Symmetric matrix real eigenvalues

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A symmetric 2x2 matrix of the form {(a,b),(b,d)} always has real eigenvalues, as demonstrated through the determinant equation derived from the characteristic polynomial. The determinant leads to the quadratic equation l^2 - (a+d)l + (ad - b^2) = 0. To show that the discriminant, (a + d)^2 - 4(ad - b^2), is always non-negative, it can be expressed as a sum of squares, which is inherently non-negative. The term -4ad can be analyzed by recognizing that (a+d)^2 - 4ad simplifies to (a-d)^2, confirming that the discriminant is always non-negative. Thus, all eigenvalues of a symmetric 2x2 matrix are guaranteed to be real.
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Homework Statement



Prove a symmetric (2x2) matrix always has real eigenvalues. The problem shows the matrix as {(a,b),(b,d)}.


Homework Equations



The problem says to use the quadratic formula.

The Attempt at a Solution



From the determinant I get (a-l)(d-l) - b^2 = 0 which expands to l^2 - (a+d)l + (ad - b^2) = 0

Using the quadratic formula I get for under the square root: (a + d)^2 - 4(ad-b^2)
How can I show that this is always positive?

Thanks for the help
 
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You can write the square root term as a sum of squares, which is always positive.
 
How do you deal with the -4ad term? I tried to factor it but couldn't figure out how
 
Can you see what (a+d)^2 - 4ad is?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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