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Suppose $A$ is a $2\times2$ matrix. How do I prove that, if $\det(A) < 0$, then $A$ is a diagonalizable matrix over $\mathbb{R}$?

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You can't, because it's false. – Chris Eagle May 3 '12 at 21:10
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The Jordan form must actually be diagonal, since a $2x2$ Jordan block has nonnegative determinant. – Grumpy Parsnip May 3 '12 at 21:17
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@ChrisEagle's comment was made before the OP was modified to include the 2x2 hypothesis. – Grumpy Parsnip May 3 '12 at 21:18
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@JimConant : If you write 2x2 in $\TeX$, with the letter $x$, then it looks like an $x$, not like a $\times$. If you write 2\times 2, then it looks like $2\times 2$. – Michael Hardy May 3 '12 at 21:53
@MichaelHardy: Yes I know, I was just being lazy. – Grumpy Parsnip May 4 '12 at 1:03

1 Answer

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Since the matrix is $2 \times 2$, its characteristic polynomial is given by $x^2 - \operatorname{tr} A \cdot x + \det A$. Since $(\operatorname{tr} A)^2 - 4 \det A > 0$ by $\det A < 0$, this polynomial has two distinct real zeroes, i.e., $A$ has two distinct real eigenvalues. Since each eigenvalue has at least one eigenvector, the geometric and algebraic multiplicities of the eigenvalues coincide, i.e., $A$ is diagonalizable.

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Also we can say that product of eigen values is the determinant of the matrix. A is $2\times2$ matrix with $\det(A)<0$ that means $\lambda_{1}\lambda_{2}<0$ , where $\lambda$ refers to the eigen values of given matrix $A$.Consequently $\lambda_{i}$ ,i=1,2 must be of opposite signs or we can say that A must have two distinct eigen values having oppsite signs. Which results in $A$ is diagonalizable. – srijan May 12 '12 at 12:06

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