# Calculus: Find the Eigenvalues, Eigenvectors, and Orthogonal Matrix that Diagonalizes the 2x2 Symmetric Matrix

**Find the eigenvalues, eigenvectors, and orthogonal matrix that diagonalizes the 2×2 symmetric matrix A**

A =

1 | 2 |

2 | 4 |

μ+ = (a+d)/2 + sqrt(b^2 + ((a-d)/2)^2) and μ- = (a+d)/2 – sqrt(b^2 + ((a-d)/2)^2)

μ+ = (1+4)/2 + sqrt(2^2 + ((1-4)/2)^2) and μ- = (1+4)/2 – sqrt(2^2 + ((1-4)/2)^2)

μ+ = ^{5}⁄_{2} + ^{5}⁄_{2} and μ- = 5/2 – ^{5}⁄_{2}

μ+ = 5 and μ- = 0

Now that we have the eigenvalues, we can find the eigenvectors of A. Because A is a 2×2 matrix, we can find these eigenvectors using the following formulas:

U1 = (1/|r1|) * vector orthogonal to r1

U2 = vector orthogonal to U1

where

A – μ+I =

r1 |

r2 |

A – μ+I =

-4 | 2 |

2 | -1 |

Now we find the first eigenvector (U1) by multiplying the vector orthogonal to r1 by 1 over the dot product of r1

1/|r1| = 1/sqrt((-4)^2 + 2^2) -> 1/(2*sqrt(5))

The vector orthogonal to r1 can be found by flipping the vector values of r1 then multiplying the top one by -1.

Vector orthogonal to r1 = (-2, -4)

U1 = (1/|r1|) * vector orthogonal to r1 = (1/(2*sqrt(5)) * (-2,-4) -> (1/sqrt(5))*(-1,-2)
U2 = vector orthogonal to U1 = (1/sqrt(5))*(2,-1)

The orthogonal matrix that diagonalizes A can be found by:

U = {U1, U2}

U = (1/sqrt(5))*

-1 | 2 |

-2 | -1 |

**Solution: Eigenvalues: μ+ = 5 and μ- = 0**

Eigenvectors: U1 = (1/sqrt(5))

Orthogonal matrix that diagonalizes A: U = (1/sqrt(5))*

Eigenvectors: U1 = (1/sqrt(5))

*(-1,-2) and U2 = (1/sqrt(5))*(2,-1)Orthogonal matrix that diagonalizes A: U = (1/sqrt(5))

-1 | 2 |

-2 | -1 |