $\vc{0}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}$.

    Pick a value for $x$:
    . What is the corresponding value of $y$?
    Write down the eigenvector that you have found:

  5. Find the eigenvalues and eigenvectors of $A=\left[\begin{matrix}1 & 2\\3 & 4\end{matrix}\right]$.
    1. Step 1. Calculate the eigenvalues.

      Write down the characteristic equation for the eigenvalues, i.e., $\det(A - \lambda I)=0$?

      Solve this equation for the eigenvalues. $\lambda_1 = $
      , $\lambda_2 = $

      (Order doesn't matter. If you round your answers, they must be correct to at least 5 significant digits.)

      Need help?
    2. Step 2. Calculate an eigenvector corresponding to the first eigenvalue you entered, $\lambda_1=＿$.

      For $\lambda_1=＿$,



      $\displaystyle A-\lambda_1 I = $





      If rounding, include at least 5 significant digits in your response.

      Write down the equation corresponding to the first row of $(A-\lambda_1I)\vc{x}=\vc{0}$, where $\vc{x} = (x,y)$ and $\vc{0}=(0,0)$.

      Pick a value for $x$:
      . What is the corresponding value of $y$?
      Write down the eigenvector that you have found:

    3. Step 3. Calculate an eigenvector corresponding to the second eigenvalue you entered, $\lambda_2=＿$.

      For $\lambda_2=＿$,



      $\displaystyle A-\lambda_2 I = $





      If rounding, include at least 5 significant digits in your response.

      Write down the equation corresponding to the first row of $(A-\lambda_2I)\vc{x}=\vc{0}$, where $\vc{x} = (x,y)$ and $\vc{0}=(0,0)$.

      Pick a value for $x$:
      . What is the corresponding value of $y$?
      Write down the eigenvector that you have found:

  6. Optional bonus problem

    Let's find the eigenvalues of the matrix $$A = \left[\begin{matrix}1 & -1\\1 & 1\end{matrix}\right].$$ The matrix looks innocent enough.

    1. First, calculate the characteristic equation as a quadratic in $\lambda$:
    2. You could try factoring the equation, but you won't get far. Better use the quadratic formula.

      The eigenvalues (in any order) are
      $\lambda_1=$

      $\lambda_2=$

      Need help?
    3. Lovely, we got complex numbers for eignevalues. What's that supposed to mean?

      We've actually seen this matrix already, in the graphical matrix-multiplication problem set. There, when we explored the action of this matrix with the following applet, we couldn't find any directions where the matrix stretched or flipped vectors, and we concluded that it didn't have any eigenvectors or eigenvalues. Our conclusion wasn't exactly correct. What we should concluded was that the matrix doesn't have any real eigenvalues or eigenvectors. It turns out that it does have complex eigenvalues. (It has complex eigenvectors, too, but we won't go there.)

      Control panel (Show)
      Matrix-vector multiplication (Show)

      When you multiply vectors by this matrix, what does the matrix $A$ do to the vectors?
      It turns out that this consistent rotation direction is due to the fact that the eigenvalues are complex.

      If you like to see how the behavior of the matrix changes as you turn the eigenvalues from complex numbers back to real numbers, you can open the control panel of applet and slowly change the upper entry of $A$ from $-1$ to $0$. Right when you reach $0$, the eigenvalues and eigenvectors become real (although there is only eigenvector at this point). At this special case, all vectors as still rotated counterclockwise except those in the direction of $(0,1)$ (which is the eigenvector). If you increase the upper right entry to a small positive number, like $0.1$, you can find two directions where the matrix does rotate (the eigenvectors). Most vectors are still rotated counterclockwise, but in between those directions, the matrix rotates the vectors in a clockwise direction.

      Need help?

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Math 2241, Spring 2023