Question

Find the cizenvalues and eigenvectors of the matrices$\left(\begin{array}{rr}5& -4\\ -4& 5\end{array}\right)$

Short answer

Eigenvalues: $9$ and $1$, Eigenvectors: $\left(\begin{array}{c}-11\end{array}\right)$ and $\left(\begin{array}{c}11\end{array}\right)$

Step-by-step solution

- Write Down the Matrix

First, look at the given matrix: $$A=\left(\begin{array}{ccc}5& -4-4& 5\end{array}\right)$$

- Set Up the Characteristic Equation

The characteristic equation is found using the formula $det(A-\lambda I)=0$ where $I$ is the identity matrix and $\lambda$ represents the eigenvalues. For matrix $A$, this becomes: $$\left|\begin{array}{ccc}5-\lambda & -4-4& 5-\lambda \end{array}\right|=0$$

- Calculate the Determinant

Calculate the determinant of the matrix $\left(\begin{array}{ccc}5-\lambda & -4-4& 5-\lambda \end{array}\right)$. Expand this determinant: $$(5-\lambda )(5-\lambda )-(-4)(-4)={\lambda }^2-10\lambda +25-16={\lambda }^2-10\lambda +9$$

- Solve the Characteristic Polynomial

Set the characteristic polynomial equal to zero and solve for $\lambda$: $${\lambda }^2-10\lambda +9=0$$ Solving this quadratic equation using the quadratic formula $\lambda =\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, where $a=1$, $b=-10$, and $c=9$. $$\lambda =\frac{10\pm \sqrt{100-36}}{2}=\frac{10\pm 8}{2}$$ Thus, the eigenvalues are: $${\lambda }_1=9\text{and}{\lambda }_2=1$$

- Find the Eigenvectors

For each eigenvalue, solve the equation $(A-\lambda I)x=0$. First, for ${\lambda }_1=9$: $$\left(\begin{array}{ccc}5-9& -4-4& 5-9\end{array}\right)\left(\begin{array}{c}{x}_1{x}_2\end{array}\right)=\left(\begin{array}{c}00\end{array}\right)$$ Simplifying: $$\left(\begin{array}{ccc}-4& -4-4& -4\end{array}\right)\left(\begin{array}{c}{x}_1{x}_2\end{array}\right)=\left(\begin{array}{c}00\end{array}\right)$$ This gives $-4x_1-4x_2=0$, which simplifies to $x_1=-x_2$. Choose $x_2=t$, then: $$x_1=-t$$ Therefore, an eigenvector for ${\lambda }_1=9$ is: $\left(\begin{array}{c}-11\end{array}\right)$. Similarly, for ${\lambda }_2=1$: $$\left(\begin{array}{ccc}5-1& -4-4& 5-1\end{array}\right)\left(\begin{array}{c}{x}_1{x}_2\end{array}\right)=\left(\begin{array}{c}00\end{array}\right)$$ Simplifying: $$\left(\begin{array}{ccc}4& -4-4& 4\end{array}\right)\left(\begin{array}{c}{x}_1{x}_2\end{array}\right)=\left(\begin{array}{c}00\end{array}\right)$$ This gives $4x_1-4x_2=0$, which simplifies to $x_1=x_2$. Choose $x_2=t$, then: $$x_1=t$$ Therefore, an eigenvector for ${\lambda }_2=1$ is: $\left(\begin{array}{c}11\end{array}\right)$.

Key concepts

matrix algebra

Matrix algebra is a branch of mathematics that deals with matrices and their operations. Matrices are rectangular arrays of numbers, symbols, or expressions arranged in rows and columns. You can add, subtract, and multiply matrices under certain conditions. In this exercise, we work with a 2x2 matrix:

• To add or subtract matrices, ensure they are of the same dimension.
• Matrix multiplication involves the dot product of rows and columns and requires the number of columns in the first matrix to equal the number of rows in the second matrix.
• The identity matrix, denoted as $I$, is a special matrix with 1s on the diagonal and 0s elsewhere.

Matrix algebra is fundamental in linear algebra, and it forms the basis for understanding concepts like eigenvalues and eigenvectors.

characteristic equation

The characteristic equation is essential for finding the eigenvalues of a matrix. It is derived from the determinant of the matrix $A-\lambda I$, where $\lambda$ represents the eigenvalues and $I$ is the identity matrix of the same size as $A$.In this exercise, we set up the characteristic equation for the given matrix as follows:$$\left|\begin{array}{ccc}5-\lambda & -4\text{-}4& 5-\lambda \end{array}\right|=0$$Calculate the determinant and expand:$$(5-\lambda )(5-\lambda )-(-4)(-4)={\lambda }^2-10\lambda +25-16$$This simplifies to the characteristic polynomial:$${\lambda }^2-10\lambda +9=0$$Solving this equation gives us the eigenvalues of the matrix.

determinant

A determinant is a scalar value associated with a square matrix and is used in various matrix operations, including finding eigenvalues via the characteristic equation. For a 2x2 matrix $\left(\begin{array}{ccc}a& bc& d\end{array}\right)$, the determinant is calculated as:$$\text{det}(A)=ad-bc$$In our exercise, the determinant of the matrix $\left(\begin{array}{ccc}5-\lambda & -4-4& 5-\lambda \end{array}\right)$ is found by calculating:$$(5-\lambda )(5-\lambda )-(-4)(-4)={\lambda }^2-10\lambda +25-16$$The result is the characteristic polynomial that we set to zero to find the eigenvalues. Determinants help determine if a matrix is invertible and play a crucial role in linear transformations.

quadratic equation

A quadratic equation is a second-order polynomial equation in a single variable $x$, with the general form $a{x}^2+bx+c=0$. Quadratic equations can be solved using methods like factoring, completing the square, or the quadratic formula:$$\lambda =\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$In our exercise, we solve the characteristic polynomial ${\lambda }^2-10\lambda +9=0$ using the quadratic formula. Here, $a=1$, $b=-10$, and $c=9$:$$\lambda =\frac{10\pm \sqrt{100-36}}{2}=\frac{10\pm 8}{2}$$This gives us the eigenvalues ${\lambda }_1=9$ and ${\lambda }_2=1$. Understanding quadratic equations is vital for solving characteristic polynomials and finding eigenvalues.