Proving that a Symmetric 2×2 Matrix Always Has Two Real Eigenvalues

Every symmetric (that is, A = A^T ) 2 × 2 matrix always has two real eigenvalues, counting multiplicities. (T/F)

True

True.

A symmetric 2×2 matrix will always have two real eigenvalues. This can be proven using the fact that for any 2×2 matrix A, the eigenvalues can be found by solving the characteristic equation, given as det(A – λI) = 0, where λ is the eigenvalue and I is the identity matrix.

For a symmetric matrix A, A = A^T, which means the matrix is equal to its transpose. Let’s assume the matrix A is as follows:
“`
| a b |
| b c |
“`

To find the eigenvalues, we set up the characteristic equation:
“`
det(A – λI) = 0
“`

Substituting the values of A and I:
“`
| a-λ b |
| b c-λ |
“`

Expanding the determinant:
“`
(a-λ)(c-λ) – b^2 = 0
“`

Simplifying:
“`
ac – aλ – cλ + λ^2 – b^2 = 0
“`

Rearranging terms:
“`
λ^2 – (a+c)λ + ac – b^2 = 0
“`

This is a quadratic equation in terms of λ. Applying the quadratic formula:
“`
λ = (-(a+c) ± sqrt((a+c)^2 – 4(ac – b^2))) / 2
“`

Simplifying further:
“`
λ = (-(a+c) ± sqrt(a^2 + 2ac + c^2 – 4ac + 4b^2)) / 2
= (-(a+c) ± sqrt(a^2 – 2ac + c^2 + 4b^2)) / 2
= (-(a+c) ± sqrt((a-c)^2 + 4b^2)) / 2
“`

We can see that in order for λ to be real, the term inside the square root must be non-negative. This means:
“`
(a-c)^2 + 4b^2 ≥ 0
“`

Since squares are always non-negative, this inequality will always hold true, ensuring that the eigenvalues λ will be real.

Hence, we can conclude that every symmetric 2×2 matrix will have two real eigenvalues, counting multiplicities.

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